Statics Session 2 Problem 1 (ENGR5004)

Statics Session 2 Problem 1TutorialCurrent spec2:28 min

ENGR5004 Statics session 2 worked example 1 (Problem 1, slides 7-9) official answers

Bar AB of length $L$ and uniform cross section $A$ is attached to rigid supports at A (top) and B (bottom) before being loaded. A downward axial load $P$ is then applied at point C, a distance $L_1$ below A and $L_2$ above B ( $L_1 + L_2 = L$ ). The bar is linear elastic with Young's modulus $E$ .

Left: vertical bar AB between rigid supports at A (top) and B (bottom), with a downward load P at C; AC = L1, CB = L2, total length L. Right: free-body diagram with upward reactions R_A at A and R_B at B.

Formulas you may need

Equilibrium of the bar: $\sum F = 0$ (learn this)
Axial deformation of a uniform segment $\delta_i = \dfrac{P_i L_i}{A_i E_i}$ , summed over segments $\delta = \sum_i \dfrac{P_i L_i}{A_i E_i}$ (learn this)
Compatibility for a bar between rigid supports: total change in length $= 0$ (learn this)
Normal stress $\sigma = P/A$ (learn this)

(a)
Draw the free-body diagram of the bar and explain why the reactions cannot be found from equilibrium alone.
(b)
Find the reactions $R_A$ and $R_B$ in terms of $P$ , $L_1$ , $L_2$ and $L$ .
(c)
Find the normal stresses in portions AC and BC, stating whether each is tensile or compressive.