Explanation Given information: The bending moment is applied alone to a solid bar with a circular cross section. It causes the distortion energy as the combination of an applied bending moment M and torque T . Calculation: Consider that the distortion energy is the bending moment applied to the solid bar. The normal stress applied to the solid bar. σ = M e c I Here, M e is the equivalent bending moment, c is the radius of the bar, and I is the moment of inertia. Find the principal stresses in complex system as shown below. σ 1 , 2 = σ x + σ y 2 ± ( σ x − σ y 2 ) 2 + τ x y 2 (1) Here, σ x is the normal stress in x direction, σ y is the normal stress in y direction, and τ x y is the shear stress. Substitute M e c I for σ x , 0 for σ y , and 0 for τ x y in Equation (1). σ 1 , 2 = M e c I + 0 2 ± ( M e c I − 0 2 ) 2 + 0 = M e c 2 I ± M e c 2 I σ 1 = M e c 2 I + M e c 2 I = M e c I σ 2 = M e c 2 I − M e c 2 I = 0 Apply the maximum distortion theory for plane stress as shown below. σ 1 2 − σ 1 σ 2 + σ 2 2 = σ Y 2 Here, σ 1 and σ 2 are the principal stresses and σ Y is the principal yield stress. Apply the strain-energy density as shown below. u d = 1 + ν 3 E ( σ 1 2 − σ 1 σ 2 + σ 2 2 ) (2) Here, E is the modulus of elasticity and ν is the Poisson’s ratio. Substitute M e c I for σ 1 and 0 for σ 2 in Equation (2). ( u d ) 1 = 1 + ν 3 E ( ( M e c I ) 2 − ( M e c I ) ( 0 ) + ( 0 ) 2 ) = 1 + ν 3 E ( M e 2 c 2 I 2 ) (3) Consider that the distortion energy is the combination of the bending moment and torque applied to the solid bar. Consider the normal stress and shear stress applied to the solid bar. σ = M c I Here, M is the bending moment. τ = T c J Here, T is the torque and J is the polar moment of inertia. Substitute σ for σ x , 0 for σ y , and τ for τ x y in Equation (1). σ 1 , 2 = σ + 0 2 ± ( σ − 0 2 ) 2 + τ 2 = σ 2 ± σ 2 4 + τ 2 σ 1 = σ 2 + σ 2 4 + τ 2 σ 2 = σ 2 − σ 2 4 + τ 2 Consider a = σ 2 and b = σ 2 4 + τ 2 . Hence, σ 1 = a + b and σ 2 = a − b . Substitute ( a + b ) for σ 1 and ( a − b ) for σ 2 in Equation (2)

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

See similar textbooks

Concept explainers

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…

Videos

Question

Chapter 10.7, Problem 10.70P

To determine

To derive: An expression for an equivalent bending moment Me.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.7, Problem 10.69P

Chapter 10.7, Problem 10.71P

Students have asked these similar questions

Can you provide me the step by step process of this? Thank you! Use the graphical method to construct the bending-moment diagram and identify the magnitude of the largest moment (consider both positive and negative). The ground reactions and shear-force diagram are provided.MA = 22 kN-mMB = 10 kN-mAy = 17.00 kNBy = 13.00 kN

A steel bar withstands simultaneously a 30 kN compressive force, a 5 kNm bending moment and a lkNm torque. Assume the cross-section is circular with diameter R. Determine, for a point at the position of greatest compressive bending stress: (a) The normal stress on a plane inclined at 30° to the shaft axis (b) The shear stress on a plane inclined at 60° to the shaft axis

A wire of circular cross-section ofdiameter 1.0 mm is bent into a circular arc of radius 1.0 m by application of pure bending moments at its ends. The Young's modulus of the material of the wire is 100 GPa. The maximum tensile stress developed in the wire is MPa.

Chapter 10 Solutions

Mechanics of Materials (10th Edition)

Ch. 10.3 - Prove that the sum of the normal strains in...Ch. 10.3 - The state of strain at the point on the arm has...Ch. 10.3 - The state of strain at the point on the pin leaf...Ch. 10.3 - The state of strain at the point on the pin leaf...Ch. 10.3 - The state of strain at the point on the leaf of...Ch. 10.3 - Use the strain transformation equations and...Ch. 10.3 - Use the strain transformation equations and...Ch. 10.3 - Use the strain transformation equations to...Ch. 10.3 - Use the strain transformation equations to...Ch. 10.3 - Use the strain- transformation equations to...

