Calculate Cantilever Beam Deflection
This solution explains how to calculate the elastic deflection at the free end of a cantilever beam subjected to a vertical point load.
Deflection Formula for Cantilever Beam
The elastic deflection ($\delta$) at the free end of a cantilever beam under a point load (P) at the free end is given by:
$ \delta = \frac{P L^3}{3 E I} $
Where:
- P is the point load
- L is the span of the beam
- E is the elastic modulus of the material
- I is the moment of inertia of the beam's cross-section
Given Data
- Load, P = 30 kN
- Span, L = 5 m
- Elastic Modulus, E = 30 GPa
- Width, b = 0.3 m
- Depth, d = 0.6 m
Unit Conversion
Convert parameters to consistent SI units (N, m, Pa):
- P = $30 \times 10^3$ N
- L = 5 m
- E = $30 \times 10^9$ Pa (N/m$^2$)
Calculate Moment of Inertia (I)
For a rectangular section, the moment of inertia about the neutral axis is:
$ I = \frac{b d^3}{12} $
Substitute the dimensions:
$ I = \frac{(0.3 \text{ m}) \times (0.6 \text{ m})^3}{12} = \frac{0.3 \times 0.216}{12} = \frac{0.0648}{12} = 0.0054 \text{ m}^4 $
Calculate Elastic Deflection ($\delta$)
Plug the values into the deflection formula:
$ \delta = \frac{(30 \times 10^3 \text{ N}) \times (5 \text{ m})^3}{3 \times (30 \times 10^9 \text{ N/m}^2) \times (0.0054 \text{ m}^4)} $
$ \delta = \frac{30,000 \times 125}{3 \times 30 \times 10^9 \times 0.0054} \text{ m} $
$ \delta = \frac{3,750,000}{486 \times 10^9} \text{ m} $
$ \delta \approx 7.716 \times 10^{-3} \text{ m} $
Convert to Millimeters
Convert meters to millimeters:
$ \delta \approx 7.716 \times 10^{-3} \text{ m} \times \frac{1000 \text{ mm}}{1 \text{ m}} \approx 7.716 \text{ mm} $
The calculated deflection of approximately 7.716 mm falls within the range of 7.65 mm to 7.75 mm.
Important Questions from Strength of Materials
- 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]
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 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 steel I-beam section is subjected to a bending moment of 96 kN-m. The moment of inertia of the beam section is $24,000 \text{ cm}^4$. The bending stress at 100 mm above the neutral axis of the beam in MPa will be ________
A simply supported beam AB has a clear span of 7 meter. The bending moment diagram (BMD) of the beam due to a single concentrated load is shown in the figure below.
The magnitude of the concentrated load in kN is __________.
