Beam Deflection by Integration — Triangular Load Plus End Couple: 04-BS-6, December 2017, Q3

Question 3: Beam Deflection by Integration — Triangular Load Plus End Couple (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Span, L	6 m
Triangular load	0 at A rising linearly to 20 kN/m at B
Applied couple at A	120 kN·m, counterclockwise
Cross-section	75 mm × 200 mm rectangle
Elastic modulus, E	200 GPa

Find. (a) the maximum deflection and (b) the slope at A, both by direct integration of $EI\,y''=M(x)$.

Triangularly distributed load and applied end couple on the simply supported beam.

Approach. Find the reactions, write $M(x)$ for the triangular load plus the jump caused by the applied couple at A, then integrate twice and apply the two zero- deflection boundary conditions at the supports to solve for both constants of integration.

Reactions. Total triangular load $=\tfrac12(20)(6)=60$ kN, acting at its centroid $\tfrac23(6)=4$ m from A. Taking moments about A (CCW+, the applied couple entering directly as +120): $$R_B(6)-60(4)+120=0 \ \Rightarrow\ R_B=20\text{ kN},\qquad R_A=60-20=40\text{ kN}$$
Bending moment. With load intensity $w(x)=\tfrac{20}{6}x$ and the applied couple contributing a $-120$ kN·m jump immediately to the right of A: $$M(x)=R_A x-\frac{w_0}{6L}x^3-120=40x-\frac{20}{36}x^3-120\ \text{kN}\cdot\text{m}$$ Check: $M(0^+)=-120$ kN·m (hogging, consistent with the applied CCW couple); $M(6)=0$, correctly matching the moment-free roller at B. An independent free body taken from the right end (unaffected by the applied couple) reproduces this identical $M(x)$, confirming both the reactions and the sign of the jump term.
Section & first integration. $$I=\frac{bh^3}{12}=\frac{(75)(200)^3}{12}=5.00\times 10^{7}\text{ mm}^4=5.00\times10^{-5}\text{ m}^4,\qquad EI=(200\times10^9)(5.00\times10^{-5})=1.00\times10^{7}\text{ N}\cdot\text{m}^2$$ $$EI\,\theta(x)=\int M(x)\,dx + C_1$$
Second integration and boundary conditions. $$EI\,y(x)=\int\!\!\int M(x)\,dx\,dx + C_1x+C_2$$ Applying $y(0)=0$ and $y(6)=0$ (both supports have zero deflection) fixes $C_1$ and $C_2$ uniquely.
Slope at A (part b). Evaluating $\theta(x)$ at $x=0$: $$\boxed{\theta_A = 0.0156\text{ rad} = 0.894^\circ}$$ (the beam rotates in the same sense as the applied couple — counterclockwise — at A).
Maximum deflection (part a). Setting $\theta(x)=0$ and scanning the span for the largest $|y(x)|$ gives the extremum at $x=1.867$ m from A: $$\boxed{y_{max}=12.49\text{ mm (upward, away from the load), at }x=1.87\text{ m from A}}$$ This upward bulge is the dominant effect of the strong end couple; the deflection remains of this sign over the region where $M(x)$ is hogging (0 to ≈3.6 m) before the beam curves back down toward zero at B.

Check: sign convention throughout is the standard $EI y''=M(x)$ with y measured upward and M sagging-positive; the "maximum deflection" reported is the largest magnitude found by scanning the closed-form $y(x)$, which for this end-loaded case occurs inside the hogging region rather than under the heaviest part of the distributed load.

Quantity	Value
R_A	40 kN
R_B	20 kN
Slope at A, θ_A	0.0156 rad = 0.894°
Maximum deflection, y_max	12.49 mm, at x = 1.87 m from A

Topic: Elastic curve by double integration, with a distributed load and a concentrated end couple.

Key relations: $EI\,y''=M(x)$; a concentrated couple $M_0$ applied at a point introduces a jump of magnitude $M_0$ (with sign) directly into the $M(x)$ expression at that location, exactly like a concentrated force introduces a jump into $V(x)$.

Why this works: Because $M(x)$ is a single smooth cubic on this span (no other point loads), it can be integrated in closed form once, rather than needing Macaulay brackets for multiple segments; the two boundary conditions (zero deflection at both supports) are exactly enough to fix both constants of integration.

Common pitfall: Forgetting the jump term from the applied couple (which shifts the whole moment diagram down by 120 kN·m relative to the "load only" solution), or assuming the maximum deflection always occurs where the distributed load is largest — here it is dominated by the end couple instead.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 8 (deflection of beams by integration).

Question 3 of 8: Beam Deflection by Integration — Triangular Load Plus End Couple

Question 3: Beam Deflection by Integration — Triangular Load Plus End Couple (20 marks)