Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
bartleby

Videos

Textbook Question
Book Icon
Chapter 7, Problem 7.6.3P

-3 An element of aluminum in the form of a rectangular parallelepiped (see figure) of dimensions a = 5.5 in., h = 4.5 in, and c = 3.5 in. is subjected to iriaxial stresses = 12,500 psi, o. = —5000 psi, and ci. = —1400 psi acting on the x,i, and z faces, respectively.

Determine the following quantities: (a) the maxim um shear stress in the material; (b) the changes ..la, .11. and 1c in the dimensions of the element:

(C) the change .IJ’ in the volume: (d) the strain energy U stored in the element: (e) the maximum value of cr1

when the change in volume must be limited to 0.021%; and (f) the required value of o when the strain energy must be 900 in.-lb. (Assume E = 10,400 ksi and v = 0.33.)

  Chapter 7, Problem 7.6.3P, -3 An element of aluminum in the form of a rectangular parallelepiped (see figure) of dimensions a =

(a)

Expert Solution
Check Mark
To determine

The maximum shear stress in the material.

Answer to Problem 7.6.3P

The maximum shear stress on the material is 8750 psi .

Explanation of Solution

Given information:

The aluminium element of length 5.5 in , width 3.5 in and height 4.5 in is subjected to triaxial stress. The stress in x direction is 12500 psi , in y direction is 5000 psi and in z direction is 1400 psi . The modulus of elasticity is 10400 ksi and the Poisson’s ratio is 0.33 .

Explanation:

Write the expression for the maximum shear stress.

   τ max = σ largest σ smallest 2 ...... (I)

Here, the maximum shear stress is τ max , the largest stress acting on the element is σ largest and the smallest stress acting on the element is σ smallest .

Calculation:

Since no shear stresses act on the parallelepiped, x , y , z directions coincide with the principal stress directions.

Substitute, 12500 psi for σ largest and 5000 psi for σ smallest in Equation (I).

   τ max = 12500 psi+5000 psi 2 = 17500 psi 2 = 8750 psi

Conclusion:

The maximum shear stress on the material is 8750 psi .

(b)

Expert Solution
Check Mark
To determine

The changes in the dimensions of the element.

Answer to Problem 7.6.3P

The change in length is 7.73 × 10 3 in .

The change in height is 3.75 × 10 3 in .

The change in width is 1.3 × 10 3 in .

Explanation of Solution

Write the expression for the strain along x axis.

   ε x = 1 E ( σ x ν ( σ y + σ z ) ) ...... (II)

Here, the strain in the x axis is ε x , the modulus of elasticity is E , the stress along x axis is σ x , stress along y axis is σ y , stress along z axis is σ z and the Poisson’s ratio is ν .

Write the expression for strain in y direction.

   ε y = 1 E ( σ y ν ( σ x + σ z ) ) ...... (III)

Here, the strain in y direction is ε y .

Write the expression for strain in z direction.

   ε z = 1 E ( σ z ν ( σ x + σ y ) ) ...... (IV)

Here, the strain in z direction is ε z .

Write the expression for the change in length.

   Δ a = a × ε x ...... (V)

Here, the length of element is a .

Write the expression for change in height.

   Δ b = b × ε y ...... (VI)

Here, the height of element is b .

Write the expression for the change in width.

   Δ c = c × ε z ...... (VII)

Here, the width of the element is c .

Calculation:

Substitute 10400 × 10 3 psi for E , 12500 psi for σ x , 5000 psi for σ y , 1400 psi for σ z and 0.33 for ν in Equation (II).

   ε x = 1 10400 × 10 3 psi ( 12500 psi 0.33 ( 5000 psi 1400 psi ) ) = 14612 × 10 3 psi 10400 psi = 1.405 × 10 3

Substitute 10400 × 10 3 psi for E , 12500 psi for σ x , 5000 psi for σ y , 1400 psi for σ z and 0.33 for ν in Equation (III).

   ε y = 1 10400 × 10 3 psi ( 5000 psi 0.33 ( 12500 psi 1400 psi ) ) = 8663 × 10 3 psi 10400 psi = 0.83298 × 10 3

Substitute 10400 × 10 3 psi for E , 12500 psi for σ x , 5000 psi for σ y , 1400 psi for σ z and 0.33 for ν in Equation (IV).

   ε z = 1 10400 × 10 3 psi ( 1400 psi 0.33 ( 12500 psi 5000 psi ) ) = 3875 × 10 3 psi 10400 psi = 0.3726 × 10 3

Substitute 1.405 × 10 3 for ε x and 5.5 in for a in Equation (V).

