B 100 lb D 60 lb -8 in.- -12 in.- -8 in.- -x Equation dV = p(x) (Eq. 5.2) dx AV Po (Eq. 54) Load Diagram Shear Diagram 1. Slope of shear diagram equals value of load Moment Diagram M V₁ Slope P₁ 2. Jump in shear equals value of concentrated load M₁ V₁ Positive V-jump V₂ M₂ V₂ M₂ 3. Change in shear equals area under distributed-load diagram V₁₂- V₁ = LP(r)dx p(x)dx (Eq. 5.6) dM dx =V(x) (Eq.5.3) (Area) M My V₂-V₁ = (Area) 4. Slope of moment diagram equals value of shear 5. Jump in moment equals (value of concentrated couple) Mo 6. Change in moment equals area under shear diagram AM -Mo (Eq. 5.5) MV M₂-My- [*V(x)dx V(x)dx (Eq.5.7) Slope - V₁ M Negative M-jump M₂ M₂ (Area)y V₂ M₂-M₁ = (Area) M₂ Fig. 1 Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for the simply-supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated load at B. we need to consider two spans, 0 <x<a and a <x<L Solution Equilibrium-Reactions: To determine the reactions A, and C,, we first draw the free-body diagram of the entire beam AC (Fig. 2). B Vix) PU- L B L-a (a) Load diagram. (3) (2) (1) (4) -2 F A P (5) (b) Shear diagram. M(x) (3) (4) (2) (1) (c) Moment diagram. (6) +C(ΣM)- - 0: +C(ΣM) - 0: Ay Fig. 2 A free-body diagram. Pa-CyL = 0, Cy = P(+) A,L-P(La) 0, Ay P(L-a) - L Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations, we can sketch V(x) progressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram (Fig. 3a). Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x = 0 is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AVA Ay m (5) Fig. 3 Shear and moment diagrams. P(La) L = Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<a, p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0. 4. Atx a there is a downward force P, so AVB = -P. dV 5. For a <x<L, p(x) = 0, so- = 0. dx Pa 6. The reaction at C causes AVC which closes the shear diagram back to zero at x=L'. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x=0 is zero [simply supported beam]. dM dx 2. For 0<x<a, Eq. 5.3 gives = V(x)=- P(L-a) L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore maximum bending moment. -0-|V(x)dx= P(L-a), = L (a). M(a) = Pa(L-a) L is the dM -Pa 4. For a<x<L, Eq. 5.3 gives - V(x)= = constant. dx 5. Equation 5.7 gives M(L) M(a) = - [V(x)d Pa V(x)dx (L-a), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (FL) are correct. If we draw finite free-body diagrams of the ends of the beam, we get Fig. 4. Therefore, the shear diagram in Fig. 36 has the correct signs according to the free-body sketches in Fig. 4 and the sign convention in Fig. 5.6. The downward force will bend the beam as shown in Fig. 5, which is consistent with the fact that the bending moment is positive everywhere. The maximum bending moment occurs at the cross section where the force P is applied and where the shear force changes sign. B Shear Fig. 4 Fig. 5 Shear Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for simply supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC in Fig. 2 to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0 < x <a and a <x<L Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. -Mo L (1) V(x) A₂ Ma Fig. 1 B Fig. 2 A free-body diagram. (2) M(x) (D) (2) (a) Load diagram. (b) Shear diagram. (3) -My-) L (c) Moment diagram. Мо +C(μM)- +C (ΣM)- = 0: -Mo - C,L=0→C, = Mo -0→A, - 0: A,L-Mo- Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x = 0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<L.p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0, V(x)= M/L = constant. 4. The reaction at C causes AVC --M/L, which closes the shear diagram back to zero at x = L". Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x = 0 is zero (simply supported beam). 2. For <x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore M(a) -0- fgv(x)dx = M₁al L. 4. At xa there is a negative jump in moment given by Eq. 5.5. So Moa MoL-a) L L M(a)=- 5. Fig. 3 Load, shear, and moment diagrams 6. Mo= For a<x<L, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. Equation 5.7 gives M(L) M(a) = V(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F. L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated. force. Determine the reactions and sketch the shear and moment diagrams for the beam shown in Fig. 1. (This beam is said to have an overhang BC.) Show all significant values (that is, maxima, minima, positions of maxima and minima, etc.) on the diagrams. 8 kN/m 16kN Fig. 1 Plan the Solution We can use a free-body diagram of the whole beam to compute the reactions. Then we can use Eqs. 5.2 through 5.7 to sketch the V(x) and M(x) diagrams, as we did in Examples 5.7 and 5.8. Solution Equilibrium-Reactions: The reactions must be determined first. Figure 2 shows the appropriate free-body diagram. 18(4) 32 kN -2m-l 16 KN 4m 2 m Ay By Fig. 2 A free-body diagram. + C(ΣM)₁ = 0 0: (8 kN/m)(4 m)(2 m) + (16 kN)(6 m) By(4 m) = 0 By= =40 kN + C(ΣM)= 0: - A,(4 m) (8 kN/m)(4 m)(2 m) + (16 kN)(2m) = 0 A, = 8 kN Check: Is Fy=0? 8 3240 160? Yes Ans. Ans. 8 kN V(KN) (2) (1) (8) 8 kN/m 40 kN (a) Load diagram 16kN (6) (5) 16 (7) x(m) It is convenient to sketch the V (Fig. 3b) and M (Fig. 3c) diagrams directly below a sketch of the beam that has all of the loads and reactions shown (Fig. 3a). Shear Diagram: The following steps are used in sketching the shear diagram (Fig. 3b). 1. V(0)=0 [no shear at end of beam]. 2. V(0) 8 kN [Eq. 5.4]. 3. dV/dx--8 kN/m [Eq. 5.2]. 4. V(4) V(0) + (-8 kN/m)(4 m) -8-32-24 kN [Eq.5.6]. 5. V(4) V(4) = 40 kN [Eq. 5.4]. 6. dV/dx 0 [Eq. 5.2]. = (3) 24H 7. V(6) V(6)-16=0. (b) Shear diagram. 8. 4 (1) kNm) -32H (2 (3) & (4) Im (6) (e) Moment diagram. 116 (8) x(m) Fig. 3 Shear and moment diagrams. 16kN Shear 8 EN Shear ப 8 kN རྒྱལ (a) Shear 16kN AB C B 24 KN 40 kN (b) Fig. 4 Since dV/dx=-8 for 0 <x<4 m, and since V(0) = 8 kN, V = 0 at x = 1 m [Eq. 5.6]. Moment Diagram: The steps employed in constructing the moment dia- gram (Fig. 3c) using Eqs. 5.2 through 5.7 will now be described: 1. M(0) = 0 [no moment at end of beam]. 2. From dM/dx = V(x) we have the slope of M(x) going from +8 kN m/m at x=0 to zero at x = x=1 m. Therefore, the mo- ment diagram for 0 < x <1 m must have the general shape. [Eq.5.3]. 3. M(x) is maximum where V(x) = 0 [Eq. 5.3]. 4. M(1) M(0)+5V(x)dx = (8 kN)(1 m) = 4 kN·m [Eq. 5.7; area of triangle]. 5. From x = 1 m to x 4 m, V(x) gets progressively more negative. Therefore, M(x) must have the general shape [Eq.5.3]. 6. M(4) M(0)+ V(x)dx=0+ (8 kN)(1 m)+(-24 kN)(3 m) = -32 kN m (Eq. 5.7; net of areas of triangles] 7. dM/dx V(x)-16 kN [Eq.5.3]. 8. M(6) M(4)+5V(x)dx = -32 kNm+ (16 kN)(2 m) = 0 [Eq. 5.7; no moment at end of beam]. The maximum shear occurs just to the left of the support at B and has a magnitude of 24 kN. The maximum positive moment occurs where V = 0 at x = 1 m and has a magnitude of 4 kN m; and the maximum 16kN negative moment occurs at the support B, and it has a magnitude of 32 kN·m. (c) Review the Solution By imagining cuts just to the right of A (Fig. 4a), just to the left of B (Fig. 4b), and just to the right of B (Fig. 4c), we can check the sign of the shear at these points. The moment diagram is best checked by seeing if the sign of the moment diagram corresponds to a reasonable deflected shape, that is, concave upward where M(x) is positive and concave downward where Deflection. M positive exaggerated M negative Fig. 5 A sketch showing the deflection of beam AC. M(x) is negative, according to the sign convention that is given Fig. 5.6c. Where M(x) = 0, the beam is locally straight, that is, it is neither concave upward nor concave downward. We are able to sketch (Fig. 5) a plausi- ble deflection curve that passes over the supports at A and B and that is concave upward where M(x) is positive and concave downward where M is negative. The distributed load between A and B and the concentrated load at C could, indeed, cause the beam to deflect as sketched. Mo Fig. 1 B Fig. 2 A free-body diagram. (a) Load diagram. Vix) L (2) (1) MIX) (b) Shear diagram. (1) (2) (3) L L (c) Moment diagram. load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0<x<a and a<x<L. Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. - 0: +C (ΣM)- +C (ΣM)= 0: -Mo -Mo-CL-0→Cy - A,L. - Mo=0→A, M Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x-L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<Lp(x) = 0. Therefore, from Eq. 5.2, dVidx = 0, V(x) = ML constant. 4. The reaction at C causes AV-M/L, which closes the shear diagram back to zero at x = L*. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction. of the moment diagram are explained and numbered. (6) 1. The moment at x = 0 is zero (simply supported beam). 2. For 0<x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa,M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a." Therefore M(a)-0fgV(x)dx = Moal L. 4. At x = a there is a negative jump in moment given by Eq.5.5. So Moa Mo(L-a) L L M(a)= 5. Fig. 3 Load, shear, and moment diagrams. 