The plane frame has a hinge and a roller support, and is loaded as shown in the figure. Both the columns have same height.
(Figure not to scale)
What is the absolute value of the maximum bending moment (in kN-m) in the frame?
Show Hint
Find the support reactions first, then work out the bending moment at the top of each column and at the point-load location; the frame's moment comes out to zero at the roller-side column top.
Step 1: Set up coordinates and identify the reactions.
Place the hinge support P at the origin \((0,0)\). Then the top of the left column is \(A(0,3)\), the load point on the beam is \(B(2,3)\), the top of the right column is \(C(4,3)\), and the roller support is \(R(4,0)\).
The hinge at P resists both a horizontal reaction \(H_P\) and a vertical reaction \(V_P\). The roller at R resists only a vertical reaction \(V_R\); it offers no horizontal resistance.
Step 2: Take moments about P to find \(V_R\).
The 50 kN horizontal force acts at A(0,3), 3 m above P, giving a moment of \(50\times3=150\) kN-m about P.
The 90 kN downward force acts at B(2,3), 2 m to the right of P, giving a moment of \(90\times2=180\) kN-m about P.
Both these moments are balanced by the reaction \(V_R\) acting at 4 m from P:
\[ V_R\times4 = 150+180 = 330 \implies V_R=\frac{330}{4}=82.5 \text{ kN} \]
Step 3: Use vertical and horizontal equilibrium to find \(V_P\) and \(H_P\).
Vertical equilibrium: \(V_P+V_R=90 \implies V_P=90-82.5=7.5\) kN.
Horizontal equilibrium: the roller gives no horizontal reaction, so the hinge alone must balance the 50 kN applied load: \(H_P=50\) kN, acting opposite to the applied force.
Step 4: Find the bending moment at the top of the left column, point A.
The left column carries a constant horizontal shear equal to \(H_P=50\) kN over its full height of 3 m, since no other horizontal load acts between P and A. The moment at P is zero (a hinge cannot carry moment), so the moment builds up linearly to
\[ M_A = H_P\times\text{height} = 50\times3=150 \text{ kN-m} \]
Step 5: Find the bending moment at the load point B, and at the top of the right column, C.
All points on the beam sit at the same height (3 m) as A, so the horizontal reaction \(H_P\) always contributes \(150\) kN-m to the moment at any point on the beam. In addition, the vertical reaction \(V_P\) contributes a moment equal to \(V_P\) times the horizontal distance from P to that point.
At B (2 m from P horizontally): \(M_B = H_P\times3 + V_P\times2 = 150+7.5\times2=150+15=165\) kN-m.
At C (4 m from P horizontally), the moment must also include the effect of the 90 kN load acting over the 2 m from B to C, which exactly cancels the buildup from \(V_P\) by C: working it out gives \(M_C=0\), meaning the right column carries no bending moment at all (only the axial force \(V_R\), which passes straight through it since the roller reaction is purely vertical and the column is vertical).
Step 6: Compare the moments and rule out other options.
The moment magnitudes found are \(M_A=150\), \(M_B=165\), \(M_C=0\). Since the moment changes continuously and linearly along each straight member with no other loads in between, these are also the extreme values along the frame, so the maximum absolute bending moment is 165 kN-m, at B.
Option (B) 150 is only the moment at A, not the true maximum. Options (C) 240 and (D) 195 would come from arithmetic slips, such as forgetting that the roller carries no horizontal reaction, or using the wrong lever arm for \(V_P\) at B.
Final Answer:
The absolute maximum bending moment in the frame is 165 kN-m.
\[ \boxed{|M|_{max}=165 \text{ kN-m}} \]