   Δ a = 5.5 in × 1.405 × 10 3 = 7.73 × 10 3 in

Substitute 0.83298 × 10 3 for ε y and 4.5 in for b in Equation (VI).

   Δ b = 4.5 in × ( 0.83298 × 10 3 ) = 3.7484 × 10 3 in

Substitute 0.3726 × 10 3 for ε z and 3.5 in for c in Equation (VII).

   Δ c = 3.5 in × ( 0.3726 × 10 3 ) = 1.3041 × 10 3 in

Conclusion:

The change in length is 7.73 × 10 3 in .

The change in height is 3.75 × 10 3 in .

The change in width is 1.3 × 10 3 in .

(c)

Expert Solution
Check Mark
To determine

The change in the volume of the element.

Answer to Problem 7.6.3P

The change in the volume is 0.0173 in 3 .

Explanation of Solution

Write the expression for the change in the volume.

   Δ V = V 0 ( ε x + ε y + ε z ) ...... (VIII)

Here, the change in volume is Δ V and the original volume is V 0 .

Calculation:

Substitute ( 5.5 in × 4.5 in × 3.5 in ) for V 0 , 1.405 × 10 3 for ε x , 0.83298 × 10 3 for ε y and 0.3726 × 10 3 for ε z in Equation (VIII).

   Δ V = ( 5.5 in × 4.5 in × 3.5 in ) ( 1.405 × 10 3 0.83298 × 10 3 0.3726 × 10 3 ) = 86.625 in 3 × 0.19942 × 10 3 = 0.0173 in 3

Conclusion:

The change in the volume is 0.0173 in 3 .

(d)

Expert Solution
Check Mark
To determine

The strain energy stored in the element.

Answer to Problem 7.6.3P

The strain energy stored in the element is 964 lb in .

Explanation of Solution

Write the expression for the strain energy.

   U = 1 2 V 0 ( σ x ε x + σ y ε y + σ z ε z ) ...... (IX)

Here, the strain energy is U .

Calculation:

Substitute 86.625 in 3 for V 0 , 12500 psi for σ x , 1.405 × 10 3 for ε x , 5000 psi for σ y , 0.83298 × 10 3 for ε y , 1400 psi for σ z and 0.3726 × 10 3 for ε z in Equation (IX).

   U = [ 86.625 in 3 2 ( ( 12500 psi × 1.405 × 10 3 ) + ( 5000 psi × 0.83298 × 10 3 ) + ( 1400 psi × 0.3726 × 10 3 ) ) ] = ( 963661.545 × 10 3 in 3 psi ) ( 1 lb / in 2 1 psi ) = 963.66 lb in 964 lb in

Conclusion:

The strain energy stored in the element is 964 lb in .

(e)

Expert Solution
Check Mark
To determine

The maximum value of normal stress along the x axis.

Answer to Problem 7.6.3P

The maximum value of normal stress along the x axis is 12824 psi .

Explanation of Solution

Given information:

The change in volume is limited to 0.021 0 0 .

Explanation:

Write the expression for the change in volume.

   Δ V V 0 = ( 1 2 ν ) E ( σ x + σ y + σ z ) ...... (X)

Calculation:

Substitute 2.1 × 10 4 for Δ V V 0 , 0.33 for ν , 10400 × 10 3 psi for E , 5000 psi for σ y and 1400 psi for σ z in Equation (X).

   2.1 × 10 4 = 0.34 10400 × 10 3 psi ( σ x 5000 psi 1400 psi ) 0.34 σ x = 4360 psi σ x = 12823.53 psi

Conclusion:

The maximum value of normal stress along the x axis is 12824 psi .

(f)

Expert Solution
Check Mark
To determine

The required value of normal stress along the x axis.

Answer to Problem 7.6.3P

The required value of the normal stress along the x axis is 11967 psi .

Explanation of Solution

Given information:

The strain energy of the system is 900 lb in .

Explanation:

Write the expression for the strain energy in terms of stresses using Hooke’s law.

   U = V 0 2 E ( σ x 2 + σ y 2 + σ z 2 2 ν ( σ x σ y + σ y σ x + σ x σ z ) ) ...... (XI)

Calculation:

Substitute 86.625 in 3 for V 0 , 900 lb in for U , 0.33 for ν , 10400 × 10 3 psi for E , 5000 psi for σ y and 1400 psi for σ z .