6. - Mo= For a <x<L, Eq. 5.3 gives dM/dx - V(x) - M/L - constant. Equation 5.7 gives M(L) M(a) = f(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated force. dV = p(x) dx dM == V(x) dx AVA = Po |AMB = MB V2 V₁ = x2 x1 p(x)dx M₂-M₁ = V(x) dx M(x) M(x) DC1 V(x) V(x) (a) Positive V and M on section "x." +x face +y face M M V V V (b) Positive shear. (c) Positive moment. (d) Positive V and M. A 2P B C D x 3a |_ 2a | 2a | За E 2P 3 Pa Equation dV = p(x) (Eq. 5.2) dx AV Po (Eq. 54) Load Diagram Shear Diagram 1. Slope of shear diagram equals value of load Moment Diagram M V₁ Slope P₁ 2. Jump in shear equals value of concentrated load M₁ V₁ Positive V-jump V₂ M₂ V₂ M₂ 3. Change in shear equals area under distributed-load diagram V₁₂- V₁ = LP(r)dx p(x)dx (Eq. 5.6) dM dx =V(x) (Eq.5.3) (Area) M My V₂-V₁ = (Area) 4. Slope of moment diagram equals value of shear 5. Jump in moment equals (value of concentrated couple) Mo 6. Change in moment equals area under shear diagram AM -Mo (Eq. 5.5) MV M₂-My- [*V(x)dx V(x)dx (Eq.5.7) Slope - V₁ M Negative M-jump M₂ M₂ (Area)y V₂ M₂-M₁ = (Area) M₂ Fig. 1 Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for the simply-supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated load at B. we need to consider two spans, 0 <x<a and a <x<L Solution Equilibrium-Reactions: To determine the reactions A, and C,, we first draw the free-body diagram of the entire beam AC (Fig. 2). B Vix) PU- B L-a (a) Load diagram. L (3) (2) (1) (4) -2 F P (5) (b) Shear diagram. M(x) (3) (4) (2) (1) (c) Moment diagram. (6) +C(ΣM)- - 0: +C(ΣM) - 0: Ay Fig. 2 A free-body diagram. Pa-CyL = 0, Cy = P(+) A,L-P(La) 0, Ay P(L-a) - L Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations, we can sketch V(x) progressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram (Fig. 3a). Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x = 0 is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AVA Ay m (5) Fig. 3 Shear and moment diagrams. P(La) L = Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<a, p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0. 4. Atx a there is a downward force P, so AVB = -P. dV 5. For a<x<L, p(x) = 0, so- = 0. dx Pa 6. The reaction at C causes AVC which closes the shear diagram back to zero at x=L'. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x=0 is zero [simply supported beam]. dM dx 2. For 0<x<a, Eq. 5.3 gives = V(x)=- P(L-a) L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore maximum bending moment. -0-|V(x)dx= P(L-a), = L (a). M(a) = Pa(L-a) L is the dM -Pa 4. For a<x<L, Eq. 5.3 gives - dx V(x)= = constant. 5. Equation 5.7 gives M(L) M(a) = - [V(x)d Pa V(x)dx (L-a), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (FL) are correct. If we draw finite free-body diagrams of the ends of the beam, we get Fig. 4. Therefore, the shear diagram in Fig. 36 has the correct signs according to the free-body sketches in Fig. 4 and the sign convention in Fig. 5.6. The downward force will bend the beam as shown in Fig. 5, which is consistent with the fact that the bending moment is positive everywhere. The maximum bending moment occurs at the cross section where the force P is applied and where the shear force changes sign. B Shear Fig. 4 Fig. 5 Shear Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for simply supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC in Fig. 2 to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0 < x <a and a <x<L Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. -Mo L (1) V(x) A₂ Ma Fig. 1 B Fig. 2 A free-body diagram. (2) M(x) (D) (2) (a) Load diagram. (b) Shear diagram. (3) -My-) L (c) Moment diagram. Мо +C(μM)- +C (ΣM)- = 0: -Mo - C,L=0→C, = Mo -0→A, - 0: A,L-Mo- Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x = 0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<L.p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0, V(x)= M/L = constant. 4. The reaction at C causes AVC --M/L, which closes the shear diagram back to zero at x = L". Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x = 0 is zero (simply supported beam). 2. For <x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore M(a) -0- fgv(x)dx = M₁al L. 4. At xa there is a negative jump in moment given by Eq. 5.5. So Moa MoL-a) L L M(a)=- 5. Fig. 3 Load, shear, and moment diagrams 6. Mo= For a<x<L, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. Equation 5.7 gives M(L) M(a) = V(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F. L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated. force. Determine the reactions and sketch the shear and moment diagrams for the beam shown in Fig. 1. (This beam is said to have an overhang BC.) Show all significant values (that is, maxima, minima, positions of maxima and minima, etc.) on the diagrams. 8 kN/m 16kN Fig. 1 Plan the Solution We can use a free-body diagram of the whole beam to compute the reactions. Then we can use Eqs. 5.2 through 5.7 to sketch the V(x) and M(x) diagrams, as we did in Examples 5.7 and 5.8. Solution Equilibrium-Reactions: The reactions must be determined first. Figure 2 shows the appropriate free-body diagram. 18(4) 32 kN -2m-l 16 KN 4m 2 m Ay By Fig. 2 A free-body diagram. + C(ΣM)₁ = 0 0: (8 kN/m)(4 m)(2 m) + (16 kN)(6 m) By(4 m) = 0 By= =40 kN + C(ΣM)= 0: - A,(4 m) (8 kN/m)(4 m)(2 m) + (16 kN)(2m) = 0 A, = 8 kN Check: Is Fy=0? 8 3240 160? Yes Ans. Ans. 8 kN V(KN) (2) (1) (8) 8 kN/m 40 kN (a) Load diagram 16kN (6) (5) 16 (7) x(m) It is convenient to sketch the V (Fig. 3b) and M (Fig. 3c) diagrams directly below a sketch of the beam that has all of the loads and reactions shown (Fig. 3a). Shear Diagram: The following steps are used in sketching the shear diagram (Fig. 3b). 1. V(0)=0 [no shear at end of beam]. 2. V(0) 8 kN [Eq. 5.4]. 3. dV/dx--8 kN/m [Eq. 5.2]. 4. V(4) V(0) + (-8 kN/m)(4 m) -8-32-24 kN [Eq.5.6]. 5. V(4) V(4) = 40 kN [Eq. 5.4]. 6. dV/dx 0 [Eq. 5.2]. = (3) 24H 7. V(6) V(6)-16=0. (b) Shear diagram. 8. 4 (1) kNm) -32H (2 (3) & (4) Im (6) (e) Moment diagram. 116 (8) x(m) Fig. 3 Shear and moment diagrams. 16kN Shear 8 EN Shear ப 8 kN རྒྱལ (a) Shear 16kN AB C B 24 KN 40 kN (b) Fig. 4 Since dV/dx=-8 for 0 <x<4 m, and since V(0) = 8 kN, V = 0 at x = 1 m [Eq. 5.6]. Moment Diagram: The steps employed in constructing the moment dia- gram (Fig. 3c) using Eqs. 5.2 through 5.7 will now be described: 1. M(0) = 0 [no moment at end of beam]. 2. From dM/dx = V(x) we have the slope of M(x) going from +8 kN m/m at x=0 to zero at x = x=1 m. Therefore, the mo- ment diagram for 0 < x <1 m must have the general shape. [Eq.5.3]. 3. M(x) is maximum where V(x) = 0 [Eq. 5.3]. 4. M(1) M(0)+5V(x)dx = (8 kN)(1 m) = 4 kN·m [Eq. 5.7; area of triangle]. 5. From x = 1 m to x 4 m, V(x) gets progressively more negative. Therefore, M(x) must have the general shape [Eq.5.3]. 6. M(4) M(0)+ V(x)dx=0+ (8 kN)(1 m)+(-24 kN)(3 m) = -32 kN m (Eq. 5.7; net of areas of triangles] 7. dM/dx V(x)-16 kN [Eq.5.3]. 8. M(6) M(4)+5V(x)dx = -32 kNm+ (16 kN)(2 m) = 0 [Eq. 5.7; no moment at end of beam]. The maximum shear occurs just to the left of the support at B and has a magnitude of 24 kN. The maximum positive moment occurs where V = 0 at x = 1 m and has a magnitude of 4 kN m; and the maximum 16kN negative moment occurs at the support B, and it has a magnitude of 32 kN·m. (c) Review the Solution By imagining cuts just to the right of A (Fig. 4a), just to the left of B (Fig. 4b), and just to the right of B (Fig. 4c), we can check the sign of the shear at these points. The moment diagram is best checked by seeing if the sign of the moment diagram corresponds to a reasonable deflected shape, that is, concave upward where M(x) is positive and concave downward where Deflection. M positive exaggerated M negative Fig. 5 A sketch showing the deflection of beam AC. M(x) is negative, according to the sign convention that is given Fig. 5.6c. Where M(x) = 0, the beam is locally straight, that is, it is neither concave upward nor concave downward. We are able to sketch (Fig. 5) a plausi- ble deflection curve that passes over the supports at A and B and that is concave upward where M(x) is positive and concave downward where M is negative. The distributed load between A and B and the concentrated load at C could, indeed, cause the beam to deflect as sketched. Mo Fig. 1 B Fig. 2 A free-body diagram. (a) Load diagram. Vix) L (2) (1) MIX) (b) Shear diagram. (1) (2) (3) L L (c) Moment diagram. load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0<x<a and a<x<L. Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. - 0: +C (ΣM)- +C (ΣM)= 0: -Mo -Mo-CL-0→Cy - A,L. - Mo=0→A, M Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x-L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<Lp(x) = 0. Therefore, from Eq. 5.2, dVidx = 0, V(x) = ML constant. 4. The reaction at C causes AV-M/L, which closes the shear diagram back to zero at x = L*. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction. of the moment diagram are explained and numbered. (6) 1. The moment at x = 0 is zero (simply supported beam). 2. For 0<x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa,M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a." Therefore M(a)-0fgV(x)dx = Moal L. 4. At x = a there is a negative jump in moment given by Eq.5.5. So Moa Mo(L-a) L L M(a)= 5. Fig. 3 Load, shear, and moment diagrams. 