Tutorial 2 Problem 3 (ENGR5004)

Tutorial 2 Problem 3Exercise sheetCurrent spec2:125 min

ENGR5004 Statics Tutorial Sheet 2, Problem 3 official answers

Consider the loaded beam shown below, subject to a triangular distributed load, constrained by a roller at its left end, and pinned at its right end.

NOTE: the beam of problem 3 is that for which we computed shearing load $V(x)$ and bending load $M(x)$ in problem 4 of Tutorial 1. Please start the solution of problem 3 (tutorial 2) by using the expression of $M(x)$ determined in problem 4 of Tutorial 1.

Beam AB of length L, roller at A (left), pin at B (right), triangular load rising from zero at A to w0 at B; x along the beam from A, y up.

Formulas you may need

Load-shear-moment relations: $\dfrac{dV}{dx} = -w$ , $\dfrac{dM}{dx} = V$ (on the formula sheet)
Elastic curve: $EI\dfrac{d^2y}{dx^2} = M(x)$ , maximum deflection where $\dfrac{dy}{dx} = 0$ (learn this; on the 2025 sheet but not the 2026 one)

(a)
Calculate the position $x^*$ along the beam axis where the maximum deflection of the beam occurs, expressing it as a fraction of the beam length $L$ ;
x* =L
(b)
calculate the absolute value of the maximum deflection of the beam $\delta = |y(x^*)|$ , expressing it as a function of the flexural rigidity $EI$ , the beam length $L$ , and the distributed load $w_0$ at beam end B (note that $E$ is Young modulus and $I$ is the second area moment past the cross section neutral axis), and state if the deflection is upwards or downwards.
delta =w0*L^4/(EI)
(c)
calculate the numerical value of the maximum deflection $\delta$ , assuming $I = 0.333\times10^{-3}\ \mathrm{m^4}$ , $w_0 = 66$ kN/m, $L = 6$ m, and $E = 200$ GPa.
x* =m
delta =mm