Calculating Steel Rod Radius of Curvature
The radius of curvature ($R$) is determined using the flexure formula, which relates the bending moment ($M$), elastic modulus ($E$), moment of inertia ($I$), and radius of curvature ($R$):
$M = \frac{EI}{R}$
Rearranging the formula to solve for $R$:
$R = \frac{EI}{M}$
Step-by-Step Calculation
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Calculate Moment of Inertia (I):
The rod has a diameter ($d$) of 100 mm, so its radius ($r$) is 50 mm. The moment of inertia for a solid circular section is:
$I = \frac{\pi r^4}{4}$
Using $\pi = 3.14$ and $r = 50$ mm:
$I = \frac{3.14 \times (50 \text{ mm})^4}{4} = \frac{3.14 \times 6250000 \text{ mm}^4}{4} = 4906250 \text{ mm}^4$
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Convert Units:
Ensure consistent units. Convert the applied moment ($M$) from kNm to Nmm:
$M = 50 \text{ kNm} = 50 \times 10^3 \text{ Nm} \times 1000 \text{ mm/m} = 50 \times 10^6 \text{ Nmm}$
The elastic modulus ($E$) is given as $2 \times 10^5$ MPa, which is equivalent to $2 \times 10^5$ N/mm$^2$.
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Calculate Radius of Curvature (R):
Substitute the values of $E$, $I$, and $M$ into the formula for $R$:
$R = \frac{EI}{M} = \frac{(2 \times 10^5 \text{ N/mm}^2) \times (4906250 \text{ mm}^4)}{50 \times 10^6 \text{ Nmm}}$
$R = \frac{9.8125 \times 10^{11} \text{ Nmm}^3}{50 \times 10^6 \text{ Nmm}} = 19625 \text{ mm}$
The calculated radius of curvature is 19625 mm, which falls within the range of 19615 mm and 19635 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 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______
- A simply-supported steel beam made of an I-section has a span of $8 \text{ m}$. The beam is carrying a uniformly distributed load of $15 \text{ kN/m}$. The overall depth of the beam is $450 \text{ mm}$. The moment of inertia of the beam section is $18000$ cm$^4$. The maximum bending stress in the beam will be _________ N/mm$^2$. [in integer]
