Math Problem Statement

A cantilever beam of cross section 150 mm x 200 mm is 2 m long. If deflection at free end is 2 mm, calculate u.d.l. that the beam can carry over whole span. Also calculate maximum slope of the beam. Take E = 210 GPa.

Solution

To solve this problem, we will use the principles of beam deflection and slope for a cantilever beam subjected to a uniformly distributed load (UDL). Let's break down the problem into the required steps.

Given Data:

  • Cross-sectional dimensions of the beam:
    • Width, b=150b = 150 mm
    • Height, h=200h = 200 mm
  • Length of the beam, L=2L = 2 m
  • Deflection at the free end, δ=2\delta = 2 mm
  • Young's Modulus, E=210E = 210 GPa = 210×103210 \times 10^3 N/mm²

Step 1: Calculate the moment of inertia (I)

The moment of inertia for a rectangular cross-section is given by:

I=bh312I = \frac{b \cdot h^3}{12}

Substituting the given values:

I=150×200312 mm4=150×8×10612 mm4=100×106 mm4I = \frac{150 \times 200^3}{12} \text{ mm}^4 = \frac{150 \times 8 \times 10^6}{12} \text{ mm}^4 = 100 \times 10^6 \text{ mm}^4

Step 2: Relate deflection to UDL

For a cantilever beam subjected to a UDL ww, the deflection at the free end is given by:

δ=wL48EI\delta = \frac{w \cdot L^4}{8 \cdot E \cdot I}

Rearranging to solve for ww:

w=8EIδL4w = \frac{8 \cdot E \cdot I \cdot \delta}{L^4}

Substituting the known values:

w=8210×103100×1062(2×103)4 N/mmw = \frac{8 \cdot 210 \times 10^3 \cdot 100 \times 10^6 \cdot 2}{(2 \times 10^3)^4} \text{ N/mm}

Simplifying:

w=8210×103100×106216×1012 N/mmw = \frac{8 \cdot 210 \times 10^3 \cdot 100 \times 10^6 \cdot 2}{16 \times 10^{12}} \text{ N/mm}

w=336×10916×1012 N/mm=21×103 N/mm=21 kN/mw = \frac{336 \times 10^{9}}{16 \times 10^{12}} \text{ N/mm} = 21 \times 10^{-3} \text{ N/mm} = 21 \text{ kN/m}

Thus, the UDL that the beam can carry is w=21w = 21 kN/m.

Step 3: Calculate the maximum slope of the beam

The maximum slope in a cantilever beam subjected to a UDL occurs at the fixed end and is given by:

θmax=wL36EI\theta_{\text{max}} = \frac{w \cdot L^3}{6 \cdot E \cdot I}

Substituting the known values:

θmax=21×103×(2×103)36210×103100×106 radians\theta_{\text{max}} = \frac{21 \times 10^3 \times (2 \times 10^3)^3}{6 \cdot 210 \times 10^3 \cdot 100 \times 10^6} \text{ radians}

Simplifying:

θmax=21×8×109126×109 radians=168×109126×109 radians\theta_{\text{max}} = \frac{21 \times 8 \times 10^9}{126 \times 10^9} \text{ radians} = \frac{168 \times 10^9}{126 \times 10^9} \text{ radians}

θmax=1.333×103 radians=1.333 mrad\theta_{\text{max}} = 1.333 \times 10^{-3} \text{ radians} = 1.333 \text{ mrad}

Summary:

  • The UDL that the beam can carry is w=21w = 21 kN/m.
  • The maximum slope of the beam is θmax=1.333\theta_{\text{max}} = 1.333 mrad.

Would you like more details or explanations on any of these steps?


Relative Questions:

  1. How would the deflection change if the length of the beam were increased?
  2. What would be the effect of changing the material's Young's modulus on the deflection and slope?
  3. How does the moment of inertia influence the deflection of a beam?
  4. What happens to the maximum slope if the UDL is doubled?
  5. How would the deflection and slope change if the beam were simply supported instead of cantilevered?

Tip: When dealing with beam deflection problems, always double-check the units of the inputs to ensure consistent and accurate calculations.

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Math Problem Analysis

Mathematical Concepts

Beam Deflection
Moment of Inertia
Young's Modulus
Slope of Beams

Formulas

Moment of Inertia for Rectangular Section
Deflection due to UDL in Cantilever Beam
Maximum Slope in Cantilever Beam

Theorems

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Suitable Grade Level

Undergraduate Engineering