Question

(1) (2)

A. The square bar shown (Figure 1) is 2 in. thick and 12 ft long and is fixed supported at both ends. A load directed leftward is applied at point C, as shown, L2 = 10.6 ft from the left end. The modulus of elasticity is E = 1.5×10⁴ ksi. If point C moves δ = 3×10⁻³ in. to the left, what is the applied force?

B. Consider a new structure, where the thickness of the bar is reduced to 1 in. from C to B (it is still square) (Figure 2) and x=10.6ft. If the applied load is F = 95 kip , then what is the reaction at A? Let a positive reaction act to the right. The total length is still 12 ft .

C. Consider the structure from Part B (Figure 2). What value of x will lead to equal reaction forces at A and B?

Answer 1

Draw the free body diagram A F4- р &g {Fa=0 RA+&g-F=0 RB = F-RA {{net=0 FAC+ 8 eB=0 -RAL RB (12-22) AE -BA (10.6) + (F-RA) (12-10.6) АЕ ЖЕ, -RA (10.6) + (*-Ra) *.4 = 0 - (0-6 RA 1 :4F- - + R = 0. - 10. 6RA - 1.4R4 =1045

- 1Q84 = 1.46 RA = lot F RA = Oillor de = PAC 3X103 - RAX 10.6x12 2.2 x 1.5x104 3x10-3 = 0.116F) x10.6x12 2x2x1-57104 TF= 12.199 tipl Efazo RA+ Rg-F=0 Rg = F-RA PACT OCB =0 CA , 4 X 2) dex2.ca na t (05-R )(+2xT2 lmi Xlo

solve for RA I RA = nofkio EFX = 0 RPP-F = 0 2R = F R= 95 = 47.5 kip Ernet = 0 fact PCB=0 RA et RB (12-2) AE ALE J – 47.5XXX 12 242X2Q 47.5 (12-2)x12 1x1 - 6 a= 9.6ft