Statically Indeterminate Tension in Wires with Deformation

Prequisite Knowledge
General Knowledge of Statics
Deformation Due To Axial Loading

Problem
A rigid bar ABC is suspended from three wires of the same material. The cross-sectional area of the wire at B is equal to one-third of the cross-sectional area of the wires at A and C. Determine the tension in each wire caused by the load P.



Solution
(This is known as a statically indeterminate problem)

The problem implies that the forces at A, B, and C shall be found in terms of P. Therefore the force P can be considered as a known value. Then, three unkowns must be found. This means that three equations are required.

The first two equations are from statics:

Equation 1 - force balance

(Eq1)    P = PA + PB + PC

Equation 2 - moment balance

∑ MC = 0 = 0.5LP = LPB + 2LPA

since Ls cancel:

(Eq2)    0.5P = PB + 2PA

Equation 3 - deformation

The third equation involves a little bit of common sense.

The third equation should be some kind of relation between the deformations in the wires. But it is not fixed anywhere, there is no pivot point. But as the bar is rigid, the deformations can be examined as such:



which then means that:

(Eq3*)    δA + δC = 2δB

And, using the equation for deformation due to axial loading:

δA =
PAL
3AE
δB =
PBL
AE
δC =
PCL
3AE

L, A, and E cancel out, so:

δA =
PA
3
δB =
PB
δC =
PC
3


substituting these values into (Eq3*) yields equation 3:

(Eq3)    0.3333PA + 0.3333PC = 2PB

Three equations, three unknowns, the problem can be solved.

Nothing but basic substitution follows:

from Eq2:

PB = 0.5P – 2PA

from Eq3:

PC = 6PB – PA

Combining the previous two equations:

PC = 3P – 12PA – PA = 3P – 13PA

Substituting for PB and PC into Eq1:

PA + 0.5P – 2PA + 3P – 13PA = P

and solving for PA in terms of P:

-14PA + 3.5P = P

14PA = 2.5P

PA = 0.17857P

PB = 0.5P – 0.35714P

PB = 0.14286P

0.17857P + 0.14286P + PC = P

PC = 0.67857P

It may be checked to see that these values satisfy the three equations.