   900 = [ 86.625 2 × 10400 × 10 3 ( σ x 2 + 5000 2 + 1400 2 2 × 0.33 ( 5000 σ x + 5000 × 1400 + σ x × 1400 ) ) ] 1800 × 10400 × 10 3 86.625 = σ x 2 + 25 × 10 6 + 1.96 × 10 6 + 3300 σ x 4.76 × 10 6 + 952 σ x σ x 2 + 4224 σ x 193762900 = 0

Solve the quadratic equation for obtaining the value of σ x .

   σ x = 4224 psi ± 4224 2 psi 2 + 4 × 193762900 psi 2 2 = 23934 2 psi = 11967 psi

Conclusion:

The required value of the normal stress along the x axis is 11967 psi .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 7 Solutions

Mechanics of Materials (MindTap Course List)

Ch. 7 - The polyethylene liner of a settling pond is...Ch. 7 - Solve the preceding problem if the norm al and...Ch. 7 - Two steel rods are welded together (see figure):...Ch. 7 - Repeat the previous problem using ? = 50° and...Ch. 7 - A rectangular plate of dimensions 3.0 in. × 5.0...Ch. 7 - Solve the preceding problem for a plate of...Ch. 7 - A simply supported beam is subjected to point load...Ch. 7 - Repeat the previous problem using sx= 12 MPa.Ch. 7 - At a point on the surface of an elliptical...Ch. 7 - Solve the preceding problem for sx= 11 MPa and...Ch. 7 - An clement m plane stress from the frame of a...Ch. 7 - Solve the preceding problem for the element shown...Ch. 7 - : A gusset plate on a truss bridge is in plane...Ch. 7 - The surface of an airplane wing is subjected to...Ch. 7 - At a point on the web of a girder on an overhead...Ch. 7 - -26 A rectangular plate of dimensions 125 mm × 75...Ch. 7 - -27 A square plate with side dimension of 2 in. is...Ch. 7 - The stresses acting on an element are x= 750 psi,...Ch. 7 - Repeat the preceding problem using sx= 5.5 MPa....Ch. 7 - An element in plane stress is subjected to...Ch. 7 - -4. - An element in plane stress is subjected to...Ch. 7 - An element in plane stress is subjected to...Ch. 7 - The stresses acting on element A in the web of a...Ch. 7 - The normal and shear stresses acting on element A...Ch. 7 - An element in plane stress from the fuselage of an...Ch. 7 - -9The stresses acting on element B in the web of a...Ch. 7 - The normal and shear stresses acting on element B...Ch. 7 - ‘7.3-11 The stresses on an element are sx= -300...Ch. 7 - - 7.3-12 A simply supported beam is subjected to...Ch. 7 - A shear wall in a reinforced concrete building is...Ch. 7 - The state of stress on an element along the...Ch. 7 - -15 Repeat the preceding problem using ??. = - 750...Ch. 7 - A propeller shaft subjected to combined torsion...Ch. 7 - 3-17 The stresses at a point along a beam...Ch. 7 - -18 through 7.3-22 An element in plane stress (see...Ch. 7 - -18 through 7.3-22 An element in plane stress (see...Ch. 7 - -18 through 7.3-22 An element in plane stress (see...Ch. 7 - -18 through 7.3-22 An element in plane stress (see...Ch. 7 - -18 through 7.3-22 An element in plane stress (see...Ch. 7 - At a point on the web of a girder on a gantry...Ch. 7 - The stresses acting on a stress element on the arm...Ch. 7 - The stresses at a point on the down tube of a...Ch. 7 - An element in plane stress on the surface of an...Ch. 7 - A simply supported wood beam is subjected to point...Ch. 7 - A simply supported wood beam is subjected to point...Ch. 7 - Prob. 7.4.1PCh. 7 - .4-2 An element in uniaxial stress is subjected to...Ch. 7 - An element on the gusset plate in Problem 7.2-23...Ch. 7 - An element on the top surface of the fuel tanker...Ch. 7 - An element on the top surface of the fuel tanker...Ch. 7 - An element in biaxial stress is subjected to...Ch. 7 - • - 7.4-7 An element on the surface of a drive...Ch. 7 - - A specimen used in a coupon test has norm al...Ch. 7 - A specimen used in a coupon test is shown in the...Ch. 7 - The rotor shaft of a helicopter (see figure part...Ch. 7 - An element in pure shear is subjected to stresses...Ch. 7 - An clement in plane stress is subjected to...Ch. 7 - Prob. 7.4.13PCh. 7 - An clement in plane stress is subjected to...Ch. 7 - An clement in plane stress is subjected to...Ch. 7 - An clement in plane stress is subjected to...Ch. 7 - Prob. 