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Chapter 5: Problem 10

A simply supported wood beam of rectangular cross section and span length \(1.2 \mathrm{m}\) carries a concentrated load \(P\) at midspan in addition to its own weight (see figure). The cross section has width \(140 \mathrm{mm}\) and height \(240 \mathrm{mm}\). The weight density of the wood is \(5.4 \mathrm{kN} / \mathrm{m}^{3}\) Calculate the maximum permissible value of the load \(P\) if (a) the allowable bending stress is \(8.5 \mathrm{MPa}\), and (b) the allowable shear stress is \(0.8 \mathrm{MPa}\)

Short Answer

Expert verified
The maximum permissible load \( P \) is determined by the lower load value from bending and shear criteria calculations.

Step by step solution

01

Understand the Problem

We have a simply supported wood beam of span length 1.2 meters with a concentrated load \( P \) at midspan. We need to find the maximum permissible value of \( P \) based on two criteria: allowable bending stress and allowable shear stress. The beam has a width of 140 mm and a height of 240 mm. The wood's weight density is 5.4 kN/m鲁.
02

Calculate the Weight of the Beam

First, we need to calculate the self-weight of the beam. The volume of the beam is given by the cross-sectional area times the length: \( \text{Volume} = 0.14 \times 0.24 \times 1.2 \ \text{m}^3 \). The weight (\( W \)) of the beam is \( \text{Weight Density} \times \text{Volume} = 5.4 \times 0.140 \times 0.240 \times 1.2 \ \text{kN} \).
03

Self-Weight Calculation

Calculating the weight: \( W = 5.4 \times 0.04032 = 0.217728 \ \text{kN} = 217.728 \ \text{N} \). This is the weight of the beam and acts as a distributed load.
04

Determine the Moment of Inertia

For bending, we need the moment of inertia \( I \) of the beam's cross-section. For a rectangular section, \( I = \frac{b h^3}{12} \). Substituting the beam's dimensions: \( I = \frac{0.14 \times 0.24^3}{12} \ \text{m}^4 \).
05

Calculate Moment of Inertia

\( I = \frac{0.14 \times 0.013824}{12} = 0.00016128 \ \text{m}^4 \).
06

Find Maximum Bending Moment

The maximum bending moment at midspan for a point load \( P \) is \( M = \frac{P \times L}{4} + \frac{w L^2}{8} \), where \( L \) is the span (1.2 m) and \( w \) is the weight per unit length of the beam. First calculate \( w = \frac{217.728}{1.2} = 181.44 \ \text{N/m} \).
07

Calculate Maximum Bending Moment

\( M = \frac{P \times 1.2}{4} + \frac{181.44 \times 1.2^2}{8} \). With the beam's self weight, \( M = \frac{P \times 1.2}{4} + 32.73 \).
08

Use Bending Stress Formula

Using the bending stress formula \( \sigma = \frac{M c}{I} \), where \( c \) is the distance from the neutral axis (\( c = \frac{h}{2} = 0.12 \ \text{m} \)): \( 8.5 = \frac{(0.3P + 32.73) \times 0.12}{0.00016128} \).
09

Calculate P from Bending Condition

Solving for \( P \), we have \( 8.5 = \frac{0.036P + 3.9276}{0.00016128} \), giving \( P = \frac{(8.5 \times 0.00016128) - 3.9276}{0.036} \).
10

Find Shear Stress

The maximum shear force \( V_{max} = \frac{P}{2} + \frac{wL}{2} \). The shear stress formula is \( \tau = \frac{V \times S}{I b} \), where \( S = \frac{bh^2}{8} \). Substitute values to find if \( P \) meets the shear criteria.
11

Calculate Shear Stress and P

Calculate \( \tau \) for \( V = \frac{P}{2} + 108.864 \) and set it to 0.8 MPa. Solve for \( P \) to ensure the shear condition is met.
12

