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ENGR5004 / Week 6
ENGR216 2024 Statics Q A1Past paperOld spec ENGR2162:125 marks30 min

ENGR216 Summer 2024 Statics Q A1[VALID]

Figure A1 shows a constrained structure of negligible mass constrained by a roller at end P and a pin at end R. End P is loaded by a horizontal concentrated load FF, as indicated in Figure A1.

Figure A1-a: frame with vertical PQ (length L, roller at P, horizontal load F at P) and horizontal QR (length 2L, pin at R). Figure A1-b: I cross-section.
Figure A1-a: frame with vertical PQ (length L, roller at P, horizontal load F at P) and horizontal QR (length 2L, pin at R). Figure A1-b: I cross-section.
Formulas you may need
  • Equilibrium of a rigid body in the plane: ∑Fx=0\sum F_x = 0, ∑Fy=0\sum F_y = 0, ∑M=0\sum M = 0 (learn this)
  • Internal forces by the method of sections; sign convention for positive shear and positive bending as pictured, dMdx=V\dfrac{dM}{dx} = V (on the formula sheet)
  • Pure bending σx=−MyI\sigma_x = -\dfrac{My}{I}, σmax=∣M∣cI\sigma_{max} = \dfrac{|M|c}{I} (on the formula sheet)
  • Combined axial force and bending (eccentric loading form) σx=NA−MyI\sigma_x = \dfrac{N}{A} - \dfrac{My}{I} (on the formula sheet)
  • Rectangle I=bh312I = \dfrac{bh^3}{12} (on the formula sheet); I-section by subtraction or by three rectangles with the parallel-axis theorem I=Ic+Ad2I = I_c + Ad^2 (learn this)
  • Elastic curve EI d2ydx2=M(x)EI\,\dfrac{d^2y}{dx^2} = M(x) for the curvature argument in (d) (learn this; on the 2025 sheet but not the 2026 one)
  1. (a)
    Draw the free-body-diagram of the entire structure, indicating only nonzero reactions and applied loads.
    [4]Third
  2. (b)
    Calculate the reactions acting on the considered structure.
    [4]2:2
  3. (c)
    Calculate the axial load distribution N(x)N(x), the sharing load distribution V(x)V(x) and the bending moment distribution M(x)M(x) along members PQ and QR.
    [6]2:2
  4. (d)
    Draw as accurately as possible the bending moment diagram M(x)M(x) along PQ and QR and the elastic curve of the whole structure, explaining why the curvature of the elastic curve is consistent with the diagram of M(x)M(x) in both PQ and QR.
    [6]
  5. (e)
    Calculate the minimum length LL required for the magnitude of the maximum normal stress in the cross section of PQ at distance 0.95L0.95L from P not to exceed 120 MPa. The beam cross section is reported in Figure A1-b. Use a=t=30a = t = 30 mm, b=270b = 270 mm and F=100F = 100 kN. The section is positioned so that the neutral axis of the bending moment alone is parallel to side AB.
    [5]