Engr Help

Deformation of a Circular Bar

*Example: Alternate Solution

Strategy: Replace the original bar with the three loads with three bars each with one load

as shown below. Then apply the relation for axial deformation, δ_i = P_iL_i/A_iE_i , for each

individual bar to determine the deformation of each bar. Again the deformations must then

sum to zero.

δ₁ = 9(L₁+ L₂ + L₃)/AE δ₂ = 15(L₁+ L₂)/AE δ₃ = F₁(L₁)/AE

Again take the positive direction as up. So

δ₁ + δ₂ + δ₃ = 0 = - 9(L₁+ L₂ + L₃)/AE - 15(L₁+ L₂)/AE + F₁(L₁)/AE

or - 9(2) - 15(1.4) + F₁ = 0 with the same result that F₁ = 39 kN

Return to Notes on Solid Mechanics