Question

A beam, 3 m long is freely supported at it's ends and carries a uniformly distributed load of 2,4 kN/m run from B to it's centre. Draw the shearing force and bending moment diagram and give the magnitude and position of the maximum bending moment

Answer 1

Answer on Question #64290-Engineering-Civil and Environmental Engineering

A beam, $3\mathrm{m}$ long is freely supported at its ends and carries a uniformly distributed load of $2.4\mathrm{kN / m}$ run from B to its center. Draw the shearing force and bending moment diagram and give the magnitude and position of the maximum bending moment

Solution

Free-Body-Diagram

The shearing force and bending moment diagram

The magnitude and position of the maximum bending moment: 1.52 kNm at 1.88 m.

1. A beam is in equilibrium when it is stationary relative to an inertial reference frame. The following conditions are satisfied when a beam, acted upon by a system of forces and moments, is in equilibrium:

$$\Sigma F _ {x} = 0: H _ {B} = 0$$

$\Sigma M_A = 0$ : The sum of the moments about a point $A$ is zero:

$$- q _ {1} 1. 5 \left(1. 5 + \frac {1 . 5}{2}\right) + R _ {B} 3 = 0$$

$\Sigma M_{B}=0$: The sum of the moments about a point B is zero:

$-R_{A}3+q_{1}1.5\left(1.5-\frac{1.5}{2}\right)=0$

2. Solve this system of equations:

$H_{B}=0\;(kN)$

Calculate reaction of pin support about point B:

$RB=\frac{q_{1}1.5\left(1.5+\frac{1.5}{2}\right)}{3}=\frac{2.4\cdot 1.5\left(1.5+\frac{1.5}{2}\right)}{3}=2.70\;(kN)$

Calculate reaction of roller support about point A:

$RA=\frac{q_{1}1.5\left(1.5-\frac{1.5}{2}\right)}{3}=\frac{2.4\cdot 1.5\left(1.5-\frac{1.5}{2}\right)}{3}=0.90\;(kN)$

3. The sum of the forces is zero:

$\Sigma F_{y}=0$: $R_{A}-q_{1}1.5+R_{B}=0.90-2.4\cdot 1.5+2.70=0$

Draw diagrams for the beam

First span of the beam: $0\leq x_{1}<1.5$

Determine the equations for the shear force (Q):

$Q(x_{1})=R_{A}$

$Q1(0)=0.90=0.90\;(kN)$

$Q1(1.50)=0.90=0.90\;(kN)$

Determine the equations for the bending moment (M):

$M(x_{1})=R_{A}(x_{1})$

$M_{1}(0)=0.90(0)=0\;(kNm)$

$M_{1}(1.50)=0.90(1.50)=1.35\;(kNm)$

Second span of the beam: $1.5\leq x_{2}<3$

Determine the equations for the shear force (Q):

$Q(x_{2})=RA-q_{1}(x_{2}-1.5)$

$Q_{2}(1.50)=0.90-2.40(1.5-1.5)=0.90\;(kN)$

$Q_{2}(3)=0.90-2.40(3-1.5)=-2.70\;(kN)$

The value of Q on this span that crosses the horizontal axis. Intersection point:

$x=0.38$

Determine the equations for the bending moment (M):

$$M(x_2) = R_A(x_2) - \frac{q_1 (x_2 - 1.5)^2}{2}$$ $$M_2(1.50) = 0.90(1.50) - \frac{2.40(1.50 - 1.5)^2}{2} = 1.35\,(\text{kNm})$$ $$M_2(3) = 0.90(3) - \frac{2.40(3 - 1.5)^2}{2} = 0\,(\text{kNm})$$

Local extremum at the point $x = 0.38$:

$$M_2(1.88) = 0.90(1.88) - \frac{2.40(1.88 - 1.5)^2}{2} = 1.52\,(\text{kNm})$$

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