Flexibility (force) method for a bent frame: 07-Str-A4, May 2014, Q7

Question 7: Flexibility (force) method for a bent frame (24 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 two-member frame encastré at ①, with a horizontal member to the knee ② and an inclined member down to a roller at ③, carrying a UDL over the whole horizontal projection.

Given data
Item	Value
Horizontal member ①–②	3 m
Inclined member ②–③	run 3 m, drop 4 m, length 5 m
Support at ①	encastré
Support at ③	roller (vertical reaction only)
Uniformly distributed load	5.2 kN/m over the 6 m projection
Flexural rigidity	the same $EI$ in both members

Find. The redundant reaction, hence the complete shear force and bending moment diagrams with their extreme ordinates.

Question 7: bent frame, encastré at ①, roller at ③.

Approach. With four reaction components and three equations the frame is once redundant. Release the roller, integrate $\int M_0m\,\mathrm{d}s/EI$ and $\int m^{2}\,\mathrm{d}s/EI$ over both members to obtain $\delta_{10}$ and $f_{11}$, and enforce $\delta_{10}+R_3f_{11}=0$.

Choose the release. Removing the roller at ③ leaves a determinate cantilever built in at ① — the cheapest primary structure available, because both moment fields are then written from the free end with no reactions to find.
Write the two moment fields in one variable. All loads are vertical, so the lever arm of everything to the right of a section is horizontal, and both fields depend only on the horizontal coordinate $x$: $M_0=-2.6(6-x)^{2}$ and $m=+(6-x)$, valid on both members.
Respect the arc length. The integrals run along the member, not along $x$: on ①–② $\mathrm{d}s=\mathrm{d}x$, but on the inclined member $\mathrm{d}s=(5/3)\,\mathrm{d}x$. Forgetting that factor is the most expensive slip available in this question.
Evaluate the flexibility coefficients. With $u=6-x$, $\delta_{10}EI=-2.6[(6^{4}-3^{4})/4]-\tfrac{5}{3}(2.6)(3^{4}/4)=-789.75-87.75=-877.5$ and $f_{11}EI=(6^{3}-3^{3})/3+\tfrac{5}{3}(3^{3}/3)=63+15=78$.
Solve the compatibility equation. The roller does not settle, so $\delta_{10}+R_3f_{11}=0$; $EI$ cancels, which is why no numerical rigidity is needed.
Recover the remaining reactions and the diagrams. Vertical equilibrium gives $V_1=5.2(6)-11.25=19.95$ kN; moment equilibrium about ① gives $M_1=-26.1$ kN·m; the moment field $M(x)=11.25(6-x)-2.6(6-x)^{2}$ then supplies every ordinate.

Compatibility at the released roller requires

$$\delta_{10}+R_3\,f_{11}=0\;\Longrightarrow\;R_3=-\frac{\delta_{10}}{f_{11}}=\frac{877.5}{78}=\boxed{11.25\text{ kN}\ \uparrow}$$

Statics then completes the solution:

$$V_1=19.95\text{ kN},\qquad H_1=0,\qquad M_1=\boxed{-26.1\text{ kN}\cdot\text{m}}$$

The horizontal reaction is exactly zero because the roller supplies none and no horizontal load acts — which also means member ①–② carries no axial force at all.

Question 7: bending moment developed along ①-②-③, sagging positive.

Along the horizontal member the shear falls linearly from $+19.95$ kN at the built-in end to $+4.35$ kN at the knee, and the moment climbs from $-26.1$ kN·m to $+10.35$ kN·m, passing through zero at $x=1.62$ m. On the inclined member the shear normal to the member is what matters: it runs from $-2.61$ kN at the knee to $+6.75$ kN at the roller, vanishing 1.39 m down the slope, where the moment reaches its maximum of $+12.17$ kN·m before returning to zero at the roller.

Result	Value
Redundant reaction $R_3$	11.25 kN upward
Vertical reaction at ①	19.95 kN upward
Horizontal reaction at ①	0
Fixing moment at ①	−26.1 kN·m (hogging)
Moment at the knee ②	+10.35 kN·m (sagging)
Maximum sagging moment	+12.17 kN·m, 1.39 m along ②–③
Shear on ①–②	+19.95 kN to +4.35 kN
Normal shear on ②–③	−2.61 kN to +6.75 kN

Topic. The flexibility (force) method with one redundant.

Key relations. $\delta_{10}+R_1f_{11}=0$ with $\delta_{10}=\int M_0m\,\mathrm{d}s/EI$ and $f_{11}=\int m^{2}\,\mathrm{d}s/EI$; $M_0$ is the moment field of the loads on the released structure and $m$ that of a unit value of the redundant. On an inclined member $\mathrm{d}s=(L/L_{\text{proj}})\,\mathrm{d}x$.

Why this works. The force method reduces an indeterminate frame to two integrals over a determinate one, and the whole art is choosing a release that makes those integrals easy. Cutting the roller here leaves a cantilever, so both moment fields are written directly from the free end — no reactions to solve for first. Because all applied loads are vertical, the lever arms are purely horizontal and one variable $x$ describes both members; only the arc-length factor distinguishes them. Constant $EI$ then cancels completely, which is what allows a numerical answer with no section properties given.

Common pitfall. Integrating with $\mathrm{d}x$ instead of $\mathrm{d}s$ on the sloping member (here it would give $R_3=12.4$ kN, about 10 per cent high), and plotting the vertical shear rather than the shear normal to the inclined member.

Source. Hibbeler, Structural Analysis, 10th ed., Ch. 10 (force method for beams and frames); Ghali, Neville and Brown, 7th ed., Ch. 4.

Question 7 of 9: Flexibility (force) method for a bent frame

Question 7: Flexibility (force) method for a bent frame (24 marks)