/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 A steel beam of length \(L=400 \... [FREE SOLUTION] | 91Ó°ÊÓ

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A steel beam of length \(L=400 \mathrm{mm}\) and cross-sectional dimensions \(b=12 \mathrm{mm}\) and \(h=50 \mathrm{mm}\) (see figure) supports a uniform load of intensity \(q=45 \mathrm{kN} / \mathrm{m}\) which includes the weight of the beam. Calculate the shear stresses in the beam (at the cross section of maximum shear force \()\) at points located \(6.25 \mathrm{mm}, 12.5 \mathrm{mm}, 18.75 \mathrm{mm},\) and \(25 \mathrm{mm}\) from the top surface of the beam. From these calculations, plot a graph showing the distribution of shear stresses from top to bottom of the beam.

Short Answer

Expert verified
Calculate shear stress at specified points and plot its distribution.

Step by step solution

01

Calculate the maximum shear force

The maximum shear force, for a uniform load distributed over a simply supported beam, occurs at the supports. First, convert the load intensity from kN/m to N/mm by multiplying by 1000 and dividing by 1000, resulting in 45 N/mm. The total load is then given by \( q \times L = 45 \times 400 = 18000 \) N. Therefore, the maximum shear force, \( V_{max} \), at each support is \( \frac{18000}{2} = 9000 \) N.
02

Calculate the moment of inertia

The moment of inertia \( I \) for a rectangular cross-section is calculated using the formula \( I = \frac{b imes h^3}{12} \). Substituting the given values, \( b = 12 \) mm and \( h = 50 \) mm, we get: \[ I = \frac{12 \times 50^3}{12} = 125000 \, \text{mm}^4. \]
03

Express shear stress in terms of distance y

Shear stress \( \tau \) at a distance \( y \) from the neutral axis is given by \( \tau = \frac{V \times Q}{I \times b} \), where \( V \) is the shear force, \( Q \) is the statical moment of area about the neutral axis, \( I \) is the moment of inertia, and \( b \) is the width of the section. The neutral axis is at \( h/2 = 25 \) mm.
04

Calculate Q for specific points

For each point (6.25 mm, 12.5 mm, 18.75 mm) from the top, calculate \( Q = A' \times y' \), where \( A' \) is the area above (or below for points below the neutral axis) the horizontal section at distance \( y \) and \( y' \) is the distance from the neutral axis to the centroid of \( A' \). Calculate and use these values for subsequent calculations.Example at 6.25 mm:- \( A' = b \times y = 12 \times 6.25 = 75 \, \text{mm}^2 \) - Distance to centroid \( y' = 25 - 3.125 = 21.875 \, \text{mm} \)- Therefore, \( Q = 75 \, \text{mm}^2 \times 21.875 = 1640.625 \, \text{mm}^3 \).Repeat for 12.5 mm, 18.75 mm, 25 mm.
05

Calculate shear stress at each point

Use the formula \( \tau = \frac{V \times Q}{I \times b} \) to find shear stress at each distance from the top.Example at 6.25 mm:- \( \tau = \frac{9000 \times 1640.625}{125000 \times 12} \approx 0.988 \mathrm{N/mm^2} \).Perform similar calculations at 12.5 mm, 18.75 mm, and 25 mm.
06

Plot the shear stress distribution

Plot the calculated shear stresses against the distances from the top surface on a graph to illustrate how the shear stress distribution varies across the depth of the beam.

