Question

A member ABCD is subjected to point loads P1, P2, P3 and P4 as shown in Fig. Calculate the force P3,necessary for equilibrium if P1 = 120 kN, P2 = 220 kN and P4 = 160 kN. Determine also the net change in length of the member. Take E = 200 GN/m2.

Answer 1

$\sum P=0$

$P_1+P_2+P_3+P_4=0$

120+220+160+P₃=0

P₃=-500kN

$∆l=\frac{\sigma l}{E}$

$=\frac{500000*25}{25*15*200*10^9}$

=1.6667*10^-7