Question

For the beam illustrated in the figure, find the locations and magnitudes of the maximum tensile bending stress due to M and the maximum shear stress due to V. The parameters are a = 7.8 in, b = 7.8 in, c = 1.1 in, h = 1.8 in, and w= 91 lbfin. w In O in in 4 psi. The moment of inertia is The maximum tensile bending stress due to Mis The maximum shear stress due to Vis psi.

Answer 1

Date Page No. Shivalal Ro rW dbflin. * B a b A RA Re and Let RA be the reactions at O and A. applying force equilibrium- Efy -RotRA - Wb=0 RA-Ro = Wb ΣΜΑ 2ο RAG - Wb (a + b) = RA= W (a+b) {a= b = 7.8" Civen w = 91 lbf/ in RA= 91. 91.17.8+ 7.8 1064.7 lbf 2. from ean 1064.7- Ro=9187.8 Ro= 354.9 lbf

Date 2 Page No. Shivalal moment of 90ea (I) Given his.8" Neutral bris will be the centroidal quis 4 . I = ch² 1.1X1.83 = 0.5346 in 12 maximum bending stress. Comon) To find o max o max first we have to find Maximum bending Mmax. moment Tog. 8 SFD- .A D 354.9 Bending moment will be maximum iat A Since shear changing its sign at A. force is Mmax = Roxq = 364.9 x 7.8 Mmax 32 295.9 abf - in -

Duotes... Unquotes Date Page No. Shivalal Maximum tensile bending Stress +M ymax max Mmax Imax I Mmax = 3229 5.9 lbf -in max 0.9 in 2 2 I 5346 int Omax 322 95.9 x 0.9 0.5346 o max 54,370.2 (psi) inz shear + from SED, it is it is clear that maximum force Umar = Tog.8 ci maximum Tmax - sheal Stress Vmax (A) I b Shear strey distribution will be parabolic with maximum at centre А 카 -y Ih shear stress distribution

Date max- 4 Page No. Shivalal Shivalal Tmax = 70 9.8x (ex h) x (6) 0.5 346 xc bie width of section at desired point? Emay = 709.8 x (LoLXlB. 13.) *41:) 0.534 6 X lil emax 2150.9 Abf (psi)