1.2 EQUILIBRIUM OF A DEFORMABLE BODY 13 EXAMPLE 1.3 Determine the resultant internal loadings acting on the cross section at G of the beam shown in Fig. 1-6a. Each joint is pin connected. B 1500 lb A 1500 lb 3 ft 3 ft G D E- 300 lb/ft 2 ft 2 ft- 6 ft- (a) Fig. 1-6 6 ft- Fac = 6200 lb E, = 6200 lb E, = 2400 lb ² (6 ft) = 4 ft-- (6 ft)(300 lb/ft) = 900 lb (b) FBD = 4650 lb (c) SOLUTION Support Reactions. Here we will consider segment AG. The free-body diagram of the entire structure is shown in Fig. 1-6b. Verify the calculated reactions at E and C. In particular, note that BC is a two-force member since only two forces act on it. For this reason the force at C must act along BC, which is horizontal as shown. Since BA and BD are also two-force members, the free-body diagram of joint B is shown in Fig. 1-6c. Again, verify the magnitudes of forces FBA and FBD- Free-Body Diagram. Using the result for FBA, the free-body diagram of segment AG is shown in Fig. 1-6d. FBA = 7750 lb 1500 lb Equations of Equilibrium. 5 ΣΕ = 0; +1ΣF, = 0; 7750 lb (3) + No = 0 NG = -6200 lb -1500 lb + 7750 lb (3) - VG = 0 VG = 3150 lb C+ΣMG = 0; MG (7750 lb)(3)(2 ft) + 1500 lb(2 ft) = 0 MG = 6300 lb-ft 7750 lb 6200 lb MG 2 ft VG Ans. (d) Ans. Ans.