Tutorial 1 Problem 1Exercise sheetCurrent spec2:210 min

ENGR5004 Statics Tutorial Sheet 1, Problem 1 official answers

A rod of length LL, cross-sectional area A1A_1, and modulus of elasticity E1E_1 has been placed inside a tube of the same length LL, but of cross-sectional area A2A_2 and modulus of elasticity E2E_2. A force PP is applied on a rigid plate attached to both tube and rod, as shown in the sketch below. Determine:

HINT: deformation of tube and rod is constrained to be the same.

Rod (A1, E1) inside a tube (A2, E2), both fixed to the wall at the left and to a rigid end plate at the right; force P pushes the plate towards the wall.
Rod (A1, E1) inside a tube (A2, E2), both fixed to the wall at the left and to a rigid end plate at the right; force P pushes the plate towards the wall.
Formulas you may need
  • Axial deformation: δ=PLAE\delta = \dfrac{PL}{AE} (learn this)
  • Equilibrium of the end plate: P1+P2=PP_1 + P_2 = P (learn this: free-body diagram)
  • Compatibility (rigid plate): δ1=δ2\delta_1 = \delta_2 (learn this)
  1. (a)
    the horizontal displacement δ\delta of the rigid plate as a function of PP, LL, A1A_1, A2A_2, E1E_1 and E2E_2;
  2. (b)
    the fixed support reactions acting on the rod (P1P_1) and tube (P2P_2) when E1=E2E_1 = E_2;
  3. (c)
    the fixed support reactions acting on the rod (P1P_1) and tube (P2P_2) when E1=2E2E_1 = 2E_2;