Question

***please be as accurate as possible this is extra credit because we did bad on test***

Determine the maximum shear stress in ksi to the nearest 0.1 ksi. P b a = 3 ft b= 4 ft C= 3 in d = 9 in P = 321 kip

Answer 1

Solution:

321 kip K 3" -3'- *4 Ra Rb Ra=32184 $2X4 = 183.43 kip. & Pb= 82723=184.57K. by T=0 166463-4) de. N.A. d. Imax.N.A. T=0. diu Here, consider section which is hatched, Ads bx (2-4). y = 1/2 ( 26-y)+Y. ty : j = d2 +/2Y = (a +y). Now, equation for shear stress po T = FA Ib Where, F= force or Load. (to kip or eb). I = moment of inertia of section whole A = Area of small sechon under considerchio Here, I=bd

:.- Fx [beda y)] x [% & ds ty] bd3 12 x 6 T=6F d2 bd² But, I man ir mazeimum at Neutral Azis, So, y=0 I= GF 6d² 6F bo T = 32 E bd = 1.5 [mean Now, f = Romas. = 183.43 kip la = .3" & d = 9" I = 312 188-43 (3x) 10. 19ksi. :: Masimum shear stress at Neutral Axels ir Iman = 10.19 ksi.