Question

Q2. A solid wooden beam (pine, with a maximum longitudinal stress 30 MPa, maximum shear stress 3 MPa and E = 9 GPa) spans a length 3L = 4.5 m (L = 1.5 m), and is supported as shown in Figure Q2. The beam is b = 10 cm wide and h = 50(1+S) cm high. The distributed force load q = 200 kN/m is applied on either side, as shown below. 21 Figure Q2 a) Derive expressions for the internal shear force V(x) and bending moment M(x) and draw the internal load diagrams as a function of x. [15 marks]

S=0.5845

Please include all steps.

Answer 1

Lum 9- 200 kN/m AK I D Asked to F.BiD: * I * 1 TB C find (derive) Vx & M(x) 122 Tet 1 2 3 Ay +84-91-220 22-200x15 - 300KN to Ay By = 600 – &MA=0 (+V)) BX3L – 24 (52) – 24 ( 1 ) = 0 3By = 5.21+ 2+ 38 = 300x5 + 300 By 900 -> By = 300 KN) bvom Ay = 600-300 =>44= 300 kN)

internal load diagram: - (bor 225x232) K t A-300 ut u , (2-22) 2(x-222 (2-24) J => ) shear bovce (intemal) V);- x=0 (1 +ve) -> V(x) + 2 (4-22) + 2L - Ay = 0 VOA) = 300-20-2) - 2L V64) = 300 - 200(1-3)-300 V(1) - - 200(4-3) 607 33*44.5