ENGR216 2025 Q A1 (ENGR5004)

ENGR216 2025 Q A1Past paperOld spec ENGR2162:125 marks30 min

ENGR216 Summer 2025 Q A1[VALID] official answers

Figure A1-a shows a structure of negligible mass constrained by a pin at end P and a roller at end R. Member PQ is loaded by a linear distributed load $w(x_v)$ , which is 0 at end P and equals $w_0$ at end Q, with $x_v$ denoting the position along member PQ. The coordinate along member QR, with origin at Q, is denoted by $x_h$ .

Figure A1-a: frame with vertical member PQ (length L, pin at P, triangular load) and horizontal member QR (length 2L, roller at R). Figure A1-b: T cross-section.

Formulas you may need

Resultant of a distributed load $P = \int w\,dx$ and its line of action $x_P = \dfrac{\int w\,x\,dx}{\int w\,dx}$ (on the formula sheet)
Equilibrium $\sum F_x = 0$ , $\sum F_y = 0$ , $\sum M = 0$ ; a pin gives two force components, a roller one force normal to its surface (learn this)
Load, shear and moment: $\dfrac{dV}{dx} = -w(x)$ , $\dfrac{dM}{dx} = V(x)$ , with the positive shear and positive bending pictures (on the formula sheet)
Transverse shear stress $\tau = \dfrac{VQ}{It}$ with $Q = \bar y\,A'$ (first moment about the neutral axis of the area on one side of the point) (learn this; on the 2025 sheet but not the 2026 one)
Radius of gyration $I = \rho^2 A$ (on the formula sheet)
Rectangle $I = \dfrac{bh^3}{12}$ (on the formula sheet); parallel-axis theorem $I = I_c + A d^2$ and centroid of a composite area $\bar y = \dfrac{\sum A_i \bar y_i}{\sum A_i}$ (learn this; only needed to check the given data)

(a)
Draw the free-body-diagram of the entire structure, indicating ONLY the nonzero reactions and applied loads.
[4]2:2
(b)
Calculate the reactions acting on the considered structure.
P_x (to the left +) =w0*L
P_y (downward +) =w0*L
R_y (upward +) =w0*L
[6]2:2
(c)
Knowing that the expression of the bending moment $M(x_v)$ along member PQ is: $M(x_v) = \frac{1}{2} w_0 L^2 \left[\frac{x_v}{L} - \frac{1}{3}\left(\frac{x_v}{L}\right)^3\right] \qquad \text{(Equation 1)}$ and the expression of the bending moment $M(x_h)$ along member QR is: $M(x_h) = \frac{1}{6} w_0 L^2 \left[2 - \frac{x_h}{L}\right] \qquad \text{(Equation 2)}$ determine the expression of the sharing load $V(x_v)$ along PQ and the shearing load $V(x_h)$ along QR.
V at P (xv = 0) =w0*L
V along QR =w0*L
[4]2:2
(d)
Draw as accurately as possible the shearing and the bending moment diagrams along PQ and QR, consistently with the expressions given or obtained above.
M at corner Q =w0*L^2
[4]2:2
(e)
Calculate the minimum length $L$ required for the magnitude of the maximum shearing stress in the cross section of member PQ at end P not to exceed 10 MPa. The cross section is reported in Figure A1-b. Use $a = 150$ mm, $b = 30$ mm and $w_0 = 40$ kN/m. The section is positioned so that the neutral axis of the bending moment is parallel to side AB. The centroid K of the section is at distance $d_K$ of 120 mm from side GH, and the radius of gyration past the neutral axis is 54.8 mm.
L (limiting length) =m
[7]