Question 7 of 8: Three-hinged frame under a point load and a horizontal distributed load (alternative 2 of 3)

Question 7: Three-hinged frame under a point load and a horizontal distributed load (alternative 2 of 3) (22 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 trapezoidal frame of 12.5 m span and 6 m height with a hinge in the top chord.

Find. The four reaction components, and shear force and bending moment diagrams for all four members with their maximum and minimum ordinates.

Approach. Four reaction components and four equations — three of overall equilibrium plus zero moment at the crown hinge — make the frame determinate; solve those, then walk along the frame accumulating shear and moment member by member.

1. Confirm determinacy and set out the equations. Two pinned bases give $r = 4$; the hinge at joint 3 supplies one equation of condition, so $r = 3 + c = 3 + 1 = 4$ and the frame is determinate. This is the classical three-hinged frame with its third hinge replaced by the pair of pinned bases.
2. Vertical reactions from overall equilibrium. The horizontal distributed load has no vertical component, so taking moments about joint 1 and remembering that its resultant of 37.5 kN acts leftward at mid-height, 3.0 m above the bases, $$\sum M_1 = 0:\quad -30(5.0) + 37.5(3.0) + 12.5\,V_5 = 0$$ $$\begin{aligned} V_5 &= \frac{150 - 112.5}{12.5} = \boxed{3.0\ \text{kN}\ (\uparrow)}, \\ V_1 &= 30 - 3.0 = \boxed{27.0\ \text{kN}\ (\uparrow)} \end{aligned}$$
3. Horizontal reactions from the hinge condition. Take the piece of frame to the right of the hinge, which carries the whole distributed load and the reactions at joint 5, and set its moment about the hinge at $(8.0,\ 6.0)$ to zero: $$-37.5(6.0 - 3.0) + 3.0(12.5 - 8.0) + 6.0\,H_5 = 0$$ $$\begin{aligned} H_5 &= \frac{112.5 - 13.5}{6.0} = \boxed{16.5\ \text{kN}\ (\rightarrow)}, \\ H_1 &= 37.5 - 16.5 = \boxed{21.0\ \text{kN}\ (\rightarrow)} \end{aligned}$$ Both bases push the frame back against the distributed load, which is what one would expect. The left-hand piece provides the independent check: $-27.0(8.0) + 21.0(6.0) + 30(8.0 - 5.0) = -216 + 126 + 90 = 0$.
4. Left leg 1 – 2. The leg rises 6 m in 2.5 m, so its length is 6.5 m and its direction cosines are 2.5/6.5 and 6.0/6.5. Resolving the base reaction $(21.0,\ 27.0)$ along and across the member, $$N = -\left[21.0\left(\tfrac{2.5}{6.5}\right) + 27.0\left(\tfrac{6.0}{6.5}\right)\right] = -33.0\ \text{kN (compression)}$$ $$V = 21.0\left(-\tfrac{6.0}{6.5}\right) + 27.0\left(\tfrac{2.5}{6.5}\right) = -9.0\ \text{kN (constant)}$$ so the moment grows linearly from zero at the pin to $$M_2 = 9.0(6.5) = \boxed{58.5\ \text{kN}\cdot\text{m at knee 2}}$$ with the outer face of the leg in tension.
5. Top chord 2 – 3. Measuring $x$ from the left reference line, the moment carried by the chord is $M = 27.0x - 21.0(6.0)$ up to the 30 kN load, so it starts at $-58.5$ kN.m at the knee (the same 58.5 kN.m, transferred round the corner), crosses zero at $$x = \frac{21.0(6.0)}{27.0} = 4.667\ \text{m}$$ and reaches $+9.0$ kN.m just left of the load. The shear over that stretch is a constant $+27.0$ kN. Beyond the load the shear falls to $27.0 - 30 = -3.0$ kN and the moment runs from $+9.0$ kN.m back to exactly zero at the hinge, which is the check that the reactions are right.
6. Short chord 3 – 4. Nothing is applied between the hinge and the right-hand knee, so the shear stays at $-3.0$ kN and the moment grows linearly from zero at the hinge to $$M_4 = 3.0(2.0) = \boxed{6.0\ \text{kN}\cdot\text{m at knee 4}}$$
7. Right leg 4 – 5, the only member with a curved diagram. Measure $y$ upward from the pin at joint 5. The portion of frame below the section carries the base reaction $(16.5,\ 3.0)$ and the part of the distributed load already passed, $6.25y$ acting leftward with its resultant at $y/2$. The moment about the section is $$M(y) = 16.5y + 3.0\left(\tfrac{2.5}{6.0}\right)y - \tfrac{6.25}{2}y^{2} = 17.75y - 3.125y^{2}$$ which is zero at the pin, as it must be. Differentiating, $$\frac{dM}{dy} = 17.75 - 6.25y = 0 \;\Rightarrow\; y = 2.84\ \text{m}$$ $$M_{max} = \frac{17.75^{2}}{2(6.25)} = \boxed{25.21\ \text{kN}\cdot\text{m}}$$ which is the largest moment anywhere on the frame after the 58.5 kN.m at knee 2. At the top, $M(6.0) = 106.5 - 112.5 = -6.0$ kN.m, agreeing in magnitude with the value carried in from member 3 – 4 and confirming joint equilibrium at knee 4. The moment changes sign at $y = 17.75/3.125 = 5.68$ m.
8. Shear on the right leg. Resolving perpendicular to the member, which is inclined at 6.0 vertical to 2.5 horizontal, $$V(y) = \left(16.5 - 6.25y\right)\left(\tfrac{6.0}{6.5}\right) + 3.0\left(\tfrac{2.5}{6.5}\right) = 16.39 - 5.77y$$ This runs from $+16.39$ kN at the pin to $-18.23$ kN at knee 4, passing through zero at $y = 2.84$ m, exactly where the moment peaks. The axial force in this leg varies from 3.58 kN of tension at the base to 10.85 kN of compression at the knee, because the horizontal load progressively reverses it — a point worth stating, since a leg one might expect to be a strut is a tie over its lower part.