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ENGR5004 / Week 4
ENGR273 2026 Q1Past paperCurrent spec2:125 marks30 min

ENGR273 Summer 2026 Q1[VALID] official answers

Figure Q1-1 shows a structure of negligible mass constrained by a fixed support at A, loaded by a distributed triangular normal load w(x)w(x) along beam DB and subjected to a concentrated moment ME=13w0L2M_E = \frac{1}{3} w_0 L^2 at the end E of beam CE. Beam DB has length 2L2L, and parts AB, BC and CE have each length LL. The expression of the distributed load is:

w(x)=w0(1−x2L)(Q1-1)w(x) = w_0 \left(1 - \frac{x}{2L}\right) \qquad \text{(Q1-1)}

where xx is the horizontal coordinate along beam DB with origin at D, and w0w_0 is the value of the distributed load at end D.

Figure Q1-1: frame fixed at A; column ABC, beam DB with triangular load w(x) (w0 at D), arm CE with moment M_E at E.
Figure Q1-1: frame fixed at A; column ABC, beam DB with triangular load w(x) (w0 at D), arm CE with moment M_E at E.
Formulas you may need
  • Resultant of a distributed load P=∫0Lw(x) dxP = \int_0^{L} w(x)\,dx and its position xP−x0=∫0Lw(x)(x−x0) dxPx_P - x_0 = \dfrac{\int_0^{L} w(x)(x - x_0)\,dx}{P} (on the formula sheet)
  • Positive shear and positive bending sign convention (pictures) (on the formula sheet)
  • dVdx=−w(x)\dfrac{dV}{dx} = -w(x), dMdx=V(x)\dfrac{dM}{dx} = V(x), VD−VC=−∫xCxDw dxV_D - V_C = -\int_{x_C}^{x_D} w\,dx, MD−MC=∫xCxDV dxM_D - M_C = \int_{x_C}^{x_D} V\,dx (on the formula sheet)
  • Maximum bending stress σmax=∣M∣cI\sigma_{max} = \dfrac{|M| c}{I} (on the formula sheet)
  • Solid circle I=πr44I = \dfrac{\pi r^4}{4} (on the formula sheet)
  • Static equilibrium ∑Fx=0\sum F_x = 0, ∑Fy=0\sum F_y = 0, ∑M=0\sum M = 0; a fixed support gives two forces and a couple (learn this)
  1. (a)
    Draw the free-body-diagram of the entire structure, indicating ONLY the nonzero reactions and applied loads.
    [6]2:2
  2. (b)
    Calculate the reactions acting on the structure in figure Q1-1.
    [6]2:2
  3. (c)
    Calculate and draw as accurately as possible the shearing load diagram along beam DB, and the bending moment diagram along column ABC.
    [8]
  4. (d)
    Calculate the minimum diameter of the circular cross section of beam DB required for the maximum normal stress in any of the beam's cross sections not to exceed 150 MPa. Use w0=40w_0 = 40 kN/m and L=5L = 5 m.
    [5]2:2