Question 3 of 8: Deflection by Integration, Triangular Load + Couple
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2017 — 04-BS-6: Mechanics of Materials (3 hours, closed book, one hand-written aid sheet permitted). Any FIVE of the eight questions constitute a complete paper on the official exam; every question is answered below. A wide-flange (W-shape) section-property table is attached at the end of the official exam; every question below supplies its own built-up or standard section directly, so the table is not needed for any of the eight solutions.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (axial members with initial gaps, shear/moment diagrams, deflection by direct integration, transformation of stress via Mohr's circle, combined axial+bending on eccentrically-loaded columns, torsion of stepped shafts, Euler column buckling, unsymmetric-section flexure of brittle materials).
Given. Pin at A (x=0), roller at B (x=9 m). Triangular load rising linearly from 0 at A to 60 kN/m at B, plus a 180 kN·m CCW couple at A. Cross-section: I-beam, 500 mm deep, 300×10 mm flanges (top and bottom), 10 mm web, E=200 GPa.
Quantity
Value
Span, L
9 m
Distributed load, w(x)
triangular, 0 at A to 60 kN/m at B
Applied couple, M0
180 kN·m, CCW, at A
Section
I-beam, d=500 mm, flanges 300×10 mm, web 10 mm
E
200 GPa
Find. Maximum deflection, slope at A, and whether the L/240 limit is satisfied.
Beam with triangular load and CCW couple at the left support, span 9 m.
I-beam cross-section: 300×10 mm flanges, 10 mm web, 500 mm overall depth.
Approach. Find the reactions from statics, write M(x) by direct integration of the triangular load with the couple's jump, then integrate $EIy″=M(x)$ twice with the two simply-supported boundary conditions to get slope and deflection.
Reactions. The triangular load resultant is $W=\tfrac12(60)(9)=270$ kN at $x=\tfrac23(9)=6$ m from A. Moments about A (CCW+, couple enters as +180):
$$180+R_B(9)-270(6)=0\ \Rightarrow\ R_B=160\text{ kN},\quad R_A=270-R_B=\boxed{110\text{ kN}}$$
M(x) by direct integration of the load. With $w(x)=\tfrac{60}{9}x=6.667x$ kN/m and the couple dropping M by 180 immediately to the right of A,
$$M(x)=R_Ax-180-\frac{w_{max}}{9}\cdot\frac{x^3}{6}=110x-180-\tfrac{10}{9}x^3\text{ kN}\!\cdot\!\text{m}$$
Check: $M(9)=110(9)-180-\tfrac{10}{9}(729)=0$, matching the zero-moment condition at the roller.
Section properties. The 500 mm deep I-section (300×10 mm flanges top and bottom, 10 mm web) has
$$I=\frac{b_fd^3-(b_f-t_w)h_w^3}{12}=4.5236e+08\text{ mm}^4$$
(b) Slope at A by integration. Integrating $EIy″=M(x)$ twice and fixing the two constants with $y(0)=y(9000\text{ mm})=0$ (simply supported) gives the slope at A
$$\boxed{\theta_0=-3.4320e-03\text{ rad}}$$
(a) Maximum deflection by integration. Setting $\theta(x)=0$, the maximum deflection occurs at $x=5039.3$ mm:
$$\boxed{y_{max}=-18.62\text{ mm (downward)}}$$
(c) Deflected shape and the L/240 check. The beam sags smoothly downward from both supports to the single maximum located just past midspan (toward the more heavily loaded end), with no reversal of curvature in between. The allowable limit is $L/240=9000/240=37.50$ mm. Since $|y_{max}|=18.62$ mm is less than this limit, the beam satisfies the deflection limit.
Exaggerated deflected shape: the beam sags downward everywhere between the supports, with maximum deflection near midspan (slightly toward the stiffer, more heavily loaded end).