Compare and Conclude

Compare the permissible \( P \) values obtained from bending and shear calculations to determine the lower value. This will be the maximum permissible load \( P \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bending Stress
Bending stress plays a crucial role in beam analysis. When a beam like the simply supported wood beam in this problem experiences a load, it tries to bend. As it bends, the material at different parts of the beam will either be in tension or compression.
Bending stress can be calculated using the formula:
  • \(蟽 = \frac{M \cdot c}{I}\)
where:
  • \(蟽\) is the bending stress,
  • \(M\) is the moment at the section,
  • \(c\) is the distance from the neutral axis to the outermost fiber (for a rectangular beam, this is half the height),
  • \(I\) is the moment of inertia.
For the simply supported beam in our problem, determining the bending stress helps in finding the maximum load it can safely support without exceeding the material's capacity. In our solution, the bending stress limit was not to exceed 8.5 MPa.
Shear Stress
Shear stress in a beam occurs when force is applied parallel or tangential to a surface. For the simply supported beam, the shear stress is essential to find because it determines whether the beam will fail at certain loads. The shear stress is calculated by the formula:
  • \(蟿 = \frac{V \cdot S}{I \cdot b}\)
where:
  • \(蟿\) is the shear stress,
  • \(V\) is the maximum shear force in the beam,
  • \(S\) is the first moment of area,
  • \(b\) is the width of the beam,
  • \(I\) is the moment of inertia.
In this calculation, the focus was on ensuring shear stress did not exceed 0.8 MPa. It's crucial to address both shear and bending stress to conclude a safe permissible load.
Moment of Inertia
The moment of inertia is a geometric property that describes how a beam's cross-section will resist bending. It depends on the distribution of material around the axis of rotation. For sections that are rectangular in shape, like our wooden beam, the moment of inertia \( I \) is calculated using the formula:
  • \(I = \frac{b \cdot h^3}{12}\)
where:
  • \(b\) is the width of the section,
  • \(h\) is the height of the section.
Understanding the moment of inertia is vital for structural engineers since it directly influences the bending stress equation. The greater the moment of inertia, the more load a beam can handle without bending excessively. In this exercise, calculating \( I \) helped determine whether the beam could handle the applied loads without structural failure.
Simply Supported Beam
A simply supported beam is a basic and widely used structural element. It typically rests on two supports and is free to move horizontally, accommodating bending and shear from applied loads. This simplicity makes it ideal for testing concepts like bending and shear stresses.In our example, the beam is simply supported and has a load \( P \) applied at its midspan, which is a common setup for calculating structural responses. Key points to understand about simply supported beams include:
  • They can experience bending moments that vary along the span of the beam.
  • The support reactions are usually vertical and equal to half the total load for symmetric loading.
These characteristics define how the beam responds to loads and inform calculations of both bending and shear stresses. The understanding of such a beam helps ensure realistic analysis results, leading to safer and more robust designs.

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Most popular questions from this chapter

A wood beam \(A B C\) with simple supports at \(A\) and \(B\) and an overhang \(B C\) has height \(h=300 \mathrm{mm}\) (see figure). The length of the main span of the beam is \(L=3.6 \mathrm{m}\) and the length of the overhang is \(L / 3=1.2 \mathrm{m}\). The beam supports a concentrated load \(3 P=18 \mathrm{kN}\) at the midpoint of the main span and a moment \(P L / 2=10.8 \mathrm{kN} \cdot \mathrm{m}\) at the free end of the overhang. The wood has weight density \(\gamma=5.5 \mathrm{kN} / \mathrm{m}^{3}\) (a) Determine the required width \(b\) of the beam based upon an allowable bending stress of \(8.2 \mathrm{MPa}\) (b) Determine the required width based upon an allowable shear stress of \(0.7 \mathrm{MPa}\)

A flying buttress transmits a load \(P=25 \mathrm{kN}\), acting at an angle of \(60^{\circ}\) to the horizontal, to the top of a vertical buttress \(A B\) (see figure). The vertical buttress has height \(h=5.0 \mathrm{m}\) and rectangular cross section of thickness \(t=1.5 \mathrm{m}\) and width \(b=1.0 \mathrm{m}\) (perpendicular to the plane of the figure). The stone used in the construction weighs \(\gamma=26 \mathrm{kN} / \mathrm{m}^{3}\), What is the required weight \(W\) of the pedestal and statue above the vertical buttress (that is, above section \(A\) ) to avoid any tensile stresses in the vertical buttress?

\mathrm{A}\( simple beam \)A B\( of span length \)L=7 \mathrm{m}\( is subjected to two wheel loads acting at distance \)d=1.5 \mathrm{m}\( apart (see figure). Each wheel transmits a load \)P=14 \mathrm{kN}\( and the carriage may occupy any position on the beam. (a) Determine the maximum bending stress \)\sigma_{\max }\( due to the wheel loads if the beam is an I-beam having section \\[ \text { modulus } S=265 \mathrm{cm}^{3} \\] (b) If \)d=1.5 \mathrm{m},\( find the required span length \)L\( to reduce the maximum stress in part (a) to \)124 \mathrm{MPa}\(. (c) If \)L=7 \mathrm{m}\(, find the required wheel spacing \)s\( to reduce the maximum stress in part (a) to \)124 \mathrm{MPa}$.

A thin, high-strength stecl rule \((E=200 \mathrm{GPa})\) having thickness \(t=4 \mathrm{mm}\) and length \(L=1.5 \mathrm{m}\) is bent by couples \(M_{0}\) into a circular are subtending a central angle \(\alpha=40^{\circ}\) (see figure) (a) What is the maximum bending stress \(\sigma_{\max }\) in the rule? (b) By what percent does the stress increase or decrease if the central angle is increased by \(10 \% ?\) (c) What percent increase or decrease in rule thickness will result in the maximum stress reaching the allowable value of \(290 \mathrm{MPa} ?\)

Each girder of the lift bridge (see figure) is \(50 \mathrm{m}\) long and simply supported at the cnds. The design load for each girder is a uniform load of intensity \(18 \mathrm{kN} / \mathrm{m}\). The girders are fabricated by welding three steel plates so as to form an I-shaped cross section (see figure) having section modulus \(S=46,000 \mathrm{cm}^{3}\) What is the maximum bending stress \(\sigma_{\max }\) in a girder due to the uniform load?
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