Question 8 of 8: Reactions and internal-force diagrams for a determinate portal frame with an internal hinge

Question 8: Reactions and internal-force diagrams for a determinate portal frame with an internal hinge (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.

Find. The four reaction components, then the shear force and bending moment diagrams for each of the three members with their maximum and minimum ordinates.

Approach. Four reaction components need four equations, so use the three global ones plus the condition that the moment about the hinge at joint 2 vanishes for the part on either side of it. Then walk the internal actions member by member.

1. Confirm determinacy. Three members, four joints, four reaction components and one internal hinge give $$i = 3m + r - 3j - c = 9 + 4 - 12 - 1 = 0$$ so the frame is determinate, and the hinge is exactly what makes the second pin admissible.
2. Resolve the distributed load. The 20 kN/m acts per metre of vertical projection over the full 6 m height of member 3–4, so its resultant is $$W = 20\,(6) = 120\ \text{kN}\ \text{to the left, acting at mid-height}, \; y = 3\ \text{m}$$ Because the force is purely horizontal, only its height matters when moments are taken; its horizontal position is irrelevant.
3. Take global moments about joint 1. Counting anticlockwise as positive, the 50 kN load at $(6.5,\,6)$ gives $-325$ kN·m, the 120 kN horizontal resultant at $y = 3$ gives $+360$ kN·m, and the vertical reaction at joint 4 acts 12.5 m away: $$-325 + 360 + 12.5\,V_{4} = 0 \;\Rightarrow\; V_{4} = \boxed{2.8\ \text{kN}\ \text{downward}}$$ $$V_{1} = 50 + 2.8 = \boxed{52.8\ \text{kN}\ \text{upward}}$$ A downward reaction at joint 4 is the frame resisting the overturning couple of the horizontal load.
4. Use the hinge for the horizontal split. Member 1–2 carries no load along its length and is attached by a pin at each end, so it is a two-force member and its end forces must lie along it. Equivalently, take moments about joint 2 for the left part: $$6H_{1} - 4.5V_{1} = 0 \;\Rightarrow\; H_{1} = \tfrac{4.5}{6}\,(52.8) = \boxed{39.6\ \text{kN}\ \text{to the right}}$$ $$H_{4} = 120 - 39.6 = \boxed{80.4\ \text{kN}\ \text{to the right}}$$ Both reactions push to the right because the applied distributed load pushes the frame to the left.
5. Member 1–2 — a strut with no shear and no moment. The resultant at joint 1 is $\sqrt{39.6^{2} + 52.8^{2}} = 66.0$ kN, and its direction cosines $39.6/66 = 0.6$ and $52.8/66 = 0.8$ are exactly those of the member, so the resultant lies along the member axis: $$\begin{aligned} N_{1\text{-}2} &= \boxed{66.0\ \text{kN compression}} \\ V &= 0 \\ M &= 0 \end{aligned}$$ This is the standard consequence of a member that is pinned at both ends and unloaded between them, and it is worth spotting before any arithmetic, because it settles a whole third of the frame at once.
6. Member 2–3 — shear and moment. The hinge delivers 39.6 kN horizontally and 52.8 kN vertically to joint 2, so the horizontal member carries a constant axial compression of 39.6 kN and a shear of $+52.8$ kN over its first 2.0 m, dropping to $+2.8$ kN beyond the 50 kN load. The moment starts at zero because joint 2 is a hinge and builds up: $$\begin{aligned} M_{\text{load}} &= 52.8\,(2.0) = \boxed{+105.6\ \text{kN}\cdot\text{m}} \\ M_{3} &= 105.6 + 2.8\,(1.5) = \boxed{+109.8\ \text{kN}\cdot\text{m}} \end{aligned}$$
7. Member 3–4 — write the moment as a function of height. Measuring $y$ upward from joint 4 and taking the free body below the section, the reaction components act at $y = 0$ while the accumulated distributed load $20y$ acts at $y/2$: $$M(y) = 80.4y - 2.1y - 10y^{2} = 78.3y - 10y^{2}$$ in which the $2.1y$ term is the moment of the 2.8 kN downward reaction about the section, whose horizontal offset is $0.75y$. Checking the ends, $M(0) = 0$ at the pin and $M(6) = 469.8 - 360 = 109.8$ kN·m at joint 3, which matches the value carried round from member 2–3 and confirms that joint 3 is in moment equilibrium.
8. Member 3–4 — extreme ordinates. Differentiating, the moment is stationary where $78.3 - 20y = 0$, that is at $y = 3.915$ m above joint 4, and $$M_{\max} = 78.3\,(3.915) - 10\,(3.915)^{2} = \boxed{+153.3\ \text{kN}\cdot\text{m}}$$ which is the largest moment anywhere in the frame. The shear measured perpendicular to the member is the slope of the moment with respect to arc length, so $$\begin{aligned} V(y) &= \frac{78.3 - 20y}{1.25} \\ V(0) &= 62.64\ \text{kN} \\ V(6) &= -33.36\ \text{kN} \end{aligned}$$ and it changes sign at the same height, 3.915 m.
9. Collect the diagram ordinates. On member 1–2 both diagrams are flat at zero. On member 2–3 the shear steps from $+52.8$ kN to $+2.8$ kN and the moment rises from zero to $+109.8$ kN·m. On member 3–4 the shear runs from $+62.64$ kN at the pin through zero at 3.915 m to $-33.36$ kN at joint 3, and the moment rises from zero to a maximum of $+153.3$ kN·m and falls back to $+109.8$ kN·m. Every moment in this frame puts the inner face of the members in tension (the underside of member 2–3 and the inside face of member 3–4, which the inward 20 kN/m load bows towards the frame), so there is no sign reversal to plot and the minimum ordinate on each member is the zero at its pinned or hinged end.