ΔA=2.946×10−4 in

Given information:

The members are pin connected at end points.

A = 3 in²

E = 29(10³) ksi

Virtual Work Method

Virtual external work done = Sum of the internal work done

   P×Δ=∑i n i N i L i EAwhere i = force membersn = force in members due to virtual unit loadN = force in members due to original loadL= length of the force members

Calculation:

The virtual and real member forces are calculated by method of joints.

Member forces when original load is applied:

Taking moment about B, we get,

   −(12×3)+(6×6)=−Cx×8Cx=0Bx+Cx=0Bx=0By=−3 kNCy=−6 kN

Joint A:

   ∑Fy=0NADsin 53.13o=−3NAD=−3.75 kN∑Fx=0−NADcos 53.13o=NABNAB=2.25 kN

Joint B:

   ∑Fy=0NBDsin 53.13o=ByNBD=3.75 kN∑Fx=0NBDcos 53.13o=−NBA−BxNBA=−2.25 kN

Joint D:

   ∑Fx=0NBDcos 53.13o=−NDC+NDAcos 53.13oNDC=0

Member forces when virtual load is applied:

Taking moment about B, we get,

   Cx=0Bx+Cx=0Bx=0By=−1 kNCy=0 kN

Joint A:

   ∑Fy=0nADsin 53.13o=−1nAD=−1.25 kN∑Fx=0−nADcos 53.13o=nABnAB=0.75 kN

Joint B:

   ∑Fy=0nBDsin 53.13o=−BynBD=1.25 kN∑Fx=0nBDcos 53.13o=−nBA−BxnBA=−0.75 kN

Joint D:

   ∑Fx=0nBDcos 53.13o=−nDC+nDAcos 53.13onDC=0

   Membern ( kN)N ( kN)L ( ft)nNLAB0.752.251220.25AD−1.25−3.751046.875BD1.253.751046.875DC0060

   P×ΔA=∑i n i N i L i EA∑i n i N i L i EA=114 kN29× 103×3=25.628 kpf29× 103×3=2.946×10−4 in1 kN ×ΔA=2.946×10−4 inΔA=2.946×10−4 in

Conclusion:

   ΔA=2.946×10−4 in