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ENGR5004 / Week 6
ENGR216 2023 Statics Q A1Past paperOld spec ENGR2162:125 marks30 min

ENGR216 Summer 2023 Statics Q A1[VALID]

Figure A1 shows a cantilevered structure of negligible mass. The vertical member AB of the structure is loaded by a parabolic distributed normal load w(x)w(x), which is given by:

w(x)=4w0[−(xL)2+(xL)](Equation 1)w(x) = 4w_0\left[-\left(\frac{x}{L}\right)^2 + \left(\frac{x}{L}\right)\right] \qquad \text{(Equation 1)}

where xx is the coordinate along the mean line of the vertical member AB (origin at A), which has length LL.

Figure A1: vertical member AB (length L, fixed at A) with horizontal parabolic load, horizontal member BC (length L).
Figure A1: vertical member AB (length L, fixed at A) with horizontal parabolic load, horizontal member BC (length L).
Formulas you may need
  • Distributed load resultant: P=∫w dxP = \int w\,dx, acting at xP=∫x w dx∫w dxx_P = \dfrac{\int x\,w\,dx}{\int w\,dx} (on the formula sheet)
  • Load-shear-moment relations: dVdx=−w\dfrac{dV}{dx} = -w, dMdx=V\dfrac{dM}{dx} = V and their integral forms (on the formula sheet)
  • Equilibrium of a rigid body: ∑Fx=0\sum F_x = 0, ∑Fy=0\sum F_y = 0, ∑MA=0\sum M_A = 0 (learn this)
  • Elastic curve: EI d2ydx2=M(x)EI\,\dfrac{d^2y}{dx^2} = M(x), slope θ=dydx\theta = \dfrac{dy}{dx}, integrate with the support conditions (learn this; on the 2025 sheet but not the 2026 one)
  • Small rotation of a straight member: transverse displacement =L θ= L\,\theta (learn this)
  • Mean normal stress: σmean=N/A\sigma_{mean} = N/A (learn this)
  • Pure bending: σx=−MyI\sigma_x = -\dfrac{My}{I}, σmax=∣M∣cI\sigma_{max} = \dfrac{|M|c}{I} (on the formula sheet)
  • Square section: I=a412I = \dfrac{a^4}{12} (from bh3/12bh^3/12) (on the formula sheet)
  1. (a)
    Draw the free-body-diagram of the entire structure, indicating only nonzero reactions and applied loads.
    [5]2:2
  2. (b)
    Calculate the reactions acting on the considered structure (3 marks), and draw the elastic curve of the member BC as accurately as possible, accounting for the suppressed degrees of freedom (2 marks).
    [5]2:2
  3. (c)
    Knowing that the bending moment M(x)M(x) along the vertical member AB is M(x)=w0L2[13(xL)4−23(xL)3+23(xL)−13](Equation 2)M(x) = w_0 L^2\left[\frac{1}{3}\left(\frac{x}{L}\right)^4 - \frac{2}{3}\left(\frac{x}{L}\right)^3 + \frac{2}{3}\left(\frac{x}{L}\right) - \frac{1}{3}\right] \qquad \text{(Equation 2)} demonstrate that the magnitude δC\delta_C of the vertical displacement of point C is: δC=110w0L4EI(Equation 3)\delta_C = \frac{1}{10}\frac{w_0 L^4}{EI} \qquad \text{(Equation 3)} where EIEI is the flexural rigidity (5 marks). State what is the mean normal stress in the cross sections of members AB and BC, and explain why (5 marks).
    [10]
  4. (d)
    Member AB has length L=2L = 2 m and features a square cross section, and w0=40w_0 = 40 kNm. Calculate the minimum length of the edge of the square cross section such that the normal stress at the midpoint of member AB does not exceed 130 MPa.
    [5]2:2