7.4.17PCh. 7 - -18 through 7.4-25 An clement in plane stress is...Ch. 7 - -18 through 7.4-25 An clement in plane stress is...Ch. 7 - Prob. 7.4.20PCh. 7 - -18 through 7.4-25 An clement in plane stress is...Ch. 7 - Through 7.4-25 An clement in plane stress is...Ch. 7 - -18 through 7.4-25 An clement in plane stress is...Ch. 7 - through 7.4-25 An clement in plane stress is...Ch. 7 - -18 through 7.4-25 An clement in plane stress is...Ch. 7 - 1 A rectangular steel plate with thickness t = 5/8...Ch. 7 - Solve the preceding problem if the thickness of...Ch. 7 - The state of stress on an element of material is...Ch. 7 - An element of a material is subjected to plane...Ch. 7 - Assume that the normal strains x and y , for an...Ch. 7 - A cast-iron plate in biaxial stress is subjected...Ch. 7 - Solve the preceding problem for a steel plate with...Ch. 7 - • - 3 A rectangular plate in biaxial stress (see...Ch. 7 - Solve the preceding problem for an aluminum plate...Ch. 7 - A brass cube of 48 mm on each edge is comp ressed...Ch. 7 - 7.5-11 in. cube of concrete (E = 4.5 X 106 psi. v...Ch. 7 - -12 A square plate of a width h and thickness t is...Ch. 7 - Solve the preceding problem for an aluminum plate...Ch. 7 - A circle of a diameter d = 200 mm is etched on a...Ch. 7 - The normal stress on an elastomeric rubber pad in...Ch. 7 - A rubber sheet in biaxial stress is subjected to...Ch. 7 - An element of aluminum is subjected to tri-axial...Ch. 7 - An element of aluminum is subjected to tri- axial...Ch. 7 - -3 An element of aluminum in the form of a...Ch. 7 - Solve the preceding problem if the element is...Ch. 7 - A cube of cast iron with sides of length a = 4.0...Ch. 7 - Solve the preceding problem if the cube is granite...Ch. 7 - An element of aluminum is subjected to iriaxial...Ch. 7 - Prob. 7.6.8PCh. 7 - A rubber cylinder R of length L and cross-...Ch. 7 - A block R of rubber is confined between plane...Ch. 7 - -11 A rubber cube R of a side L = 3 in. and cross-...Ch. 7 - A copper bar with a square cross section is...Ch. 7 - A solid spherical ball of magnesium alloy (E = 6.5...Ch. 7 - A solid steel sphere (E = 210 GPa, v = 0.3) is...Ch. 7 - Prob. 7.6.15PCh. 7 - An element of material in plain strain has the...Ch. 7 - An clement of material in plain strain has the...Ch. 7 - An element of material in plain strain is...Ch. 7 - An element of material in plain strain is...Ch. 7 - A thin rectangular plate in biaxial stress is...Ch. 7 - Prob. 7.7.6PCh. 7 - A thin square plate in biaxial stress is subjected...Ch. 7 - Prob. 7.7.8PCh. 7 - An clement of material subjected to plane strain...Ch. 7 - Solve the preceding problem for the following...Ch. 7 - The strains for an element of material in plane...Ch. 7 - Solve the preceding problem for the following...Ch. 7 - An clement of material in plane strain (see...Ch. 7 - Solve the preceding problem for the following...Ch. 7 - A brass plate with a modulus of elastici ty E = 16...Ch. 7 - Solve the preceding problem if the plate is made...Ch. 7 - An element in plane stress is subjected to...Ch. 7 - Prob. 7.7.18PCh. 7 - During a test of an airplane wing, the strain gage...Ch. 7 - A strain rosette (see figure) mounted on the...Ch. 7 - A solid circular bar with a diameter of d = 1.25...Ch. 7 - A cantilever beam with a rectangular cross section...Ch. 7 - Solve the preceding problem if the cross-...Ch. 7 - A 600 strain rosette, or delta rosette, consists...Ch. 7 - On the surface of a structural component in a...Ch. 7 - - 7.2-26 The strains on the surface of an...Ch. 7 - Solve Problem 7.7-9 by using Mohr’s circle for...Ch. 7 - 7.7-28 Solve Problem 7.7-10 by using Mohr’s circle...Ch. 7 - Solve Problem 7.7-11 by using Mohr’s circle for...Ch. 7 - Solve Problem 7.7-12 by using Mohr’s circle for...Ch. 7 - Solve Problem 7.7-13 by using Mohr’s circle for...Ch. 7 - Prob. 7.7.32P
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
An Introduction to Stress and Strain; Author: The Efficient Engineer;https://www.youtube.com/watch?v=aQf6Q8t1FQE;License: Standard YouTube License, CC-BY