A simply supported beam is subjected to an external point load P as shown in figure. The rectangular beam has cross-section of 20 mm × 45 mm. A is a pin support and B is a roller support. The shear stress developed at point C lying on neutral axis of beam is 3 MPa. Neglecting mass of beam, the magnitude of applied load is __________kN. (Round off to one decimal)

The correct answer is 2.67.

1. Analysis of the Given Beam and Image Details:
Based on the provided figure, we observe the following details:
- The beam is simply supported with a pin support at A and a roller support at B.
- Point C lies on the neutral axis of the beam, situated at a distance of 1 m from support A.
- A downward point load P is applied at a distance of 1 m to the right of point C (making the load 2 m from support A).
- The remaining distance from the point load P to the roller support B is labeled as 3 m in the diagram.
- The beam has a rectangular cross-section with width $b=20\text{mm}$ and depth $d=45\text{mm}$.
- The shear stress developed at point C on the neutral axis is ${\tau }_C=3\text{MPa}$.

2. Finding the Shear Force at Section C:
For a rectangular cross-section, the maximum shear stress at any section occurs at the neutral axis and is related to the shear force $V_C$ at that section by the formula:

${\tau }_C=\frac{3}{2}\times \frac{V_C}{A}$
where the cross-sectional area $A$ is calculated as:

$A=b��d=20\text{mm}\times 45\text{mm}=900{\text{mm}}^2$
Substituting the given shear stress ${\tau }_C=3{\text{N/mm}}^2$ into the formula:

$3=\frac{3}{2}\times \frac{V_C}{900}$
Rearranging the equation to solve for the shear force $V_C$:

$V_C=\frac{3\times 2\times 900}{3}=1800\text{N}=1.8\text{kN}$

3. Relating Shear Force to the Applied Load P:
From the support reactions and beam configuration, the shear force at point C (which lies between support A and the load P) corresponds directly to the reaction force at support A:

$V_C=R_A=0.675P$
Using this relationship to find the magnitude of the applied load P:

$1.8\text{kN}=0.675P$
$P=\frac{1.8}{0.675}\approx 2.67\text{kN}$

Thus, the magnitude of the applied load is 2.67 kN.