Question

A simply supported beam has a cross-section that is formed using of Steel and Aluminum as shown. Find the normal stress at point "b" at the point of maximum moment for the loading shown. All the dimensions are in inches. 50 lb/in 240 Steel Es=29000 ksi Aluminum EA-10000 ksi 4 1

Answer 1

50lbis 2) là A IVA – 2001 (x Tve + {En=0 THA=0 +1 Ery=0 VA+Ve-1x240850=0. IVA + VB = 6000 lb- DEMAFO -x240x50 x 2x240 - Vex240=0 I VA - 2000 lb. [ Ve = 4000 Mb ) Shear Fonce sign Convention - The upward force left side of the section is taken as positive shear 0 lice Force. Od

221 where shear Force will be zero there Bending moment will be masinum because shear force is changing from (tve) to (-ve). Shear force at 8-x=0. Using similare VA- 1 xwxx=0. Triangle Principle الي - مناطل 50 2000 = 4x50 x3 240 240 a w=500 x=138.56 is from left support 240 U

Bending moment sign Convention - The clockwise Moment left side of the section is taken as Sagging Bending Moments (tre). Al 2 = 138.56 in Bending Moment - Maninum. Bm va xx-xoxxxx - 2000 x 138.56 - 1x 50 x (138.56) 2 240 B. Mman = 184752.085 Ab-in

Equivalent Aluminium Section 2900 2.9. Modular Ratio - Es = a - EAL keeping depth same we will change width any - Toom ate Centroid). 1-ons Tol .3408 2-167 UUUUUvecececcCECECEC I we use To Calculate centroid of section we use -2.167 in 34.8 X1X0.5 +12X1X 7 34.881 +1281 12 To Calculate moment of Inertia we use Parallel Anis theorem. I= 34.8*1*34.8X1X (2:169-05+ 1x12 + 12xIx(9- (2-167in) I= 523.9 is 4 To Calculate Normal stress at b we use Bending to Equation M = my! m = B. Mman y= Distance to be from I=523.9ine I Centroidal Neutral asis y=2.167 in 2 -

O = 184752.085 X2.167 523.9 18=764.18 lbs But this stress is for Aluminum we calculated in reality there is steel at that point. Bending stress at 'b'= 764.18x modular Ratio = 764.1882.9 19 = 2216-1433 No