6. - Mo= For a <x<L, Eq. 5.3 gives dM/dx - V(x) - M/L - constant. Equation 5.7 gives M(L) M(a) = f(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated force. dV = p(x) dx dM == V(x) dx AVA = Po |AMB = MB V2 V₁ = x2 x1 p(x)dx M₂-M₁ = V(x) dx M(x) M(x) DC1 V(x) V(x) (a) Positive V and M on section "x." +x face +y face M M V V V (b) Positive shear. (c) Positive moment. (d) Positive V and M. A 1 kip 2 kips B E C 2- 2 ft 2 ft 2 ft 2 ft D Equation dV = p(x) (Eq. 5.2) dx AV Po (Eq. 54) Load Diagram Shear Diagram 1. Slope of shear diagram equals value of load Moment Diagram M V₁ Slope P₁ 2. Jump in shear equals value of concentrated load M₁ V₁ Positive V-jump V₂ M₂ V₂ M₂ 3. Change in shear equals area under distributed-load diagram V₁₂- V₁ = LP(r)dx p(x)dx (Eq. 5.6) dM dx =V(x) (Eq.5.3) (Area) M My V₂-V₁ = (Area) 4. Slope of moment diagram equals value of shear 5. Jump in moment equals (value of concentrated couple) Mo 6. Change in moment equals area under shear diagram AM -Mo (Eq. 5.5) MV M₂-My- [*V(x)dx V(x)dx (Eq.5.7) Slope - V₁ M Negative M-jump M₂ M₂ (Area)y V₂ M₂-M₁ = (Area) M₂ Fig. 1 Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for the simply-supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated load at B. we need to consider two spans, 0 <x<a and a <x<L Solution Equilibrium-Reactions: To determine the reactions A, and C,, we first draw the free-body diagram of the entire beam AC (Fig. 2). B Vix) PU- B L-a (a) Load diagram. L (3) (2) (1) (4) -2 F P (5) (b) Shear diagram. M(x) (3) (4) (2) (1) (c) Moment diagram. (6) +C(ΣM)- - 0: +C(ΣM) - 0: Ay Fig. 2 A free-body diagram. Pa-CyL = 0, Cy = P(+) A,L-P(La) 0, Ay P(L-a) - L Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations, we can sketch V(x) progressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram (Fig. 3a). Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x = 0 is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AVA Ay m (5) Fig. 3 Shear and moment diagrams. P(La) L = Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<a, p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0. 4. Atx a there is a downward force P, so AVB = -P. dV 5. For a <x<L, p(x) = 0, so- = 0. dx Pa 6. The reaction at C causes AVC which closes the shear diagram back to zero at x=L'. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x=0 is zero [simply supported beam]. dM dx 2. For 0<x<a, Eq. 5.3 gives = V(x)=- P(L-a) L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore maximum bending moment. -0-|V(x)dx= P(L-a), = L (a). M(a) = Pa(L-a) L is the dM -Pa 4. For a<x<L, Eq. 5.3 gives - V(x)= = constant. dx 5. Equation 5.7 gives M(L) M(a) = - [V(x)d Pa V(x)dx (L-a), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (FL) are correct. If we draw finite free-body diagrams of the ends of the beam, we get Fig. 4. Therefore, the shear diagram in Fig. 36 has the correct signs according to the free-body sketches in Fig. 4 and the sign convention in Fig. 5.6. The downward force will bend the beam as shown in Fig. 5, which is consistent with the fact that the bending moment is positive everywhere. The maximum bending moment occurs at the cross section where the force P is applied and where the shear force changes sign. B Shear Fig. 4 Fig. 5 Shear Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for simply supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC in Fig. 2 to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0 < x <a and a <x<L Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. -Mo L (1) V(x) A₂ Ma Fig. 1 B Fig. 2 A free-body diagram. (2) M(x) (D) (2) (a) Load diagram. (b) Shear diagram. (3) -My-) L (c) Moment diagram. Мо +C(μM)- +C (ΣM)- = 0: -Mo - C,L=0→C, = Mo -0→A, - 0: A,L-Mo- Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x = 0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<L.p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0, V(x)= M/L = constant. 4. The reaction at C causes AVC --M/L, which closes the shear diagram back to zero at x = L". Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x = 0 is zero (simply supported beam). 2. For <x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore M(a) -0- fgv(x)dx = M₁al L. 4. At xa there is a negative jump in moment given by Eq. 5.5. So Moa MoL-a) L L M(a)=- 5. Fig. 3 Load, shear, and moment diagrams 6. Mo= For a<x<L, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. Equation 5.7 gives M(L) M(a) = V(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F. L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated. force. Determine the reactions and sketch the shear and moment diagrams for the beam shown in Fig. 1. (This beam is said to have an overhang BC.) Show all significant values (that is, maxima, minima, positions of maxima and minima, etc.) on the diagrams. 8 kN/m 16kN Fig. 1 Plan the Solution We can use a free-body diagram of the whole beam to compute the reactions. Then we can use Eqs. 5.2 through 5.7 to sketch the V(x) and M(x) diagrams, as we did in Examples 5.7 and 5.8. Solution Equilibrium-Reactions: The reactions must be determined first. Figure 2 shows the appropriate free-body diagram. 18(4) 32 kN -2m-l 16 KN 4m 2 m Ay By Fig. 2 A free-body diagram. + C(ΣM)₁ = 0 0: (8 kN/m)(4 m)(2 m) + (16 kN)(6 m) By(4 m) = 0 By= =40 kN + C(ΣM)= 0: - A,(4 m) (8 kN/m)(4 m)(2 m) + (16 kN)(2m) = 0 A, = 8 kN Check: Is Fy=0? 8 3240 160? Yes Ans. Ans. 8 kN V(KN) (2) (1) (8) 8 kN/m 40 kN (a) Load diagram 16kN (6) (5) 16 (7) x(m) It is convenient to sketch the V (Fig. 3b) and M (Fig. 3c) diagrams directly below a sketch of the beam that has all of the loads and reactions shown (Fig. 3a). Shear Diagram: The following steps are used in sketching the shear diagram (Fig. 3b). 1. V(0)=0 [no shear at end of beam]. 2. V(0) 8 kN [Eq. 5.4]. 3. dV/dx--8 kN/m [Eq. 5.2]. 4. V(4) V(0) + (-8 kN/m)(4 m) -8-32-24 kN [Eq.5.6]. 5. V(4) V(4) = 40 kN [Eq. 5.4]. 6. dV/dx 0 [Eq. 5.2]. = (3) 24H 7. V(6) V(6)-16=0. (b) Shear diagram. 8. 4 (1) kNm) -32H (2 (3) & (4) Im (6) (e) Moment diagram. 116 (8) x(m) Fig. 3 Shear and moment diagrams. 16kN Shear 8 EN Shear ப 8 kN རྒྱལ (a) Shear 16kN AB C B 24 KN 40 kN (b) Fig. 4 Since dV/dx=-8 for 0 <x<4 m, and since V(0) = 8 kN, V = 0 at x = 1 m [Eq. 5.6]. Moment Diagram: The steps employed in constructing the moment dia- gram (Fig. 3c) using Eqs. 5.2 through 5.7 will now be described: 1. M(0) = 0 [no moment at end of beam]. 2. From dM/dx = V(x) we have the slope of M(x) going from +8 kN m/m at x=0 to zero at x = x=1 m. Therefore, the mo- ment diagram for 0 < x <1 m must have the general shape. [Eq.5.3]. 3. M(x) is maximum where V(x) = 0 [Eq. 5.3]. 4. M(1) M(0)+5V(x)dx = (8 kN)(1 m) = 4 kN·m [Eq. 5.7; area of triangle]. 5. From x = 1 m to x 4 m, V(x) gets progressively more negative. Therefore, M(x) must have the general shape [Eq.5.3]. 6. M(4) M(0)+ V(x)dx=0+ (8 kN)(1 m)+(-24 kN)(3 m) = -32 kN m (Eq. 5.7; net of areas of triangles] 7. dM/dx V(x)-16 kN [Eq.5.3]. 8. M(6) M(4)+5V(x)dx = -32 kNm+ (16 kN)(2 m) = 0 [Eq. 5.7; no moment at end of beam]. The maximum shear occurs just to the left of the support at B and has a magnitude of 24 kN. The maximum positive moment occurs where V = 0 at x = 1 m and has a magnitude of 4 kN m; and the maximum 16kN negative moment occurs at the support B, and it has a magnitude of 32 kN·m. (c) Review the Solution By imagining cuts just to the right of A (Fig. 4a), just to the left of B (Fig. 4b), and just to the right of B (Fig. 4c), we can check the sign of the shear at these points. The moment diagram is best checked by seeing if the sign of the moment diagram corresponds to a reasonable deflected shape, that is, concave upward where M(x) is positive and concave downward where Deflection. M positive exaggerated M negative Fig. 5 A sketch showing the deflection of beam AC. M(x) is negative, according to the sign convention that is given Fig. 5.6c. Where M(x) = 0, the beam is locally straight, that is, it is neither concave upward nor concave downward. We are able to sketch (Fig. 5) a plausi- ble deflection curve that passes over the supports at A and B and that is concave upward where M(x) is positive and concave downward where M is negative. The distributed load between A and B and the concentrated load at C could, indeed, cause the beam to deflect as sketched. Mo Fig. 1 B Fig. 2 A free-body diagram. (a) Load diagram. Vix) L (2) (1) MIX) (b) Shear diagram. (1) (2) (3) L L (c) Moment diagram. load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0<x<a and a<x<L. Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. - 0: +C (ΣM)- +C (ΣM)= 0: -Mo -Mo-CL-0→Cy - A,L. - Mo=0→A, M Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x-L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<Lp(x) = 0. Therefore, from Eq. 5.2, dVidx = 0, V(x) = ML constant. 4. The reaction at C causes AV-M/L, which closes the shear diagram back to zero at x = L*. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction. of the moment diagram are explained and numbered. (6) 1. The moment at x = 0 is zero (simply supported beam). 2. For 0<x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa,M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a." Therefore M(a)-0fgV(x)dx = Moal L. 4. At x = a there is a negative jump in moment given by Eq.5.5. So Moa Mo(L-a) L L M(a)= 5. Fig. 3 Load, shear, and moment diagrams. 