Statics Session 2 Problem 2 (ENGR5004)

Statics Session 2 Problem 2TutorialCurrent spec2:212 min

ENGR5004 Statics session 2 worked example 2 (Problem 2, slides 11-13 and Appendix 1 slide 16) official answers

A vertical steel bar ( $E = 200$ GPa) is fixed to rigid supports at A (top) and B (bottom), with a close fit at both supports before the loads are applied. Going down from A the bar passes points D, C and K: $l_{AD} = l_{DC} = l_{CK} = l_{KB} = 0.15$ m. Portion AC has area $A_{AD} = A_{DC} = 250\ \mathrm{mm^2}$ and portion CB has area $A_{CK} = A_{KB} = 400\ \mathrm{mm^2}$ . Downward loads $P_D = 300$ kN at D and $P_K = 600$ kN at K are applied.

Determine the reactions at A and B using the superposition method, treating the reaction at B as redundant.

Stepped vertical bar between rigid supports at A (top) and B (bottom). Points D, C, K divide it into four 0.15 m lengths; area A_AC above C, larger area A_CB below C; downward loads P_D at D and P_K at K.

Formulas you may need

Axial deformation of a stepped bar $\delta = \sum_i \dfrac{P_i l_i}{A_i E}$ (learn this)
Superposition: release the redundant support, find the deformation due to the loads $\delta_L$ and due to the redundant reaction $\delta_R$ , impose $\delta_L + \delta_R = 0$ (learn this)
Equilibrium $\sum F_y = 0$ (learn this)

(a)
With support B removed, find the elongation $\delta_L$ of the bar due to $P_D$ and $P_K$ .
delta_L =mm
(b)
Find the elongation $\delta_R$ due to an upward force $R_B$ alone at B (as a multiple of $R_B$ ), and hence $R_B$ and $R_A$ .
R_B =kN
R_A =kN