Question

To solve for forces in statically indeterminate bars with axial loads.

When the number of reaction forces is greater than the number of equilibrium equations, the system is statically indeterminate. Solving for the reactions requires some additional equations. These additional equations come from considering compatibility relationships (i.e.) continuity of displacements and relationships between displacements and loads).

For an axially loaded member, the compatibility relationship for the deflections can be written by setting the total relative axial displacement between the ends of the member to a known value. The load-displacement relationship δ=∑(NL/AE) gives another equation for the deflections. Once the internal normal force of each segment is written in terms of the end reactions and applied loads, there is enough information to solve for the reactions.

Part A - Force with a known deflection

The square bar shown (Figure 1 ) is 2.7 in. thick and 15 ft long and is fixed supported at both ends. A load directed leftward is applied at point C, as shown. L2=9.7 ft from the left end. The modulus of elasticity is E=1.5 × 10^{4} ksi. If point C moves δ=8 × 10^{-3} in. to the left. what is the applied force?

Part B - Reaction with a known load

Consider a new structure, where the thickness of the bar is reduced to 1.35 in. from C to B (it is still square) (Figure 2) and x=9.7 ft. If the applied load is F=85 kip, then what is the reaction at A ? Let a positive reaction act to the right. The total length is still 15 ft.

Part C - Load point for equal forces

Consider the structure from Part B (Figure 2). What value of x will lead to equal reaction forces at A and B ?

Answer 1