6. - Mo= For a <x<L, Eq. 5.3 gives dM/dx - V(x) - M/L - constant. Equation 5.7 gives M(L) M(a) = f(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated force. dV = p(x) dx dM == V(x) dx AVA = Po |AMB = MB V2 V₁ = x2 x1 p(x)dx M₂-M₁ = V(x) dx M(x) M(x) DC1 V(x) V(x) (a) Positive V and M on section "x." +x face +y face M M V V V (b) Positive shear. (c) Positive moment. (d) Positive V and M. A 150 N B D E C 300 N -100 10 mm-|-100 m 100 mm- -100 mm- -150 mm- Equation dV = p(x) (Eq. 5.2) dx AV Po (Eq. 54) Load Diagram Shear Diagram 1. Slope of shear diagram equals value of load Moment Diagram M V₁ Slope P₁ 2. Jump in shear equals value of concentrated load M₁ V₁ Positive V-jump V₂ M₂ V₂ M₂ 3. Change in shear equals area under distributed-load diagram V₁₂- V₁ = LP(r)dx p(x)dx (Eq. 5.6) dM dx =V(x) (Eq.5.3) (Area) M My V₂-V₁ = (Area) 4. Slope of moment diagram equals value of shear 5. Jump in moment equals (value of concentrated couple) Mo 6. Change in moment equals area under shear diagram AM -Mo (Eq. 5.5) MV M₂-My- [*V(x)dx V(x)dx (Eq.5.7) Slope - V₁ M Negative M-jump M₂ Ma (Area)y V₂ M₂-M₁ = (Area) M₂ Fig. 1 Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for the simply-supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated load at B. we need to consider two spans, 0 <x<a and a <x<L Solution Equilibrium-Reactions: To determine the reactions A, and C,, we first draw the free-body diagram of the entire beam AC (Fig. 2). B Vix) PU- B L-a (a) Load diagram. L (3) (2) (1) (4) -2 F P (5) (b) Shear diagram. M(x) (3) (4) (2) (1) (c) Moment diagram. (6) +C(ΣM)- - 0: +C(ΣM) - 0: Ay Fig. 2 A free-body diagram. Pa-CyL = 0, Cy = P(+) A,L-P(La) 0, Ay P(L-a) - L Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations, we can sketch V(x) progressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram (Fig. 3a). Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x = 0 is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AVA Ay m (5) Fig. 3 Shear and moment diagrams. P(La) L = Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<a, p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0. 4. Atx a there is a downward force P, so AVB = -P. dV 5. For a <x<L, p(x) = 0, so- = 0. dx Pa 6. The reaction at C causes AVC which closes the shear diagram back to zero at x=L'. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x=0 is zero [simply supported beam]. dM dx 2. For 0<x<a, Eq. 5.3 gives = V(x)=- P(L-a) L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore maximum bending moment. -0-|V(x)dx= P(L-a), = L (a). M(a) = Pa(L-a) L is the dM -Pa 4. For a<x<L, Eq. 5.3 gives - V(x)= = constant. dx 5. Equation 5.7 gives M(L) M(a) = - [V(x)d Pa V(x)dx (L-a), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (FL) are correct. If we draw finite free-body diagrams of the ends of the beam, we get Fig. 4. Therefore, the shear diagram in Fig. 36 has the correct signs according to the free-body sketches in Fig. 4 and the sign convention in Fig. 5.6. The downward force will bend the beam as shown in Fig. 5, which is consistent with the fact that the bending moment is positive everywhere. The maximum bending moment occurs at the cross section where the force P is applied and where the shear force changes sign. B Shear Fig. 4 Fig. 5 Shear Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for simply supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC in Fig. 2 to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0 < x <a and a <x<L Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. -Mo L (1) V(x) A₂ Ma Fig. 1 B Fig. 2 A free-body diagram. (2) M(x) (D) (2) (a) Load diagram. (b) Shear diagram. (3) -My-) L Мо +C(μM)- +C (ΣM)- = 0: -Mo - C,L=0→C, = Mo -0→A, - 0: A,L-Mo- Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x = 0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<L.p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0, V(x)= M/L = constant. 4. The reaction at C causes AVC --M/L, which closes the shear diagram back to zero at x = L". Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x = 0 is zero (simply supported beam). 2. For <x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore M(a) -0- fgv(x)dx = M₁al L. 4. At xa there is a negative jump in moment given by Eq. 5.5. So Moa MoL-a) L L M(a)=- (c) Moment diagram. 5. Fig. 3 Load, shear, and moment diagrams 6. Mo= For a<x<L, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. Equation 5.7 gives M(L) M(a) = V(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F. L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated. force. Determine the reactions and sketch the shear and moment diagrams for the beam shown in Fig. 1. (This beam is said to have an overhang BC.) Show all significant values (that is, maxima, minima, positions of maxima and minima, etc.) on the diagrams. 8 kN/m 16kN Fig. 1 Plan the Solution We can use a free-body diagram of the whole beam to compute the reactions. Then we can use Eqs. 5.2 through 5.7 to sketch the V(x) and M(x) diagrams, as we did in Examples 5.7 and 5.8. Solution Equilibrium-Reactions: The reactions must be determined first. Figure 2 shows the appropriate free-body diagram. 18(4) 32 kN -2m-l 16 KN 4m 2 m Ay By Fig. 2 A free-body diagram. + C(ΣM)₁ = 0 0: (8 kN/m)(4 m)(2 m) + (16 kN)(6 m) By(4 m) = 0 By= =40 kN + C(ΣM)= 0: - A,(4 m) (8 kN/m)(4 m)(2 m) + (16 kN)(2m) = 0 A, = 8 kN Check: Is Fy=0? 8 3240 160? Yes Ans. Ans. 8 kN V(KN) (2) (1) (8) 8 kN/m 40 kN (a) Load diagram 16kN (6) (5) 16 (7) x(m) It is convenient to sketch the V (Fig. 3b) and M (Fig. 3c) diagrams directly below a sketch of the beam that has all of the loads and reactions shown (Fig. 3a). Shear Diagram: The following steps are used in sketching the shear diagram (Fig. 3b). 1. V(0)=0 [no shear at end of beam]. 2. V(0) 8 kN [Eq. 5.4]. 3. dV/dx--8 kN/m [Eq. 5.2]. 4. V(4) V(0) + (-8 kN/m)(4 m) -8-32-24 kN [Eq.5.6]. 5. V(4) V(4) = 40 kN [Eq. 5.4]. 6. dV/dx 0 [Eq. 5.2]. = (3) 24H 7. V(6) V(6)-16=0. (b) Shear diagram. 8. 4 (1) kNm) -32H (2 (3) & (4) Im (6) (e) Moment diagram. 116 (8) x(m) Fig. 3 Shear and moment diagrams. 16kN Shear 8 EN Shear ப 8 kN རྒྱལ (a) Shear 16kN AB C B 24 KN 40 kN (b) Fig. 4 Since dV/dx=-8 for 0 <x<4 m, and since V(0) = 8 kN, V = 0 at x = 1 m [Eq. 5.6]. Moment Diagram: The steps employed in constructing the moment dia- gram (Fig. 3c) using Eqs. 5.2 through 5.7 will now be described: 1. M(0) = 0 [no moment at end of beam]. 2. From dM/dx = V(x) we have the slope of M(x) going from +8 kN m/m at x=0 to zero at x = x=1 m. Therefore, the mo- ment diagram for 0 < x <1 m must have the general shape. [Eq.5.3]. 3. M(x) is maximum where V(x) = 0 [Eq. 5.3]. 4. M(1) M(0)+5V(x)dx = (8 kN)(1 m) = 4 kN·m [Eq. 5.7; area of triangle]. 5. From x = 1 m to x 4 m, V(x) gets progressively more negative. Therefore, M(x) must have the general shape [Eq.5.3]. 6. M(4) M(0)+ V(x)dx=0+ (8 kN)(1 m)+(-24 kN)(3 m) = -32 kN m (Eq. 5.7; net of areas of triangles] 7. dM/dx V(x)-16 kN [Eq.5.3]. 8. M(6) M(4)+5V(x)dx = -32 kNm+ (16 kN)(2 m) = 0 [Eq. 5.7; no moment at end of beam]. The maximum shear occurs just to the left of the support at B and has a magnitude of 24 kN. The maximum positive moment occurs where V = 0 at x = 1 m and has a magnitude of 4 kN m; and the maximum 16kN negative moment occurs at the support B, and it has a magnitude of 32 kN·m. (c) Review the Solution By imagining cuts just to the right of A (Fig. 4a), just to the left of B (Fig. 4b), and just to the right of B (Fig. 4c), we can check the sign of the shear at these points. The moment diagram is best checked by seeing if the sign of the moment diagram corresponds to a reasonable deflected shape, that is, concave upward where M(x) is positive and concave downward where Deflection. M positive exaggerated M negative Fig. 5 A sketch showing the deflection of beam AC. M(x) is negative, according to the sign convention that is given Fig. 5.6c. Where M(x) = 0, the beam is locally straight, that is, it is neither concave upward nor concave downward. We are able to sketch (Fig. 5) a plausi- ble deflection curve that passes over the supports at A and B and that is concave upward where M(x) is positive and concave downward where M is negative. The distributed load between A and B and the concentrated load at C could, indeed, cause the beam to deflect as sketched. Mo Fig. 1 B Fig. 2 A free-body diagram. (a) Load diagram. Vix) L (2) (1) MIX) (b) Shear diagram. (1) (2) (3) L L (c) Moment diagram. load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0<x<a and a<x<L. Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. - 0: +C (ΣM)- +C (ΣM)= 0: -Mo -Mo-CL-0→Cy - A,L. - Mo=0→A, M Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x-L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<Lp(x) = 0. Therefore, from Eq. 5.2, dVidx = 0, V(x) = ML constant. 4. The reaction at C causes AV-M/L, which closes the shear diagram back to zero at x = L*. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction. of the moment diagram are explained and numbered. (6) 1. The moment at x = 0 is zero (simply supported beam). 2. For 0<x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa,M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a." Therefore M(a)-0fgV(x)dx = Moal L. 4. At x = a there is a negative jump in moment given by Eq.5.5. So Moa Mo(L-a) L L M(a)= 5. Fig. 3 Load, shear, and moment diagrams. 6. - Mo= For a <x<L, Eq. 5.3 gives dM/dx - V(x) - M/L - constant. Equation 5.7 gives M(L) M(a) = f(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated force. dV = p(x) dx dM == V(x) dx AVA = Po |AMB = MB V2 V₁ = x2 x1 p(x)dx M₂-M₁ = V(x) dx M(x) M(x) כו C1 V(x) V(x) (a) Positive V and M on section "x." +x face +y face M M V V V (b) Positive shear. (c) Positive moment. (d) Positive V and M. 10 kN 20 kN 20 kN B D E ||1m|1m|1m| -2 m- Equation dV = p(x) (Eq. 5.2) dx AV Po (Eq. 54) Load Diagram Shear Diagram 1. Slope of shear diagram equals value of load Moment Diagram M V₁ Slope P₁ 2. Jump in shear equals value of concentrated load M₁ V₁ Positive V-jump V₂ M₂ V₂ M₂ 3. Change in shear equals area under distributed-load diagram V₁₂- V₁ = LP(r)dx p(x)dx (Eq. 5.6) dM dx =V(x) (Eq.5.3) (Area) M My V₂-V₁ = (Area) 4. Slope of moment diagram equals value of shear 5. Jump in moment equals (value of concentrated couple) Mo 6. Change in moment equals area under shear diagram AM -Mo (Eq. 5.5) MV M₂-My- [*V(x)dx V(x)dx (Eq.5.7) Slope - V₁ M Negative M-jump M₂ Ma (Area)y V₂ M₂-M₁ = (Area) M₂ Fig. 1 Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for the simply-supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated load at B. we need to consider two spans, 0 <x<a and a <x<L Solution Equilibrium-Reactions: To determine the reactions A, and C,, we first draw the free-body diagram of the entire beam AC (Fig. 2). B Vix) PU- L B L-a (a) Load diagram. (3) (2) (1) (4) -2 F A P (5) (b) Shear diagram. M(x) (3) (4) (2) (1) (c) Moment diagram. (6) +C(ΣM)- - 0: +C(ΣM) - 0: Ay Fig. 2 A free-body diagram. Pa-CyL = 0, Cy = P(+) A,L-P(La) 0, Ay P(L-a) - L Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations, we can sketch V(x) progressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram (Fig. 3a). Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x = 0 is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AVA Ay m (5) Fig. 3 Shear and moment diagrams. P(La) L = Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<a, p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0. 4. Atx a there is a downward force P, so AVB = -P. dV 5. For a <x<L, p(x) = 0, so- = 0. dx Pa 6. The reaction at C causes AVC which closes the shear diagram back to zero at x=L'. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x=0 is zero [simply supported beam]. dM dx 2. For 0<x<a, Eq. 5.3 gives = V(x)=- P(L-a) L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore maximum bending moment. -0-|V(x)dx= P(L-a), = L (a). M(a) = Pa(L-a) L is the dM -Pa 4. For a<x<L, Eq. 5.3 gives - V(x)= = constant. dx 5. Equation 5.7 gives M(L) M(a) = - [V(x)d Pa V(x)dx (L-a), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (FL) are correct. If we draw finite free-body diagrams of the ends of the beam, we get Fig. 4. Therefore, the shear diagram in Fig. 36 has the correct signs according to the free-body sketches in Fig. 4 and the sign convention in Fig. 5.6. The downward force will bend the beam as shown in Fig. 5, which is consistent with the fact that the bending moment is positive everywhere. The maximum bending moment occurs at the cross section where the force P is applied and where the shear force changes sign. B Shear Fig. 4 Fig. 5 Shear Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for simply supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC in Fig. 2 to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0 < x <a and a <x<L Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. -Mo L (1) V(x) A₂ Ma Fig. 1 B Fig. 2 A free-body diagram. (2) M(x) (D) (2) (a) Load diagram. (b) Shear diagram. (3) -My-) L (c) Moment diagram. Мо +C(μM)- +C (ΣM)- = 0: -Mo - C,L=0→C, = Mo -0→A, - 0: A,L-Mo- Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x = 0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<L.p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0, V(x)= M/L = constant. 4. The reaction at C causes AVC --M/L, which closes the shear diagram back to zero at x = L". Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x = 0 is zero (simply supported beam). 2. For <x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore M(a) -0- fgv(x)dx = M₁al L. 4. At xa there is a negative jump in moment given by Eq. 5.5. So Moa MoL-a) L L M(a)=- 5. Fig. 3 Load, shear, and moment diagrams 6. Mo= For a<x<L, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. Equation 5.7 gives M(L) M(a) = V(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F. L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated. force. Determine the reactions and sketch the shear and moment diagrams for the beam shown in Fig. 1. (This beam is said to have an overhang BC.) Show all significant values (that is, maxima, minima, positions of maxima and minima, etc.) on the diagrams. 8 kN/m 16kN Fig. 1 Plan the Solution We can use a free-body diagram of the whole beam to compute the reactions. Then we can use Eqs. 5.2 through 5.7 to sketch the V(x) and M(x) diagrams, as we did in Examples 5.7 and 5.8. Solution Equilibrium-Reactions: The reactions must be determined first. Figure 2 shows the appropriate free-body diagram. 18(4) 32 kN -2m-l 16 KN 4m 2 m Ay By Fig. 2 A free-body diagram. + C(ΣM)₁ = 0 0: (8 kN/m)(4 m)(2 m) + (16 kN)(6 m) By(4 m) = 0 By= =40 kN + C(ΣM)= 0: - A,(4 m) (8 kN/m)(4 m)(2 m) + (16 kN)(2m) = 0 A, = 8 kN Check: Is Fy=0? 8 3240 160? Yes Ans. Ans. 8 kN V(KN) (2) (1) (8) 8 kN/m 40 kN (a) Load diagram 16kN (6) (5) 16 (7) x(m) It is convenient to sketch the V (Fig. 3b) and M (Fig. 3c) diagrams directly below a sketch of the beam that has all of the loads and reactions shown (Fig. 3a). Shear Diagram: The following steps are used in sketching the shear diagram (Fig. 3b). 1. V(0)=0 [no shear at end of beam]. 2. V(0) 8 kN [Eq. 5.4]. 3. dV/dx--8 kN/m [Eq. 5.2]. 4. V(4) V(0) + (-8 kN/m)(4 m) -8-32-24 kN [Eq.5.6]. 5. V(4) V(4) = 40 kN [Eq. 5.4]. 6. dV/dx 0 [Eq. 5.2]. = (3) 24H 7. V(6) V(6)-16=0. (b) Shear diagram. 8. 4 (1) kNm) -32H (2 (3) & (4) Im (6) (e) Moment diagram. 116 (8) x(m) Fig. 3 Shear and moment diagrams. 16kN Shear 8 EN Shear ப 8 kN རྒྱལ (a) Shear 16kN AB C B 24 KN 40 kN (b) Fig. 4 Since dV/dx=-8 for 0 <x<4 m, and since V(0) = 8 kN, V = 0 at x = 1 m [Eq. 5.6]. Moment Diagram: The steps employed in constructing the moment dia- gram (Fig. 3c) using Eqs. 5.2 through 5.7 will now be described: 1. M(0) = 0 [no moment at end of beam]. 2. From dM/dx = V(x) we have the slope of M(x) going from +8 kN m/m at x=0 to zero at x = x=1 m. Therefore, the mo- ment diagram for 0 < x <1 m must have the general shape. [Eq.5.3]. 3. M(x) is maximum where V(x) = 0 [Eq. 5.3]. 4. M(1) M(0)+5V(x)dx = (8 kN)(1 m) = 4 kN·m [Eq. 5.7; area of triangle]. 5. From x = 1 m to x 4 m, V(x) gets progressively more negative. Therefore, M(x) must have the general shape [Eq.5.3]. 6. M(4) M(0)+ V(x)dx=0+ (8 kN)(1 m)+(-24 kN)(3 m) = -32 kN m (Eq. 5.7; net of areas of triangles] 7. dM/dx V(x)-16 kN [Eq.5.3]. 8. M(6) M(4)+5V(x)dx = -32 kNm+ (16 kN)(2 m) = 0 [Eq. 5.7; no moment at end of beam]. The maximum shear occurs just to the left of the support at B and has a magnitude of 24 kN. The maximum positive moment occurs where V = 0 at x = 1 m and has a magnitude of 4 kN m; and the maximum 16kN negative moment occurs at the support B, and it has a magnitude of 32 kN·m. (c) Review the Solution By imagining cuts just to the right of A (Fig. 4a), just to the left of B (Fig. 4b), and just to the right of B (Fig. 4c), we can check the sign of the shear at these points. The moment diagram is best checked by seeing if the sign of the moment diagram corresponds to a reasonable deflected shape, that is, concave upward where M(x) is positive and concave downward where Deflection. M positive exaggerated M negative Fig. 5 A sketch showing the deflection of beam AC. M(x) is negative, according to the sign convention that is given Fig. 5.6c. Where M(x) = 0, the beam is locally straight, that is, it is neither concave upward nor concave downward. We are able to sketch (Fig. 5) a plausi- ble deflection curve that passes over the supports at A and B and that is concave upward where M(x) is positive and concave downward where M is negative. The distributed load between A and B and the concentrated load at C could, indeed, cause the beam to deflect as sketched. Mo Fig. 1 B Fig. 2 A free-body diagram. (a) Load diagram. Vix) L (2) (1) MIX) (b) Shear diagram. (1) (2) (3) L L (c) Moment diagram. load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0<x<a and a<x<L. Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. - 0: +C (ΣM)- +C (ΣM)= 0: -Mo -Mo-CL-0→Cy - A,L. - Mo=0→A, M Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x-L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<Lp(x) = 0. Therefore, from Eq. 5.2, dVidx = 0, V(x) = ML constant. 4. The reaction at C causes AV-M/L, which closes the shear diagram back to zero at x = L*. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction. of the moment diagram are explained and numbered. (6) 1. The moment at x = 0 is zero (simply supported beam). 2. For 0<x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa,M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a." Therefore M(a)-0fgV(x)dx = Moal L. 4. At x = a there is a negative jump in moment given by Eq.5.5. So Moa Mo(L-a) L L M(a)= 5. Fig. 3 Load, shear, and moment diagrams. 6. - Mo= For a <x<L, Eq. 5.3 gives dM/dx - V(x) - M/L - constant. Equation 5.7 gives M(L) M(a) = f(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated force. dV = p(x) dx dM == V(x) dx AVA = Po |AMB = MB V2 V₁ = x2 x1 p(x)dx M₂-M₁ = V(x) dx M(x) M(x) DC1 V(x) V(x) (a) Positive V and M on section "x." +x face +y face M M V V V (b) Positive shear. (c) Positive moment. (d) Positive V and M. E 500 lb B C -24 in.- 40 in.