Question

1.51 In the steel structure shown, a 6-mm-diameter pin is used at C and 10-mm-diameter pins are used at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a factor of safety of 3.0 is desired, determine the largest load P that can be applied at A. Note that ink BD is not reinforced around the pin holes Front view 18 mm Side view 160 ının 120 mm Top view Fig. P1.51

Answer 1

taking the moment about C we have

$\sum$Mc=0

0.280P-0.120Fbd=0

P=0.120Fbd/0.280

P=0.429Fbd............1)

taking the moment at B

$\sum$Mb=0

0.160P-0.120C=0

P=0.75C......................2)

now we have tension on BD

Fbd=$\sigma$uAnet/Fos=400*10⁶*6*10^-3*8*10^-3/3=6.40 kN

shear force in pins at points B and D

Fbd=$\sigma$uApin/Fos=150*10⁶*pi*(10*10^-3)²/4*3=3.93kN

we to choose the smaller value of Fbd

so Fbd=3.93kN

so using eqn 1

P=0.429*3.93kN=1.68kN

shear force in pin at point C

C=2$\tau$Apin=2*$\tau$uApin/Fos=2*150*10⁶*3.14*(6*10^-3)²/3*4=2.83kN

so using equation 2)

P=0.75C=0.75*2.83=2.12kN

we have to choose the smaller value of P

so P=1.68kN

so the answer is 1.68kN.