Question

P10.037 (GO Tutorial) For the beams and loadings shown, determine the beam deflection at point H. Assume that EI = 9.0 × 10-a kN-m2 is constant for each beam (a) Beam 1 Assume LHA = 2.75 m, LAB 7.00 m, LBC = 2.00 m, MC 200 kN·m. HA AB Answer: VH = (b) Beam 2 Assume IHA = 1.50 m, LAD = 4.75 m, w = 8 kN/m. AB Answer: VH - (c) Beam 3 Assume LAH = 48-lge* 3.50 m, ρ = 40 kN LHB AT Answer:VH (d) Beam 4 Assume LAB- 8.75 m, LeH-3.75 m, P 15 kN AB Answer: VH

Answer 1

a) We apply beam deflection formula to get beam rotation at A: It may be noted that we translate moment M_C to B without affection rotations at A and B

The deflection at H is given by

$v _H=\theta_AL_H_A=(0.002593)(2.75)=0.00713\: m$

b) We apply the formul $v_H=\frac{wL_A_B^3}{24EI}[4(L_H_A+L_A_B)-L_A_B]=\frac{8000(4.75^3)}{(24)(9\times10^7)}[4(4.75+1.5)-4.75]=0.008038\: m$

c) We define

We apply the formula

$v_H=\frac{PL_B_CL_A_H}{6LEI}(L^2-L_A_H^2-L_B_C^2)=\frac{(40000)((3.5)(3.5))}{(6)(10.5)(9\times10^7)}(10.5^2-3.5^2-3.5^2)=0.00741\: m$ d) We use the formula

$v_H=\frac{PL_B_H^2}{3EI}(L_A_B+L_B_H)=\frac{(15000)(3.75^2)}{(3(9\times10^7))}(8.75+3.75)=0.009766\: m$