Least work analysis of a propped beam: 16-Civ-B1 Advanced Structural Analysis, May 2014, Q3

Question 3: Least work analysis of a propped beam (16 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. A beam of constant $EI$, roller-supported at ① and encastré at ③, with an unloaded span ①–② and a uniformly distributed load over ②–③.

Given data
Quantity	Symbol	Value
Unloaded span, roller to point ③/2	$a$	3 m
Loaded span, point ② to the fixed end	$b$	6 m
Total span	$L=a+b$	9 m
Uniformly distributed load on $b$	$w$	1.8 kN/m
Flexural rigidity	$EI$	constant

Find. The bending moment and the shear force carried by the beam at the built-in support ③.

Question 3: propped cantilever, UDL over the outer 6 m only.

Approach. The beam has four reaction components and three equations, so it is once redundant. Take the roller reaction $R_1$ as the redundant, write $M(x)$ in terms of it, and impose the least-work condition $\partial U/\partial R_1=0$, which is exactly the statement that the roller does not deflect.

Choose the redundant and write the moment field. Measuring $x$ from the roller and taking the left free body, the sagging moment is $M=R_1x$ for $0\le x\le a$ and $M=R_1x-\tfrac{w}{2}(x-a)^{2}$ for $a\le x\le L$, so $\partial M/\partial R_1=x$ everywhere.
Impose least work. Because the roller is unyielding, the strain energy is stationary with respect to $R_1$:
Evaluate the two integrals. The first is $\int_0^{9}x^{2}\,\mathrm{d}x=243$ and the second, substituting $u=x-3$, is $\tfrac{w}{2}\int_0^{6}u^{2}(u+3)\,\mathrm{d}u =0.9\,(324+216)=486$.
Solve for the redundant. Dividing gives $R_1=486/243$, a pleasantly round result that is the tell-tale that the algebra is right.
Recover the actions at the fixed end by statics. The total applied load is $wb=10.8$ kN, so vertical equilibrium gives $R_3=10.8-2.0=8.8$ kN, and the shear immediately to the left of ③ is the algebraic sum of everything to its left.
Evaluate the moment at the fixed end. Substituting $x=9$ m in the second branch of $M(x)$ gives a hogging moment; the beam therefore sags over the first two-thirds of its length and hogs into the support.

The least-work condition is

$$\frac{\partial U}{\partial R_1}=\int_0^{L}\frac{M}{EI}\frac{\partial M}{\partial R_1}\,\mathrm{d}x=0\;\Longrightarrow\;R_1\int_0^{L}x^{2}\,\mathrm{d}x=\int_a^{L}\frac{w}{2}(x-a)^{2}x\,\mathrm{d}x$$

and substituting the two evaluated integrals,

$$R_1=\frac{486}{243}=\boxed{2.00\text{ kN}\ \uparrow}$$

With the redundant known the remaining actions follow from statics alone. At the built-in end,

$$M_3=R_1L-\frac{wb^{2}}{2}=2.00(9)-0.9(36)=\boxed{-14.4\text{ kN}\cdot\text{m}}$$ $$V_3=R_1-wb=2.00-10.8=\boxed{-8.8\text{ kN}}$$

The negative signs mean hogging and a downward-to-the-left shear respectively, i.e. the support hogs the beam by 14.4 kN·m and carries 8.8 kN of the 10.8 kN applied. Along the span the moment peaks at $+7.11$ kN·m at $x=4.11$ m and passes through zero at $x=6.92$ m.

Question 3: bending moment diagram, sagging positive.

Result	Value
Redundant roller reaction $R_1$	2.00 kN upward
Reaction at the fixed end $R_3$	8.80 kN upward
Bending moment at ③	−14.4 kN·m (hogging)
Shear on the beam at ③	−8.80 kN (8.80 kN magnitude)
Moment at ②	+6.00 kN·m sagging
Maximum sagging moment	+7.11 kN·m at $x=4.11$ m
Point of contraflexure	$x=6.92$ m from the roller

Topic. Castigliano's second theorem used as the theorem of least work.

Key relations. $\Delta_i=\partial U/\partial P_i$ with $U=\int M^{2}\,\mathrm{d}x/2EI$; for a redundant force $R$ acting at an unyielding support, $\partial U/\partial R=0$, i.e. $\int_0^{L}M\,(\partial M/\partial R)\,\mathrm{d}x/EI=0$.

Why this works. Least work is nothing more than Castigliano's theorem applied to a reaction whose corresponding displacement is known to be zero. The redundant enters $M(x)$ linearly, so $\partial M/\partial R$ is simply the moment field of a unit load at the redundant, and the condition collapses to the familiar flexibility statement $\delta_{10}+R\,f_{11}=0$. Because $EI$ is constant it cancels completely, which is why the answer is a clean 2.00 kN independent of the section. Splitting the integral at the start of the UDL is essential: the moment expression changes form there, and integrating one expression across the discontinuity is the commonest way to lose the question.

Common pitfall. Differentiating $M^{2}$ before splitting the integral at $x=a$, and forgetting that the $x$ in the second integrand is the full distance from the roller rather than the distance from the start of the UDL. Both give a redundant that is roughly right and therefore hard to spot.

Source. Hibbeler, Structural Analysis, 10th ed., §9.8–9.9 (Castigliano's theorem for beams and frames); Kassimali, Structural Analysis, 6th ed., §7.6 and Ch. 13.

Question 3 of 9: Least work analysis of a propped beam

Question 3: Least work analysis of a propped beam (16 marks)