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The stresses acting on element B on the web of a train rail (see figure part a of Problem 7.2-5) arc found to be 5700 psi in compression in the horizontal direction and 2300 psi in compression in the vertical direction (see figure). Also, shear stresses of magnitude 2500 psi act in the directions shown. Determine the stresses acting on an element oriented at a counterclockwise angle of 50° from the horizontal. Show these stresses on a sketch of an element oriented at this angle.

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Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347

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Chapter
Section
BuyFindarrow_forward

Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 7, Problem 7.2.7P
Textbook Problem
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The stresses acting on element B on the web of a train rail (see figure part a of Problem 7.2-5) arc found to be 5700 psi in compression in the horizontal direction and 2300 psi in compression in the vertical direction (see figure). Also, shear stresses of magnitude 2500 psi act in the directions shown.

Determine the stresses acting on an element oriented at a counterclockwise angle of 50° from the horizontal. Show these stresses on a sketch of an element oriented at this angle.

  Chapter 7, Problem 7.2.7P, The stresses acting on element B on the web of a train rail (see figure part a of Problem 7.2-5) arc

To determine

Stress acting on an element.

Te sketch of the element.

Explanation of Solution

Write the expression for normal stress acting along the x direction

σx=(σx+σy2)+(σxσy2)cos2θ+τxysin2θ        …… (I)

Here, stress acting along the x direction is σx, stress is acting along y direction σy,shear stress acting along xy is τxy, angle oriented on an element is θ.

Write the expression for normal stress acting on the y1axis inclined to the original axis.

σy=(σx+σy2)(σxσy2)cos2θτxysin2θ        …… (II)

Write the expression for shear stress acting on the x1y1plane inclined to the original axis.

τxy=(σxσy2)sin2θ+τxycos2θ        …… (III)

Calculation:

Substitute, 5700psifor σx, 2300psifor σy, and 2500psifor τxy, 50°for θin the Equation (I)

σx=(5700psi2300psi2)+(5700psi+2300psi2)cos(2×50°)+(2500psi)sin(2×50°)=(8000psi2)+(3400psi2)×(0.173648178)+2500psi(0.98480295726)=4000psi+295.2019psi+2462.01938psi=1242.78psi

The normal stress acting along in x direction is 1242.78psi

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Chapter 7 Solutions

Mechanics of Materials (MindTap Course List)
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