Question

Q1. (45p) Use Castigliano's Theorem to find the vertical deflection at point B for the given member under the load of P. The member is fixed at A and free at the other end. 450 Divide the member in two sections AB and BC. Use the figures below if necessary BF a=45° Use the values below to compare the contribution of each component to total deflection. R = 75 mm, h = 30 mm, b = 15 mm. P = 10 kN, E = 201 GPa, G = 80 GPa.

Answer 1

Actually for vertical detection, we derivate the the to total Strain energy with vertical force, so here we Tust convert the pinto holizontal and vertical components and again we convert the horizontal component interms of vertical component. prin 450 Phy Du = Ph Pows uso = Pv K 450 Pr all are in Pro 22 at lost come in terms of then we derivate (U) Olv) Br. the ber

2 ROOS 45- RUOSO ba Rsino & sin 45 70 Resultant Moment M = Puxa - Puxb Transferving forces foom B to another point - Pv Ca-b) Pvc - BLOSO - Rain 0 + B Pu ( S2 R-RCOSO - Reino) Da length of clement at point D = ds = R do strain energy ( m2 ds 2 EL es einenensy - (12 ds 0 - 180° ol (PvR O VR-CODO - lino,] R do 2 EI 0- 450

] do P R S [2 + cos²O + sinzo + 252 coso +22050 sino -252 sino REI 78 - 180° (20 + 0 = 252 sino - cos 20 + 252 coso ) I BURB 2 EI 0=450 - PuzR3 [( 311 - < =252) - (35-2-0 +221 221 - 1 P R3 [ 3.740156] - U and 26D JU PV R = Avertical) (3.740156) - OP B EI

Due to shear suhren 8S Kl ve J RAG du 1 = JP 2 482 I 3 x P2R [ 9 P²R LAG 11 ] Shear fdce everywhere con Aan - P 18 Pu ²R 16 2 AG odoto hii K - 3 to non circulas OU 9 PUR 8 AG da z R do ӘР,

1.2 12 ID 663 30x153 cm = 8437.5 mmy A=bxh = 30x15=450 R = 75 mm, p= 10x 103 N, E= 2018103 Moa, G= 80 x 10 mpa Pr = f / 2 = 552 KN Dv = (Ovi) + au), - PR3 13.74015) + / MG EI 175897 + 0.01657 a PUR 1.77554 mm (total Vettical defrection