Math Problem Statement

Assume that the section shown in question 2 is of a simply suppported beam spanning 10 metres.  If the beam was subjected to a moment of 100 kNm about its x axis, what normal compressive stress would be induced at a distance of 50mm above the neutral axis?  

Express your answer to the nearest 0.1 MPa (absolute value), but do not show the units (eg: an answer of 19.38 MPa should be shown as 19.4)

Solution

To solve the problem, we need to calculate the normal compressive stress induced at a distance of 50 mm above the neutral axis of the section due to a bending moment of 100 kNm about the x-axis.

Step 1: Determine the Second Moment of Area (Iₓ)

From the image provided, we can see the section is a composite of rectangular areas. The second moment of area about the x-axis (IxI_x) can be calculated using the parallel axis theorem:

Ix=I1+I2+I3I_x = I_1 + I_2 + I_3

Where:

  • I1I_1, I2I_2, I3I_3 are the second moments of area of each rectangular segment about the x-axis.

Step 2: Calculate Individual Moments of Inertia

For each rectangular segment, I=bh312+Ad2I = \frac{b \cdot h^3}{12} + A \cdot d^2, where:

  • bb is the width.
  • hh is the height.
  • A=bhA = b \cdot h is the area.
  • dd is the distance from the centroid of the rectangle to the neutral axis.

Let's break down the calculation.

1. Top Flange (horizontal segment at the top):

  • b1=100mmb_1 = 100 \, \text{mm}
  • h1=10mmh_1 = 10 \, \text{mm}
  • Distance from the centroid of this section to the neutral axis (d1d_1) = 380mmh12=3805=375mm380 \, \text{mm} - \frac{h_1}{2} = 380 - 5 = 375 \, \text{mm}

I1x=100×10312+100×10×(375)2I_{1x} = \frac{100 \times 10^3}{12} + 100 \times 10 \times (375)^2

2. Vertical Web:

  • b2=10mmb_2 = 10 \, \text{mm}
  • h2=380mmh_2 = 380 \, \text{mm}
  • Distance from the centroid of this section to the neutral axis (d2d_2) = 00 (since it is at the neutral axis).

I2x=10×380312I_{2x} = \frac{10 \times 380^3}{12}

3. Bottom Flange (horizontal segment at the bottom):

  • b3=120mmb_3 = 120 \, \text{mm}
  • h3=20mmh_3 = 20 \, \text{mm}
  • Distance from the centroid of this section to the neutral axis (d3d_3) = h22+h32=190+10=200mm\frac{h_2}{2} + \frac{h_3}{2} = 190 + 10 = 200 \, \text{mm}

I3x=120×20312+120×20×(200)2I_{3x} = \frac{120 \times 20^3}{12} + 120 \times 20 \times (200)^2

Step 3: Sum the Moments of Inertia

Ix=I1x+I2x+I3xI_x = I_{1x} + I_{2x} + I_{3x}

Step 4: Calculate the Bending Stress

Once we have IxI_x, we can calculate the stress using:

σ=MyIx\sigma = \frac{M \cdot y}{I_x}

Where:

  • M=100×103N\cdotpmM = 100 \times 10^3 \, \text{N·m}
  • y=50mm=0.05my = 50 \, \text{mm} = 0.05 \, \text{m}

Let's calculate

I'll compute these step-by-step.The normal compressive stress induced at a distance of 50 mm above the neutral axis due to a moment of 100 kNm about the x-axis is 17.7 MPa.


Relative Questions:

  1. How would the stress change if the moment was applied about the y-axis instead of the x-axis?
  2. What is the significance of the second moment of area in structural analysis?
  3. How does the distance from the neutral axis affect the induced stress in a beam?
  4. What would be the effect on stress if the beam span was reduced to 5 meters?
  5. How does the shape and size of the cross-section influence the beam's bending strength?

Tip: Always ensure to calculate the correct centroid and moment of inertia when dealing with composite sections to obtain accurate stress results.

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Math Problem Analysis

Mathematical Concepts

Mechanics of Materials
Bending Stress
Second Moment of Area
Neutral Axis

Formulas

Second Moment of Area calculation
Bending stress formula

Theorems

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Suitable Grade Level

College