Ch. 10.3 - Use the strain transformation equations to...Ch. 10.3 - Determine the equivalent state of strain on an...Ch. 10.3 - Determine the equivalent state of strain which...Ch. 10.3 - Use the strain transformation equations to...Ch. 10.3 - Determine the equivalent state of strain, which...Ch. 10.3 - Solve Prob.103 using Mohrs circle. 103. The state...Ch. 10.3 - using Mohrs circle. 103. The state of strain at...Ch. 10.3 - Solve Prob.105 using Mohrs circle. 105. The state...Ch. 10.3 - Solve Prob.108 using Mohrs circle 108. The state...Ch. 10.3 - using Mohrs circle. 106. The state of strain at a...Ch. 10.5 - The strain at point A on the bracket has...Ch. 10.5 - Determine (a) the principal strains at A, (b) the...Ch. 10.5 - Determine (a) the principal strains at A, in the...Ch. 10.5 - The following readings are obtained for each gage:...Ch. 10.5 - The following readings are obtained for each gage:...Ch. 10.5 - The following readings are obtained for each gage:...Ch. 10.5 - The following readings are obtained from each...Ch. 10.6 - For the case of plane stress, show that Hookes law...Ch. 10.6 - to develop the strain tranformation equations....Ch. 10.6 - Determine the modulus of elasticity and Polssons...Ch. 10.6 - If it is subjected to an axial load of 15 N such...Ch. 10.6 - If it has the original dimensions shown, determine...Ch. 10.6 - If it has the original dimensions shown, determine...Ch. 10.6 - A strain gage having a length of 20 mm Is attached...Ch. 10.6 - Determine the bulk modulus for each of the...Ch. 10.6 - The strain gage is placed on the surface of the...Ch. 10.6 - Determine the associated principal stresses at the...Ch. 10.6 - Determine the applied load P. What is the shear...Ch. 10.6 - If a load of P = 3 kip is applied to the A-36...Ch. 10.6 - The cube of aluminum is subjected to the three...Ch. 10.6 - The principal strains at a point on the aluminum...Ch. 10.6 - A uniform edge load of 500 lb/in. and 350 lb/in....Ch. 10.6 - Prob. 10.45P Ch. 10.6 - A single strain gage, placed in the vertical plane...Ch. 10.6 - A single strain gage, placed in the vertical plane...Ch. 10.6 - If the material is graphite for which Eg = 800 ksi...Ch. 10.6 - Determine the normal stresses x and y in the plate...Ch. 10.6 - The steel shaft has a radius of 15 mm. Determine...Ch. 10.6 - Prob. 10.51P Ch. 10.6 - The A-36 steel pipe is subjected to the axial...Ch. 10.6 - Air is pumped into the steel thin-walled pressure...Ch. 10.6 - Air is pumped into the steel thin-walled pressure...Ch. 10.6 - Prob. 10.55P Ch. 10.6 - The thin-walled cylindrical pressure vessel of...Ch. 10.6 - The thin-walled cylindrical pressure vessel of...Ch. 10.6 - Prob. 10.58P Ch. 10.7 - A material is subjected to plane stress. Express...Ch. 10.7 - A material is subjected to plane stress. Express...Ch. 10.7 - The yield stress for a zirconium-magnesium alloy...Ch. 10.7 - Solve Prob. 1061 using the maximum distortion...Ch. 10.7 - If a machine part is made of tool L2 steel and a...Ch. 10.7 - Solve Prob.1063 using the maximum distortion...Ch. 10.7 - Prob. 10.65P Ch. 10.7 - If a shaft is made of a material for which y = 75...Ch. 10.7 - Solve Prob.1066 using the maximum shear stress...Ch. 10.7 - If the material is machine steel having a yield...Ch. 10.7 - The short concrete cylinder having a diameter of...Ch. 10.7 - Prob. 10.70P Ch. 10.7 - The plate is made of Tobin bronze, which yields at...Ch. 10.7 - The plate is made of Tobin bronze, which yields at...Ch. 10.7 - An aluminum alloy is to be used for a solid drive...Ch. 10.7 - If a machine part is made of titanium (TI-6A1-4V)...Ch. 10.7 - The components of plane stress at a critical point...Ch. 10.7 - The components of plane stress at a critical point...Ch. 10.7 - The 304-stainless-steel cylinder has an inner...Ch. 10.7 - The 304-stainless-steel cylinder has an inner...Ch. 10.7 - If the 2-in diameter shaft is made from brittle...Ch. 10.7 - If the 2-in diameter shaft is made from cast iron...Ch. 10.7 - If Y = 50 ksi, determine the factor of safety for...Ch. 10.7 - Prob. 10.82P Ch. 10.7 - If the yield stress for steel is Y = 36 ksi,...Ch. 10.7 - Prob. 10.84P Ch. 10.7 - The state of stress acting at a critical point on...Ch. 10.7 - The shaft consists of a solid segment AB and a...Ch. 10.7 - The shaft consists of a solid segment AB and a...Ch. 10.7 - Prob. 10.88P Ch. 10.7 - If Y = 50 ksi, determine the factor of safety for...Ch. 10.7 - The gas tank is made from A-36 steel and has an...Ch. 10.7 - The internal loadings at a critical section along...Ch. 10.7 - If the material is machine steel having a yield...Ch. 10.7 - If the material is machine steel having a yield...Ch. 10 - In the case of plane stress, where the in-plane...Ch. 10 - The plate is made of material having a modulus of...Ch. 10 - If the material is machine steel having a yield...Ch. 10 - Determine if yielding has occurred on the basis of...Ch. 10 - The 60 strain rosette is mounted on a beam. The...Ch. 10 - Use the strain transformation equations to...Ch. 10 - If the strain gages a and b at points give...Ch. 10 - Use the strain-transformation equations and...Ch. 10 - Use the strain transformation equations to...Ch. 10 - Specify the orientation of the corresponding...

Knowledge Booster

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.

To derive: An expression for an equivalent bending moment M e .

Solution Summary: The author explains that the distortion energy is the bending moment applied to the solid bar with a circular cross section.

Concept explainers

Videos

To derive: An expression for an equivalent bending moment Me.

Want to see the full answer?

Chapter 10 Solutions