can I get this solved 2 The steel channel showm is to be wel

can I get this solved?

2) The steel channel showm is to be welded a steel plate on the three outer surfaces to a) What is the direct shear load on the weld, b) What is the torque (twisting) moment camed by the weld in-lbf? c) What is the bending moment camed by the weld, in-abr d) What is the shear stress per inch of weld at point e) Select the smallest sandard fillet weld size h necessary to support the loads shown if using an E7018 electrode, the maximum shear at point What is the factor of safety against failure of the steel channel and plate if both materials are ASTM A 36 steel with S. 36 kpsi and S. kpsi? P, 1200 lb 3000 lb 6.5 in 21.5 in

Solution

solution:

1)here given weld is subjected to direct,torsional and bending stresses and resultant stress would be algebric sum of all

2)here reaction or shear load at weld is

Fy=0

Ry=1200+3000=4200 lbf

3)here twisting about z axis if x and y axis are weld axis,then twisting is

T=torque=P2(12+4.5)=19800 lb in

4)here bending moment about x axis is

M=3000*21.5+1200*21.5

M=90300 lb in

5)here moment of inertia around x and y axis are

where x\'=l2/2=9/2=4.5\'\'

l1=l3=6.5\'\'

y\'=A1y1+A2y2+A3y3/A1+A2+A3

y\'=2(6.5)(6.5/2)t/(2*6.5*t+9*t)

y\'=1.92 \'\'

hence Ixx=2[(1/12)(6.3)^3t+6.5t(4.5-1.92)^20+9*1.92^2*t]

Ixx=165.48 t in4

where iyy=[(1/12)(l2^3t)+2(0+l1t+(x\')^2)]=324t

here J=Ixx +Iyy=489.48 t

6)hence direct stress at A

td=Ry/area=4200/22t=190.9/t

bending stress

tb=My/Ixx=90300*1.92/Ixx=1047.70/t

torsional stress

tt=Tx\'/J=182.02/t

hence resultant of torsional and direct stress is

angle m=44.49=arctan(4.5/6.5-1.92)

tr1=(td^2+tt^2+2*td*tt*cosm)^.5

tr1=345.17/t

where resultant of tb and tr1 is

here m1=90-m=45.51

tr2=(tb^2+tr1^2+2*tb*tr1*cosm1)^.5

tr2=1312.88/t

where factor of safety Nf=5

tall=.5Syt=18000

tall1=18000/5=3600 psi

hence here

tr2=tall1

t=.3646 in

and shear stress is

tr2=3600 psi for weld thickness t=.3646 in and depend on weld thickness