8 kN 8 kN/m 4 m Fig. 1 40 kN -2 m- 16kN For the beam in Fig. 1 of Example Problem 5.9, (a) use discontinuity functions to obtain expressions for p(x), V(x), and M(x), and (b) use the discontinuity functions from Part (a) to construct shear and moment diagrams for the beam, indicating the contribution of each term in the discontinuity-function expressions. The loads and reactions from Example Problem 5.9 are given in Fig. 1. Plan the Solution We can refer to Cases 2 and 3 of Table 5.2 to construct the load function p(x) and then to perform the required integrations to get V(x) and M(x). Solution (a) By referring to the Load column for Cases 2 and 3 in Table 5.2, we can write - p(x) (8 kN)(x 0m)-1-(8 kN/m)[(x -0 m)-(x-4 m)] +(40 kN) (x -4 m) (16 kN) (x-6 m)¹ Ans. (1) TABLE 5.2 Case A Summary of Loads, Shear, and Moment Represented by Discontinuity Functions Load Shear Ma 1 2 3 PX-- TL 5 P2 6 Po (2)x-as (3)α-973 Po- -Max-a- a Moment M -Mix- Mo Mo M M M M M ( a Pox- Pa-a² (2)α-a³ A B 1.5 kips/ft 8 ft- x 20 kips 12 ft C Figure 1 shows the simply supported beam of Example 5.2, including the reactions. The expressions for V(x) and M(x) obtained in Example 5.2 are V₁ (220 40x) lb, 0<x6ft. V₁ =[-140+(12-x)²] lb. 6≤x<12ft = M₁ (220x20x)lb ft, 0<xs6 ft My [140(12 x)-(12-x)] lb ft, 6x < 12 ft = (a) Using the above expressions for V(x) and M(x), plot shear and moment diagrams for this simply supported beam. (b) Determine the location of the section of maximum bending moment, and calculate the value of the maximum moment. 40 lb/ft A 220 lb 140 lb -6ft 6ft Fig. 1 Plan the Solution It is straightforward to plot the shear and moment diagrams from the given expressions (e.g., using a computer). From the shear diagram, the location of the section where V(x) = 0 can be deter- mined. Then the appropriate moment equation can be used to determine the value of the moment at this critical section. Since there is more load over the left half of the beam than over the right half, the maximum mo- ment should occur to the left of x = 6 ft. Solution (a) Plot the shear diagram and the moment diagram. Figure 2 shows the plots of shear and bending moment. (a) The shear diagram. уль) 220 12 -x(ft) -140 M(lb-ft) M-605 lb-ft (b) The moment diagram. x(ft) Fig. 2 Shear and moment diagrams. (b) Determine the maximum bending moment. The maximum moment occurs where the shear vanishes. From the shear diagram and the equa- tion for V₁(x), it can be observed that V(x) = 0 in the interval 0<x<6 ft. Therefore, V₁(x)=220-40x = 0→x = 5.50 ft Then, the maximum moment is Max M (5.50 ft) = 605 lb ft Ans. (b) Review the Solution The downward load on this simply supported beam bends it downward, so it is concave upward everywhere. This is consistent with the fact that the bending moment is positive for the en- tire length of the beam. As expected, since there is more load over the left half of the beam than over the right half, the maximum moment does occur to the left of x = 6 ft. This is an example of a maximum moment that occurs where V = dM/dx = 0. Derive expressions for V(x) and M(x) for the cantilever beam with lin- early varying load shown in Fig. 1. Use these expressions to plot shear and moment diagrams for this beam. Po pl), pla) MOO Fig. 1 Plan the Solution We can use a finite free-body diagram to determine the required expressions for V(x) and M(x). Solution The triangle in the free-body diagram in Fig. 2 is similar to the triangle in the problem statement; so, by similar triangles, Po P(x) = P(x) = x Vix) +1ΣFy=0: Fig. 2 A free-body diagram. L (+)P(x) - V(x) = 0 Aur V(x)-2 21 Po Ans -M(x)-0 +C(M)-- 20000 P 1.00 0.75- 0.50 0.25- Quadratic- curve 0.00 0.2 0.4 06 08 14 (a) The shear diagram POR? 1.00 075 0.50- Cubic curve 0.25 0.00 0.2 0.4 0.6 0.8 1.0 (b) The moment diagram Fig. 3 The shear diagram and the moment diagram. M(x)= 6L Ans To plot these expressions for the shear force V(x) and bending mo- ment M(x), we first calculate V(L.) and M(L). Figure 3 shows the plots of V(x) and M(x). V(L)-PM(L)-P Review the Solution The shear is positive everywhere, as we expect from the free-body diagram. The maximum shear occurs at the cantilever support at B and is equal to the total area under the load curve in Fig. 1. The upward load will bend the beam upward, so it will be concave upward everywhere. This is consistent with the fact that the bending moment is positive for the entire length of the beam. The maximum bending moment occurs at B. Finally, the shear and moment have the proper dimensions (F) and (F L), respectively.