The plane frame has a hinge and a roller support, and is loaded as shown in the figure. Both the columns have same height. What is the absolute value of the maximum bending moment (inkN-m) in the frame?

The correct option is 165.

Here is the detailed, step-by-step structural analysis and calculation to find the maximum bending moment in the frame:

1. Identify the Frame Geometry and Loading:
Let the base of the left column be support $A$ (hinged/pin support) and the base of the right column be support $D$ (roller support).
Let the top-left joint be $B$ and the top-right joint be $C$.
• Height of both columns ($AB$ and $CD$): $h=3\text{m}$
• Length of the beam ($BC$): $L=2\text{m}+2\text{m}=4\text{m}$
• Horizontal load acting at joint $B$ towards the right: $P_H=50\text{kN}$
• Vertical point load acting at the midspan of beam $BC$: $P_V=90\text{kN}$ (at a distance of $2\text{m}$ from both $B$ and $C$)

2. Determine Support Reactions:
Let the vertical and horizontal reactions at the hinged support $A$ be $V_A$ (upwards) and $H_A$ (towards the left).
Let the vertical reaction at the roller support $D$ be $V_D$ (upwards). Since $D$ is a roller on a horizontal surface, there is no horizontal reaction at $D$ ($H_D=0$).

Applying the equations of static equilibrium:

• Horizontal Equilibrium:
$\sum F_x=0\Rightarrow 50-H_A=0\Rightarrow H_A=50\text{kN}$ (acting to the left)

• Moment Equilibrium about Point $A$:
Taking counter-clockwise moments about $A$ as positive:
$\sum M_A=0$

$4V_D-180-150=0$
$4V_D=330\Rightarrow V_D=82.5\text{kN}$

• Vertical Equilibrium:
$\sum F_y=0\Rightarrow V_A+V_D-90=0$
$V_A+82.5-90=0\Rightarrow V_A=7.5\text{kN}$

3. Calculate Bending Moments in the Members:

• Column $AB$:
At any height $y$ from base $A$ (where $0\le y\le 3\text{m}$):
$M\left(y\right)=H_A\times y=50y$
At the base $A$ ($y=0$): $M_A=0$
At the joint $B$ ($y=3\text{m}$): $M_B=50\times 3=150\text{kN-m}$

• Column $CD$:
Since support $D$ is a roller, the horizontal reaction is zero ($H_D=0$). Therefore, there is no bending moment anywhere in the column $CD$:
$M_{CD}=0$

• Beam $BC$:
We can analyze the bending moment along the beam $BC$ from both ends towards the center:
1. Starting from the right joint $C$ towards the midspan (let $z$ be the distance from $C$, where $0\le z\le 2\text{m}$):
$M\left(z\right)=V_D\times z=82.5z$
Under the vertical point load (at $z=2\text{m}$):
$M_{\text{midspan}}=82.5\times 2=165\text{kN-m}$

2. Starting from the left joint $B$ towards the midspan (let $x$ be the distance from $B$, where $0\le x\le 2\text{m}$):
$M\left(x\right)=M_B+V_A\times x=150+7.5x$
Under the vertical point load (at $x=2\text{m}$):
$M_{\text{midspan}}=150+(7.5\times 2)=150+15=165\text{kN-m}$

4. Determine the Maximum Bending Moment:
Comparing the bending moments across the entire frame:
• Maximum moment in column $AB$ is $150\text{kN-m}$ at joint $B$.
• Bending moment in column $CD$ is $0$.
• Maximum moment in beam $BC$ is $165\text{kN-m}$ at the midspan.

Therefore, the absolute value of the maximum bending moment in the frame is 165 kN-m.