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in structural engineering, especially when analyzing beams under load. It measures how a beam's cross-section resists bending. The higher the moment of inertia, the better a beam can resist bending forces. For a rectangular cross-section, like the steel beam in our problem, the moment of inertia \( I \) is calculated using the formula:
  • \( I = \frac{b \times h^3}{12} \)
where \( b \) is the width of the beam, and \( h \) is the height of the beam.
The units for \( I \) are \( mm^4 \), which reflect the dimensions raised to the fourth power. In our example with a beam width of 12 mm and height of 50 mm, the calculated moment of inertia is 125,000 \( mm^4 \). This large number indicates the beam's relatively high capability to resist bending deflection. Understanding how the moment of inertia works helps in the design and analysis of structural components to ensure they can withstand the expected loads safely.
Statical Moment of Area
The statical moment of area, often simply called the first moment of area \( Q \), is crucial in calculating the shear stress within a beam. It relates to the distribution of a beam's area concerning a reference axis and is used specifically in shear stress calculations for determining how the parts of the cross-section contribute to the shear capacity of the beam.
The formula to compute \( Q \) at a distance \( y \) from the top or bottom of the beam is:
  • \( Q = A' \times y' \)
where \( A' \) is the area of the portion of the cross-section being considered and \( y' \) is the distance from the centroid of this area to the neutral axis of the whole cross-section. For instance, for a section \( 6.25 \) mm from the top of the beam, \( A' \) is the area from the top down to this point, and \( y' \) is the distance from its centroid to the beam's central axis located at \( h/2 \). This concept helps engineers design beams that can efficiently carry shear loads without failing or excessive deformation.
Rectangular Cross-Section
Rectangular cross-sections are common in structural beams due to their simplicity and ease of fabrication. They provide distinct analytical benefits when performing shear stress calculations. In the context of the given problem, understanding the rectangular cross-section allows for straightforward computations for both the moment of inertia and statical moment of area.
The dimensions of the cross-section directly affect how calculations are performed:
  • The width (\( b \)) contributes linearly to both calculations, affecting the area moment calculations and the statical moment of area.
  • The height (\( h \)) significantly impacts these calculations, especially in the moment of inertia, as it is cubed, amplifying its influence on how the beam resists bending.
By leveraging these dimensions, structural engineers can predict how a beam will behave under specific loading conditions, ensuring that the design remains safe and effective for its intended use.
Shear Force in Beams
Shear force in beams is the internal force parallel to the beam's cross-section arising from external loads or reactions. It dramatically influences the distribution of shear stress throughout the beam’s depth. In simple beam structures, such as the one in this example with a uniform load, the maximum shear force usually occurs at the supports.
  • Understanding shear force helps in calculating shear stresses that can lead to shear failure if not adequately accounted for.
The shear stress \( \tau \) distribution can be calculated using:
  • \( \tau = \frac{V \times Q}{I \times b} \)
where \( V \) is the shear force, \( Q \) is the statical moment of area at the point of interest, \( I \) is the moment of inertia, and \( b \) is the width of the beam.
In this problem, the calculated shear stresses at various points across the beam's depth help illustrate how these forces are distributed, providing valuable insights into the beam’s performance under loading conditions. This understanding is crucial in structural engineering to prevent shear-related failures and ensure longevity and safety in structural applications.

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Most popular questions from this chapter

A copper wire having a diameter of \(d=4 \mathrm{mm}\) is bent into a circle and held with the ends just touching (see figure) (a) If the maximum permissible strain in the copper is \(\varepsilon_{\max }=0.0024,\) what is the shortest length \(L\) of wire that can be used? (b) If \(L=5.5 \mathrm{m},\) what is the maximum acceptable diameter of the wire if the maximum normal strain must remain below yield? Assume \(E=120 \mathrm{GPa}\) and \\[ \sigma_{Y}=300 \mathrm{MPa} \\]

A cantilever beam \(A B\) of isosceles trapezoidal cross section has length \(L=0.8 \mathrm{m},\) dimensions \(b_{1}=80 \mathrm{mm}\) \(b_{2}=90 \mathrm{mm},\) and height \(h=110 \mathrm{mm}\) (see figure). The beam is made of brass weighing \(85 \mathrm{kN} / \mathrm{m}^{3}\) (a) Determine the maximum tensile stress \(\sigma_{t}\) and \(\max\) imum compressive stress \(\sigma_{c}\) due to the beam's own weight. (b) If the width \(b_{1}\) is doubled, what happens to the stresses? (c) If the height \(h\) is doubled, what happens to the stresses?

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?

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 \(0.8 \mathrm{MPa}\)

A two-axle carriage that is part of an overhead traveling crane in a testing laboratory moves slowly across a simple beam \(A B\) (see figure). The load transmitted to the beam from the front axle is \(9 \mathrm{kN}\) and from the rear axle is \(18 \mathrm{kN} .\) The weight of the beam itself may be disregarded. (a) Determine the minimum required section modulus \(S\) for the beam if the allowable bending stress is \(110 \mathrm{MPa}\) the length of the beam is \(5 \mathrm{m},\) and the wheelbase of the carriage is \(1.5 \mathrm{m}\) (b) Select the most economical I-beam (IPN shape) from Table E-2, Appendix E.

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