A simply supported beam is under a uniformly distributed load (UDL) along the full span. The mid-span deflection is measured as 24 mm. If the length and depth of the beam is doubled while keeping other parameters unchanged, the mid-span deflection is _____________ mm. (answer in integer)
Problem Analysis:
- We are given a simply supported beam under a Uniformly Distributed Load (UDL).
- Initial mid-span deflection ($\delta_1$) is 24 mm.
- The beam's length (L) and depth (D) are doubled ($L_2 = 2L_1$, $D_2 = 2D_1$).
- Other parameters (UDL intensity 'w', Young's Modulus 'E', beam width 'b') remain constant.
- We need to find the new mid-span deflection ($\delta_2$).
Deflection Formula Explanation
The mid-span deflection ($\delta$) for a simply supported beam under a UDL (w) over its full span (L) is given by the formula:
$ \delta = \frac{5 w L^4}{384 E I} $
Where:
- $L$ = Beam Length
- $E$ = Young's Modulus of the beam material
- $I$ = Moment of Inertia of the beam's cross-section
For a rectangular cross-section with width '$b$' and depth '$D$', the Moment of Inertia is:
$ I = \frac{b D^3}{12} $
Substituting $I$ into the deflection formula:
$ \delta = \frac{5 w L^4}{384 E \left( \frac{b D^3}{12} \right)} = \frac{5 \times 12 w L^4}{384 E b D^3} = \frac{60 w L^4}{384 E b D^3} $
Since $w, E, b$ are constant, the deflection is proportional to $\frac{L^4}{D^3}$:
$ \delta \propto \frac{L^4}{D^3} $
Deflection Calculation Steps
- Initial State ($\delta_1$): Let the initial length be $L_1$ and initial depth be $D_1$. $ \delta_1 \propto \frac{L_1^4}{D_1^3} $ We are given $\delta_1 = 24$ mm.
- Final State ($\delta_2$): The new length is $L_2 = 2L_1$ and the new depth is $D_2 = 2D_1$. $ \delta_2 \propto \frac{L_2^4}{D_2^3} $
- Substituting new dimensions: $ \delta_2 \propto \frac{(2L_1)^4}{(2D_1)^3} $ $ \delta_2 \propto \frac{16 L_1^4}{8 D_1^3} $ $ \delta_2 \propto 2 \left( \frac{L_1^4}{D_1^3} \right) $
- Relating $\delta_2$ to $\delta_1$: Since $\delta_1 \propto \frac{L_1^4}{D_1^3}$, we can write: $ \delta_2 \propto 2 \left( \frac{L_1^4}{D_1^3} \right) $ Therefore, $\delta_2$ is twice the proportionality factor that gives $\delta_1$. $ \delta_2 = 2 \times \delta_1 $
- Final Calculation: $ \delta_2 = 2 \times 24 \text{ mm} = 48 \text{ mm} $
Conclusion
By doubling the length and depth of the beam, while keeping other parameters constant, the mid-span deflection increases by a factor of 2. The new mid-span deflection is 48 mm.
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______
