ENGR216 2017 Q A2 (ENGR5004)

ENGR216 2017 Q A2Past paperOld spec ENGR2162:125 marks30 min

ENGR216 Summer 2017 Q A2[VALID] official answers

Answer ALL parts (a) - (e)

Section A Appendix (as printed with the paper):

The differential relationship between shear load $V(x)$ and distributed load $w(x)$ acting on a beam is:

$\frac{dV}{dx} = -w(x)$

Consider the statically determinate beam with flexural rigidity $EI$ and length $L$ depicted in Figure A2, subject to the distributed load $w(x)$ , where $x$ denotes the coordinate along the horizontal axis depicted in the figure with origin at end A of the beam.

Figure A2: beam AB of length L with a roller support at A and a pin support at B. A downward distributed load w(x) rises from zero at A to its largest value at B (concave-down curve, flat at B). x axis along the beam from A to the right, y axis upwards with origin at A.

Formulas you may need

Elastic curve $\dfrac{d^2y}{dx^2} = \dfrac{M}{EI}$ (given in the question; learn this; on the 2025 sheet but not the 2026 one)
$dV/dx = -w(x)$ (given in the question; also on the formula sheet); $dM/dx = V(x)$ and the positive shear / positive bending sign convention (on the formula sheet)
Resultant of a distributed load $P = \int w\,dx$ and its position $x_P = \dfrac{\int x\,w\,dx}{\int w\,dx}$ (on the formula sheet)
Equilibrium $\sum F_x = 0$ , $\sum F_y = 0$ , $\sum M = 0$ ; a roller gives one reaction normal to its surface, a pin gives two (learn this)
Boundary conditions at simple supports: $y = 0$ ; slope $\theta = dy/dx$ (rotation of the section, small angles) (learn this)

(a)
Knowing that the bending moment $M(x)$ is positive over the entire beam, and noting that the relationship between $M(x)$ and the local deflection $y(x)$ is: $\frac{d^2y}{dx^2} = \frac{M}{EI} \qquad (1)$ where the y axis is depicted in Figure A2 and has origin at end A of the beam, sketch the deformed shape of the beam mean line. For your sketch, use the same Cartesian (xy) axes system as defined in Figure A2.
[3]2:2
(b)
Sketch the free body diagram of the loaded beam and determine the magnitude and orientation of the reactions acting on its two ends if the distributed load $w(x)$ is given by: $w(x) = w_0\left(\frac{2x}{L} - \frac{x^2}{L^2}\right) \qquad (2)$ where $w_0$ is a constant distributed load.
R_A (up +) =w0*L
R_B (up +) =w0*L
[5]2:2
(c)
Using the differential relationship between shear load $V(x)$ and distributed load $w(x)$ provided in the Appendix, demonstrate that, for the considered problem, the shear load along the beam is: $V(x) = w_0\left(\frac{L}{4} - \frac{x^2}{L} + \frac{x^3}{3L^2}\right) \qquad (3)$
[5]2:2
(d)
Knowing that the equation of the deformed mean line of the beam is: $y(x) = \frac{w_0}{EI}\left(\frac{x^6}{360L^2} - \frac{x^5}{60L} + \frac{Lx^3}{24}\right) + C_1 x + C_2 \qquad (4)$ determine the constants $C_1$ and $C_2$ by imposing the compatibility condition of the beam deformation expressed by Equation (4) with the constraints acting on the beam depicted in Figure A2.
C1 =w0*L^3/EI
C2 =w0*L^4/EI
[6]2:2
(e)
Determine the rotation (magnitude and direction) of the beam sections at ends A and B as functions of $w_0$ , $EI$ and $L$ .
theta_A (CCW +) =w0*L^3/EI
theta_B (CCW +) =w0*L^3/EI
[6]