- -48 in. D 50 in. x Equation dV = p(x) (Eq. 5.2) dx AV Po (Eq. 54) Load Diagram Shear Diagram 1. Slope of shear diagram equals value of load Moment Diagram M V₁ Slope P₁ 2. Jump in shear equals value of concentrated load M₁ V₁ Positive V-jump V₂ M₂ V₂ M₂ 3. Change in shear equals area under distributed-load diagram V₁₂- V₁ = LP(r)dx p(x)dx (Eq. 5.6) dM dx =V(x) (Eq.5.3) (Area) M My V₂-V₁ = (Area) 4. Slope of moment diagram equals value of shear 5. Jump in moment equals (value of concentrated couple) Mo 6. Change in moment equals area under shear diagram AM -Mo (Eq. 5.5) MV M₂-My- [*V(x)dx V(x)dx (Eq.5.7) Slope - V₁ M Negative M-jump M₂ M₂ (Area)y V₂ M₂-M₁ = (Area) M₂ Fig. 1 Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for the simply-supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated load at B. we need to consider two spans, 0 <x<a and a <x<L Solution Equilibrium-Reactions: To determine the reactions A, and C,, we first draw the free-body diagram of the entire beam AC (Fig. 2). B Vix) PU- B L-a (a) Load diagram. L (3) (2) (1) (4) -2 F P (5) (b) Shear diagram. M(x) (3) (4) (2) (1) (c) Moment diagram. (6) +C(ΣM)- - 0: +C(ΣM) - 0: Ay Fig. 2 A free-body diagram. Pa-CyL = 0, Cy = P(+) A,L-P(La) 0, Ay P(L-a) - L Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations, we can sketch V(x) progressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram (Fig. 3a). Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x = 0 is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AVA Ay m (5) Fig. 3 Shear and moment diagrams. P(La) L = Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<a, p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0. 4. Atx a there is a downward force P, so AVB = -P. dV 5. For a <x<L, p(x) = 0, so- = 0. dx Pa 6. The reaction at C causes AVC which closes the shear diagram back to zero at x=L'. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x=0 is zero [simply supported beam]. dM dx 2. For 0<x<a, Eq. 5.3 gives = V(x)=- P(L-a) L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore maximum bending moment. -0-|V(x)dx= P(L-a), = L (a). M(a) = Pa(L-a) L is the dM -Pa 4. For a<x<L, Eq. 5.3 gives - V(x)= = constant. dx 5. Equation 5.7 gives M(L) M(a) = - [V(x)d Pa V(x)dx (L-a), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (FL) are correct. If we draw finite free-body diagrams of the ends of the beam, we get Fig. 4. Therefore, the shear diagram in Fig. 36 has the correct signs according to the free-body sketches in Fig. 4 and the sign convention in Fig. 5.6. The downward force will bend the beam as shown in Fig. 5, which is consistent with the fact that the bending moment is positive everywhere. The maximum bending moment occurs at the cross section where the force P is applied and where the shear force changes sign. B Shear Fig. 4 Fig. 5 Shear Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for simply supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC in Fig. 2 to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0 < x <a and a <x<L Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. -Mo L (1) V(x) A₂ Ma Fig. 1 B Fig. 2 A free-body diagram. (2) M(x) (D) (2) (a) Load diagram. (b) Shear diagram. (3) -My-) L (c) Moment diagram. Мо +C(μM)- +C (ΣM)- = 0: -Mo - C,L=0→C, = Mo -0→A, - 0: A,L-Mo- Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x = 0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<L.p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0, V(x)= M/L = constant. 4. The reaction at C causes AVC --M/L, which closes the shear diagram back to zero at x = L". Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x = 0 is zero (simply supported beam). 2. For <x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore M(a) -0- fgv(x)dx = M₁al L. 4. At xa there is a negative jump in moment given by Eq. 5.5. So Moa MoL-a) L L M(a)=- 5. Fig. 3 Load, shear, and moment diagrams 6. Mo= For a<x<L, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. Equation 5.7 gives M(L) M(a) = V(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F. L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated. force. Determine the reactions and sketch the shear and moment diagrams for the beam shown in Fig. 1. (This beam is said to have an overhang BC.) Show all significant values (that is, maxima, minima, positions of maxima and minima, etc.) on the diagrams. 8 kN/m 16kN Fig. 1 Plan the Solution We can use a free-body diagram of the whole beam to compute the reactions. Then we can use Eqs. 5.2 through 5.7 to sketch the V(x) and M(x) diagrams, as we did in Examples 5.7 and 5.8. Solution Equilibrium-Reactions: The reactions must be determined first. Figure 2 shows the appropriate free-body diagram. 18(4) 32 kN -2m-l 16 KN 4m 2 m Ay By Fig. 2 A free-body diagram. + C(ΣM)₁ = 0 0: (8 kN/m)(4 m)(2 m) + (16 kN)(6 m) By(4 m) = 0 By= =40 kN + C(ΣM)= 0: - A,(4 m) (8 kN/m)(4 m)(2 m) + (16 kN)(2m) = 0 A, = 8 kN Check: Is Fy=0? 8 3240 160? Yes Ans. Ans. 8 kN V(KN) (2) (1) (8) 8 kN/m 40 kN (a) Load diagram 16kN (6) (5) 16 (7) x(m) It is convenient to sketch the V (Fig. 3b) and M (Fig. 3c) diagrams directly below a sketch of the beam that has all of the loads and reactions shown (Fig. 3a). Shear Diagram: The following steps are used in sketching the shear diagram (Fig. 3b). 1. V(0)=0 [no shear at end of beam]. 2. V(0) 8 kN [Eq. 5.4]. 3. dV/dx--8 kN/m [Eq. 5.2]. 4. V(4) V(0) + (-8 kN/m)(4 m) -8-32-24 kN [Eq.5.6]. 5. V(4) V(4) = 40 kN [Eq. 5.4]. 6. dV/dx 0 [Eq. 5.2]. = (3) 24H 7. V(6) V(6)-16=0. (b) Shear diagram. 8. 4 (1) kNm) -32H (2 (3) & (4) Im (6) (e) Moment diagram. 116 (8) x(m) Fig. 3 Shear and moment diagrams. 16kN Shear 8 EN Shear ப 8 kN རྒྱལ (a) Shear 16kN AB C B 24 KN 40 kN (b) Fig. 4 Since dV/dx=-8 for 0 <x<4 m, and since V(0) = 8 kN, V = 0 at x = 1 m [Eq. 5.6]. Moment Diagram: The steps employed in constructing the moment dia- gram (Fig. 3c) using Eqs. 5.2 through 5.7 will now be described: 1. M(0) = 0 [no moment at end of beam]. 2. From dM/dx = V(x) we have the slope of M(x) going from +8 kN m/m at x=0 to zero at x = x=1 m. Therefore, the mo- ment diagram for 0 < x <1 m must have the general shape. [Eq.5.3]. 3. M(x) is maximum where V(x) = 0 [Eq. 5.3]. 4. M(1) M(0)+5V(x)dx = (8 kN)(1 m) = 4 kN·m [Eq. 5.7; area of triangle]. 5. From x = 1 m to x 4 m, V(x) gets progressively more negative. Therefore, M(x) must have the general shape [Eq.5.3]. 6. M(4) M(0)+ V(x)dx=0+ (8 kN)(1 m)+(-24 kN)(3 m) = -32 kN m (Eq. 5.7; net of areas of triangles] 7. dM/dx V(x)-16 kN [Eq.5.3]. 8. M(6) M(4)+5V(x)dx = -32 kNm+ (16 kN)(2 m) = 0 [Eq. 5.7; no moment at end of beam]. The maximum shear occurs just to the left of the support at B and has a magnitude of 24 kN. The maximum positive moment occurs where V = 0 at x = 1 m and has a magnitude of 4 kN m; and the maximum 16kN negative moment occurs at the support B, and it has a magnitude of 32 kN·m. (c) Review the Solution By imagining cuts just to the right of A (Fig. 4a), just to the left of B (Fig. 4b), and just to the right of B (Fig. 4c), we can check the sign of the shear at these points. The moment diagram is best checked by seeing if the sign of the moment diagram corresponds to a reasonable deflected shape, that is, concave upward where M(x) is positive and concave downward where Deflection. M positive exaggerated M negative Fig. 5 A sketch showing the deflection of beam AC. M(x) is negative, according to the sign convention that is given Fig. 5.6c. Where M(x) = 0, the beam is locally straight, that is, it is neither concave upward nor concave downward. We are able to sketch (Fig. 5) a plausi- ble deflection curve that passes over the supports at A and B and that is concave upward where M(x) is positive and concave downward where M is negative. The distributed load between A and B and the concentrated load at C could, indeed, cause the beam to deflect as sketched. Mo Fig. 1 B Fig. 2 A free-body diagram. (a) Load diagram. Vix) L (2) (1) MIX) (b) Shear diagram. (1) (2) (3) L L (c) Moment diagram. load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0<x<a and a<x<L. Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. - 0: +C (ΣM)- +C (ΣM)= 0: -Mo -Mo-CL-0→Cy - A,L. - Mo=0→A, M Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x-L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<Lp(x) = 0. Therefore, from Eq. 5.2, dVidx = 0, V(x) = ML constant. 4. The reaction at C causes AV-M/L, which closes the shear diagram back to zero at x = L*. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction. of the moment diagram are explained and numbered. (6) 1. The moment at x = 0 is zero (simply supported beam). 2. For 0<x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa,M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a." Therefore M(a)-0fgV(x)dx = Moal L. 4. At x = a there is a negative jump in moment given by Eq.5.5. So Moa Mo(L-a) L L M(a)= 5. Fig. 3 Load, shear, and moment diagrams. 6. - Mo= For a <x<L, Eq. 5.3 gives dM/dx - V(x) - M/L - constant. Equation 5.7 gives M(L) M(a) = f(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated force. dV = p(x) dx dM == V(x) dx AVA = Po |AMB = MB V2 V₁ = x2 x1 p(x)dx M₂-M₁ = V(x) dx M(x) M(x) כו C1 V(x) V(x) (a) Positive V and M on section "x." +x face +y face M M V V V (b) Positive shear. (c) Positive moment. (d) Positive V and M. 22 2P 2P P B Equation dV = p(x) (Eq. 5.2) dx AV Po (Eq. 54) Load Diagram Shear Diagram 1. Slope of shear diagram equals value of load Moment Diagram M V₁ Slope P₁ 2. Jump in shear equals value of concentrated load M₁ V₁ Positive V-jump V₂ M₂ V₂ M₂ 3. Change in shear equals area under distributed-load diagram V₁₂- V₁ = LP(r)dx p(x)dx (Eq. 5.6) dM dx =V(x) (Eq.5.3) (Area) M My V₂-V₁ = (Area) 4. Slope of moment diagram equals value of shear 5. Jump in moment equals (value of concentrated couple) Mo 6. Change in moment equals area under shear diagram AM -Mo (Eq. 5.5) MV M₂-My- [*V(x)dx V(x)dx (Eq.5.7) Slope - V₁ M Negative M-jump M₂ M₂ (Area)y V₂ M₂-M₁ = (Area) M₂ Fig. 1 Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for the simply-supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated load at B. we need to consider two spans, 0 <x<a and a <x<L Solution Equilibrium-Reactions: To determine the reactions A, and C,, we first draw the free-body diagram of the entire beam AC (Fig. 2). B Vix) PU- B L-a (a) Load diagram. L (3) (2) (1) (4) -2 F P (5) (b) Shear diagram. M(x) (3) (4) (2) (1) (c) Moment diagram. (6) +C(ΣM)- - 0: +C(ΣM) - 0: Ay Fig. 2 A free-body diagram. Pa-CyL = 0, Cy = P(+) A,L-P(La) 0, Ay P(L-a) - L Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations, we can sketch V(x) progressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram (Fig. 3a). Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x = 0 is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AVA Ay m (5) Fig. 3 Shear and moment diagrams. P(La) L = Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<a, p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0. 4. Atx a there is a downward force P, so AVB = -P. dV 5. For a<x<L, p(x) = 0, so- = 0. dx Pa 6. The reaction at C causes AVC which closes the shear diagram back to zero at x=L'. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x=0 is zero [simply supported beam]. dM dx 2. For 0<x<a, Eq. 5.3 gives = V(x)=- P(L-a) L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore maximum bending moment. -0-|V(x)dx= P(L-a), = L (a). M(a) = Pa(L-a) L is the dM -Pa 4. For a<x<L, Eq. 5.3 gives - dx V(x)= = constant. 5. Equation 5.7 gives M(L) M(a) = - [V(x)d Pa V(x)dx (L-a), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (FL) are correct. If we draw finite free-body diagrams of the ends of the beam, we get Fig. 4. Therefore, the shear diagram in Fig. 36 has the correct signs according to the free-body sketches in Fig. 4 and the sign convention in Fig. 5.6. The downward force will bend the beam as shown in Fig. 5, which is consistent with the fact that the bending moment is positive everywhere. The maximum bending moment occurs at the cross section where the force P is applied and where the shear force changes sign. B Shear Fig. 4 Fig. 5 Shear Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for simply supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC in Fig. 2 to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0 < x <a and a <x<L Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. -Mo L (1) V(x) A₂ Ma Fig. 1 B Fig. 2 A free-body diagram. (2) M(x) (D) (2) (a) Load diagram. (b) Shear diagram. (3) -My-) L (c) Moment diagram. Мо +C(μM)- +C (ΣM)- = 0: -Mo - C,L=0→C, = Mo -0→A, - 0: A,L-Mo- Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x = 0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<L.p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0, V(x)= M/L = constant. 4. The reaction at C causes AVC --M/L, which closes the shear diagram back to zero at x = L". Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x = 0 is zero (simply supported beam). 2. For <x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore M(a) -0- fgv(x)dx = M₁al L. 4. At xa there is a negative jump in moment given by Eq. 5.5. So Moa MoL-a) L L M(a)=- 5. Fig. 3 Load, shear, and moment diagrams 6. Mo= For a<x<L, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. Equation 5.7 gives M(L) M(a) = V(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F. L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated. force. Determine the reactions and sketch the shear and moment diagrams for the beam shown in Fig. 1. (This beam is said to have an overhang BC.) Show all significant values (that is, maxima, minima, positions of maxima and minima, etc.) on the diagrams. 8 kN/m 16kN Fig. 1 Plan the Solution We can use a free-body diagram of the whole beam to compute the reactions. Then we can use Eqs. 5.2 through 5.7 to sketch the V(x) and M(x) diagrams, as we did in Examples 5.7 and 5.8. Solution Equilibrium-Reactions: The reactions must be determined first. Figure 2 shows the appropriate free-body diagram. 18(4) 32 kN -2m-l 16 KN 4m 2 m Ay By Fig. 2 A free-body diagram. + C(ΣM)₁ = 0 0: (8 kN/m)(4 m)(2 m) + (16 kN)(6 m) By(4 m) = 0 By= =40 kN + C(ΣM)= 0: - A,(4 m) (8 kN/m)(4 m)(2 m) + (16 kN)(2m) = 0 A, = 8 kN Check: Is Fy=0? 8 3240 160? Yes Ans. Ans. 8 kN V(KN) (2) (1) (8) 8 kN/m 40 kN (a) Load diagram 16kN (6) (5) 16 (7) x(m) It is convenient to sketch the V (Fig. 3b) and M (Fig. 3c) diagrams directly below a sketch of the beam that has all of the loads and reactions shown (Fig. 3a). Shear Diagram: The following steps are used in sketching the shear diagram (Fig. 3b). 1. V(0)=0 [no shear at end of beam]. 2. V(0) 8 kN [Eq. 5.4]. 3. dV/dx--8 kN/m [Eq. 5.2]. 4. V(4) V(0) + (-8 kN/m)(4 m) -8-32-24 kN [Eq.5.6]. 5. V(4) V(4) = 40 kN [Eq. 5.4]. 6. dV/dx 0 [Eq. 5.2]. = (3) 24H 7. V(6) V(6)-16=0. (b) Shear diagram. 8. 4 (1) kNm) -32H (2 (3) & (4) Im (6) (e) Moment diagram. 116 (8) x(m) Fig. 3 Shear and moment diagrams. 16kN Shear 8 EN Shear ப 8 kN རྒྱལ (a) Shear 16kN AB C B 24 KN 40 kN (b) Fig. 4 Since dV/dx=-8 for 0 <x<4 m, and since V(0) = 8 kN, V = 0 at x = 1 m [Eq. 5.6]. Moment Diagram: The steps employed in constructing the moment dia- gram (Fig. 3c) using Eqs. 5.2 through 5.7 will now be described: 1. M(0) = 0 [no moment at end of beam]. 2. From dM/dx = V(x) we have the slope of M(x) going from +8 kN m/m at x=0 to zero at x = x=1 m. Therefore, the mo- ment diagram for 0 < x <1 m must have the general shape. [Eq.5.3]. 3. M(x) is maximum where V(x) = 0 [Eq. 5.3]. 4. M(1) M(0)+5V(x)dx = (8 kN)(1 m) = 4 kN·m [Eq. 5.7; area of triangle]. 5. From x = 1 m to x 4 m, V(x) gets progressively more negative. Therefore, M(x) must have the general shape [Eq.5.3]. 6. M(4) M(0)+ V(x)dx=0+ (8 kN)(1 m)+(-24 kN)(3 m) = -32 kN m (Eq. 5.7; net of areas of triangles] 7. dM/dx V(x)-16 kN [Eq.5.3]. 8. M(6) M(4)+5V(x)dx = -32 kNm+ (16 kN)(2 m) = 0 [Eq. 5.7; no moment at end of beam]. The maximum shear occurs just to the left of the support at B and has a magnitude of 24 kN. The maximum positive moment occurs where V = 0 at x = 1 m and has a magnitude of 4 kN m; and the maximum 16kN negative moment occurs at the support B, and it has a magnitude of 32 kN·m. (c) Review the Solution By imagining cuts just to the right of A (Fig. 4a), just to the left of B (Fig. 4b), and just to the right of B (Fig. 4c), we can check the sign of the shear at these points. The moment diagram is best checked by seeing if the sign of the moment diagram corresponds to a reasonable deflected shape, that is, concave upward where M(x) is positive and concave downward where Deflection. M positive exaggerated M negative Fig. 5 A sketch showing the deflection of beam AC. M(x) is negative, according to the sign convention that is given Fig. 5.6c. Where M(x) = 0, the beam is locally straight, that is, it is neither concave upward nor concave downward. We are able to sketch (Fig. 5) a plausi- ble deflection curve that passes over the supports at A and B and that is concave upward where M(x) is positive and concave downward where M is negative. The distributed load between A and B and the concentrated load at C could, indeed, cause the beam to deflect as sketched. Mo Fig. 1 B Fig. 2 A free-body diagram. (a) Load diagram. Vix) L (2) (1) MIX) (b) Shear diagram. (1) (2) (3) L L (c) Moment diagram. load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0<x<a and a<x<L. Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. - 0: +C (ΣM)- +C (ΣM)= 0: -Mo -Mo-CL-0→Cy - A,L. - Mo=0→A, M Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x-L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<Lp(x) = 0. Therefore, from Eq. 5.2, dVidx = 0, V(x) = ML constant. 4. The reaction at C causes AV-M/L, which closes the shear diagram back to zero at x = L*. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction. of the moment diagram are explained and numbered. (6) 1. The moment at x = 0 is zero (simply supported beam). 2. For 0<x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa,M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a." Therefore M(a)-0fgV(x)dx = Moal L. 4. At x = a there is a negative jump in moment given by Eq.5.5. So Moa Mo(L-a) L L M(a)= 5. Fig. 3 Load, shear, and moment diagrams. 6. - Mo= For a <x<L, Eq. 5.3 gives dM/dx - V(x) - M/L - constant. Equation 5.7 gives M(L) M(a) = f(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated force. dV = p(x) dx dM == V(x) dx AVA = Po |AMB = MB V2 V₁ = x2 x1 p(x)dx M₂-M₁ = V(x) dx M(x) M(x) כו C1 V(x) V(x) (a) Positive V and M on section "x." +x face +y face M M V V V (b) Positive shear. (c) Positive moment. (d) Positive V and M. 12 kip.ft A 4 kips -4 ft- 6 kips B 4 ft- Equation dV = p(x) (Eq. 5.2) dx AV Po (Eq. 54) Load Diagram Shear Diagram 1. Slope of shear diagram equals value of load Moment Diagram M V₁ Slope P₁ 2. Jump in shear equals value of concentrated load M₁ V₁ Positive V-jump V₂ M₂ V₂ M₂ 3. Change in shear equals area under distributed-load diagram V₁₂- V₁ = LP(r)dx p(x)dx (Eq. 5.6) dM dx =V(x) (Eq.5.3) (Area) M My V₂-V₁ = (Area) 4. Slope of moment diagram equals value of shear 5. Jump in moment equals (value of concentrated couple) Mo 6. Change in moment equals area under shear diagram AM -Mo (Eq. 5.5) MV M₂-My- [*V(x)dx V(x)dx (Eq.5.7) Slope - V₁ M Negative M-jump M₂ M₂ (Area)y V₂ M₂-M₁ = (Area) M₂ Fig. 1 Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for the simply-supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated load at B. we need to consider two spans, 0 <x<a and a <x<L Solution Equilibrium-Reactions: To determine the reactions A, and C,, we first draw the free-body diagram of the entire beam AC (Fig. 2). B Vix) PU- L B L-a (a) Load diagram. (3) (2) (1) (4) -2 F A P (5) (b) Shear diagram. M(x) (3) (4) (2) (1) (c) Moment diagram. (6) +C(ΣM)- - 0: +C(ΣM) - 0: Ay Fig. 2 A free-body diagram. Pa-CyL = 0, Cy = P(+) A,L-P(La) 0, Ay P(L-a) - L Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations, we can sketch V(x) progressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram (Fig. 3a). Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x = 0 is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AVA Ay m (5) Fig. 3 Shear and moment diagrams. P(La) L = Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<a, p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0. 4. Atx a there is a downward force P, so AVB = -P. dV 5. For a <x<L, p(x) = 0, so- = 0. dx Pa 6. The reaction at C causes AVC which closes the shear diagram back to zero at x=L'. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x=0 is zero [simply supported beam]. dM dx 2. For 0<x<a, Eq. 5.3 gives = V(x)=- P(L-a) L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore maximum bending moment. -0-|V(x)dx= P(L-a), = L (a). M(a) = Pa(L-a) L is the dM -Pa 4. For a<x<L, Eq. 5.3 gives - V(x)= = constant. dx 5. Equation 5.7 gives M(L) M(a) = - [V(x)d Pa V(x)dx (L-a), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (FL) are correct. If we draw finite free-body diagrams of the ends of the beam, we get Fig. 4. Therefore, the shear diagram in Fig. 36 has the correct signs according to the free-body sketches in Fig. 4 and the sign convention in Fig. 5.6. The downward force will bend the beam as shown in Fig. 5, which is consistent with the fact that the bending moment is positive everywhere. The maximum bending moment occurs at the cross section where the force P is applied and where the shear force changes sign. B Shear Fig. 4 Fig. 5 Shear Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for simply supported beam shown in Fig. 1. Plan the Solution We can use a free-body diagram of the beam AC in Fig. 2 to determine the reactions at A and C. Since there is no distributed load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0 < x <a and a <x<L Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. -Mo L (1) V(x) A₂ Ma Fig. 1 B Fig. 2 A free-body diagram. (2) M(x) (D) (2) (a) Load diagram. (b) Shear diagram. (3) -My-) L Мо +C(μM)- +C (ΣM)- = 0: -Mo - C,L=0→C, = Mo -0→A, - 0: A,L-Mo- Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x = L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x = 0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<L.p(x) = 0. Therefore, from Eq. 5.2, dV/dx = 0, V(x)= M/L = constant. 4. The reaction at C causes AVC --M/L, which closes the shear diagram back to zero at x = L". Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction of the moment diagram are explained and numbered. 1. The moment at x = 0 is zero (simply supported beam). 2. For <x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa, M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a. Therefore M(a) -0- fgv(x)dx = M₁al L. 4. At xa there is a negative jump in moment given by Eq. 5.5. So Moa MoL-a) L L M(a)=- (c) Moment diagram. 5. Fig. 3 Load, shear, and moment diagrams 6. Mo= For a<x<L, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. Equation 5.7 gives M(L) M(a) = V(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F. L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated. force. Determine the reactions and sketch the shear and moment diagrams for the beam shown in Fig. 1. (This beam is said to have an overhang BC.) Show all significant values (that is, maxima, minima, positions of maxima and minima, etc.) on the diagrams. 8 kN/m 16kN Fig. 1 Plan the Solution We can use a free-body diagram of the whole beam to compute the reactions. Then we can use Eqs. 5.2 through 5.7 to sketch the V(x) and M(x) diagrams, as we did in Examples 5.7 and 5.8. Solution Equilibrium-Reactions: The reactions must be determined first. Figure 2 shows the appropriate free-body diagram. 18(4) 32 kN -2m-l 16 KN 4m 2 m Ay By Fig. 2 A free-body diagram. + C(ΣM)₁ = 0 0: (8 kN/m)(4 m)(2 m) + (16 kN)(6 m) By(4 m) = 0 By= =40 kN + C(ΣM)= 0: - A,(4 m) (8 kN/m)(4 m)(2 m) + (16 kN)(2m) = 0 A, = 8 kN Check: Is Fy=0? 8 3240 160? Yes Ans. Ans. 8 kN V(KN) (2) (1) (8) 8 kN/m 40 kN (a) Load diagram 16kN (6) (5) 16 (7) x(m) It is convenient to sketch the V (Fig. 3b) and M (Fig. 3c) diagrams directly below a sketch of the beam that has all of the loads and reactions shown (Fig. 3a). Shear Diagram: The following steps are used in sketching the shear diagram (Fig. 3b). 1. V(0)=0 [no shear at end of beam]. 2. V(0) 8 kN [Eq. 5.4]. 3. dV/dx--8 kN/m [Eq. 5.2]. 4. V(4) V(0) + (-8 kN/m)(4 m) -8-32-24 kN [Eq.5.6]. 5. V(4) V(4) = 40 kN [Eq. 5.4]. 6. dV/dx 0 [Eq. 5.2]. = (3) 24H 7. V(6) V(6)-16=0. (b) Shear diagram. 8. 4 (1) kNm) -32H (2 (3) & (4) Im (6) (e) Moment diagram. 116 (8) x(m) Fig. 3 Shear and moment diagrams. 16kN Shear 8 EN Shear ப 8 kN རྒྱལ (a) Shear 16kN AB C B 24 KN 40 kN (b) Fig. 4 Since dV/dx=-8 for 0 <x<4 m, and since V(0) = 8 kN, V = 0 at x = 1 m [Eq. 5.6]. Moment Diagram: The steps employed in constructing the moment dia- gram (Fig. 3c) using Eqs. 5.2 through 5.7 will now be described: 1. M(0) = 0 [no moment at end of beam]. 2. From dM/dx = V(x) we have the slope of M(x) going from +8 kN m/m at x=0 to zero at x = x=1 m. Therefore, the mo- ment diagram for 0 < x <1 m must have the general shape. [Eq.5.3]. 3. M(x) is maximum where V(x) = 0 [Eq. 5.3]. 4. M(1) M(0)+5V(x)dx = (8 kN)(1 m) = 4 kN·m [Eq. 5.7; area of triangle]. 5. From x = 1 m to x 4 m, V(x) gets progressively more negative. Therefore, M(x) must have the general shape [Eq.5.3]. 6. M(4) M(0)+ V(x)dx=0+ (8 kN)(1 m)+(-24 kN)(3 m) = -32 kN m (Eq. 5.7; net of areas of triangles] 7. dM/dx V(x)-16 kN [Eq.5.3]. 8. M(6) M(4)+5V(x)dx = -32 kNm+ (16 kN)(2 m) = 0 [Eq. 5.7; no moment at end of beam]. The maximum shear occurs just to the left of the support at B and has a magnitude of 24 kN. The maximum positive moment occurs where V = 0 at x = 1 m and has a magnitude of 4 kN m; and the maximum 16kN negative moment occurs at the support B, and it has a magnitude of 32 kN·m. (c) Review the Solution By imagining cuts just to the right of A (Fig. 4a), just to the left of B (Fig. 4b), and just to the right of B (Fig. 4c), we can check the sign of the shear at these points. The moment diagram is best checked by seeing if the sign of the moment diagram corresponds to a reasonable deflected shape, that is, concave upward where M(x) is positive and concave downward where Deflection. M positive exaggerated M negative Fig. 5 A sketch showing the deflection of beam AC. M(x) is negative, according to the sign convention that is given Fig. 5.6c. Where M(x) = 0, the beam is locally straight, that is, it is neither concave upward nor concave downward. We are able to sketch (Fig. 5) a plausi- ble deflection curve that passes over the supports at A and B and that is concave upward where M(x) is positive and concave downward where M is negative. The distributed load between A and B and the concentrated load at C could, indeed, cause the beam to deflect as sketched. Mo Fig. 1 B Fig. 2 A free-body diagram. (a) Load diagram. Vix) L (2) (1) MIX) (b) Shear diagram. (1) (2) (3) L L (c) Moment diagram. load on the beam, p(x) = 0 everywhere. Because of the concentrated couple at B, we need to consider two spans, 0<x<a and a<x<L. Solution Equilibrium-Reactions: We first determine the reactions A, and Cy. - 0: +C (ΣM)- +C (ΣM)= 0: -Mo -Mo-CL-0→Cy - A,L. - Mo=0→A, M Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Using these equations and the load diagram in Fig. 3a, we can sketch V(x) pro- gressively from x = 0 to x-L. It is convenient to sketch the shear and moment diagrams directly below the load diagram. Each step involved in sketching V(x) is numbered in Fig. 3b. 1. The shear at x=0" is zero. 2. The shear at x=0 is determined from Eq. 5.4, that is, AV₁ = A, M/L. Note that, because of the sign convention for shear, an upward concentrated force causes an upward jump in the shear diagram. 3. For 0<x<Lp(x) = 0. Therefore, from Eq. 5.2, dVidx = 0, V(x) = ML constant. 4. The reaction at C causes AV-M/L, which closes the shear diagram back to zero at x = L*. Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M(x) and can be (4) used to sketch the moment diagram in Fig. 3c. Steps in the construction. of the moment diagram are explained and numbered. (6) 1. The moment at x = 0 is zero (simply supported beam). 2. For 0<x<a, Eq. 5.3 gives dM/dx = V(x) = M/L = constant. 3. Atxa,M(a) can be determined from Eq. 5.7 as the area of the rectangle under the shear curve from x = 0 to x = a." Therefore M(a)-0fgV(x)dx = Moal L. 4. At x = a there is a negative jump in moment given by Eq.5.5. So Moa Mo(L-a) L L M(a)= 5. Fig. 3 Load, shear, and moment diagrams. 6. - Mo= For a <x<L, Eq. 5.3 gives dM/dx - V(x) - M/L - constant. Equation 5.7 gives M(L) M(a) = f(x)dx = (M/L)(La), which closes the moment diagram back to zero at x = L, as it should [simple support at C]. Review the Solution The dimensions on the shear diagram (F) and the moment diagram (F L) are correct. Note that both the maximum moment, Moall, and the minimum bending moment, -Mo(La)/L, occur at the cross section where the concentrated couple acts. Compare this example with the previous one, where there was a concentrated force. dV = p(x) dx dM == V(x) dx AVA = Po |AMB = MB V2 V₁ = x2 x1 p(x)dx M₂-M₁ = V(x) dx 3 kips/ft- 6 ft. 3 ft-