Beam Load and Reaction Analysis
The problem describes a simply supported beam of span L. The load is distributed such that the first half of the span (0 to L/2) has a total weight W, and the second half (L/2 to L) has a total weight 2W. We need to find the maximum shear force.
First, let's determine the intensity of the distributed loads:
- Intensity for the first half ($w_1$): $w_1 \times \frac{L}{2} = W \implies w_1 = \frac{2W}{L}$
- Intensity for the second half ($w_2$): $w_2 \times \frac{L}{2} = 2W \implies w_2 = \frac{4W}{L}$
The total load on the beam is $W_{total} = W + 2W = 3W$. Let the reactions at the supports A (left) and B (right) be $R_A$ and $R_B$, respectively.
Calculating Support Reactions
Using the principle of moments about support A:
$ R_B \times L = \left(W \times \frac{L}{4}\right) + \left(2W \times \frac{3L}{4}\right) $
The term $W \times \frac{L}{4}$ represents the moment due to the load W (whose centroid is at L/4). The term $2W \times \frac{3L}{4}$ represents the moment due to the load 2W (whose centroid is at L/2 + L/4 = 3L/4).
$ R_B \times L = \frac{WL}{4} + \frac{6WL}{4} = \frac{7WL}{4} $
$ R_B = \frac{7W}{4} = 1.75W $
Now, using the equilibrium equation for vertical forces ($\sum F_y = 0$):
$ R_A + R_B = W_{total} $
$ R_A + 1.75W = 3W $
$ R_A = 3W - 1.75W = 1.25W $
Determining Shear Force Values
The shear force $V(x)$ is calculated as follows:
- At the left support (x=0): Shear Force $V_A$ = $R_A = 1.25W$.
- In the first half span ($0 < x < L/2$): $V(x) = R_A - w_1 \times x = 1.25W - \frac{2W}{L} \times x$.
- At the midpoint (x=L/2): $V(L/2) = 1.25W - \frac{2W}{L} \times \frac{L}{2} = 1.25W - W = 0.25W$.
- In the second half span ($L/2 < x < L$): $V(x) = R_A - W - w_2 \times (x - L/2) = 1.25W - W - \frac{4W}{L} \times (x - L/2) = 0.25W - \frac{4W}{L} \times (x - L/2)$.
- At the right support (x=L): $V(L) = 0.25W - \frac{4W}{L} \times (L - L/2) = 0.25W - \frac{4W}{L} \times \frac{L}{2} = 0.25W - 2W = -1.75W$. This equals $-R_B$, as expected.
Maximum Shear Force Identification
The shear forces at critical points are:
- $V_A = 1.25W$
- $V_{L/2} = 0.25W$
- $V_B = -1.75W$
The maximum absolute shear force is the largest magnitude among these values:
$ \text{Max Shear Force} = \max(|1.25W|, |0.25W|, |-1.75W|) = 1.75W $
Important Questions from Strength of Materials
- A steel wire of $5.65\text{ mm}$ diameter and $50\text{ m}$ length is used for a hoisting crane. The wire is used to vertically lift a weight of $200\text{ kg}$ attached to its lowest end. Assume the Young's Modulus of Elasticity of Steel as $2 \times 10^5\text{ N/mm}^2$ and gravitational acceleration as $10\text{ m/sec}^2$. The elongation of the steel wire (in mm ) will be ______ [rounded off to two decimal places].
A rectangular beam section of size 300 mm (width) X 500 mm (depth) is loaded with a shear force of 600 kN. The maximum shear stress on the section in N/mm² is ___________
- A load of 30 kN is applied vertically downward at the free end of a cantilever of span 5 m. If the elastic modulus of the cantilever is 30 GPa and the section has a width of 0.3 m and a depth of 0.6 m, then, the elastic deflection (in mm) is ________
- A solid straight steel rod of diameter 100 mm is bent in single curvature into a circular arc by a moment of 50 kNm applied at its ends. If elastic modulus, E, for steel is $2 \times 10^5$ MPa, the radius of curvature (mm) of the arc assuming $\pi=3.14$ is ________
A simply supported RCC beam of cross section $0.4 \text{ m} \times 0.6 \text{ m}$ covers a span of $8 \text{ m}$. It is subjected to a uniformly distributed load of $30 \text{ kN/m}$. If the unit weight of concrete is $24 \text{ kN/m}^3$, the tensile stress (in $N/mm^2$, rounded off to two decimal places) at the bottom of the beam at mid-span is______
