Explanation Given information: The diameter of the shaft is d = 100 mm . Assumptions: Young’s modulus of the 2014-T6 aluminum shaft is E = 73.1 × 10 6 kN/m 2 . Explanation: Virtual-work equation for the entire beam is: 1 ⋅ Δ = ∫ 0 L m M E I d x (1) Here, 1 is external virtual unit load acting on the beam in the direction of Δ , Δ is displacement caused by the real loads on the beam, m is internal virtual moment in the beam, expressed as a function of x and caused by the external virtual unit load, M is internal moment in the beam, expressed as a function of x and caused by the real loads, E is modulus of elasticity of a member, and I is moment of inertia of the cross-sectional area about the neutral axis. Internal moment in the beam ( M ): Determine the reactions: Entire shaft: Show the free-body diagram of the entire shaft as in Figure 1. Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 B y ( 3 ) − 8 ( 2 ) − 8 ( 1 ) = 0 (2) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y + B y − 8 − 8 = 0 (3) Show the calculation of values as follows: Conclusion: Solve Equation (2). B y ( 3 ) − 16 − 8 = 0 B y = 8 kN Substitute 8 kN for B y in Equation (3). A y + 8 − 8 − 8 = 0 A y = 8 kN Region 0 ≤ x < 1 m : Show the free body diagram of the shaft for the region 0 ≤ x < 1 m as in Figure 2. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 1 = 0 M 1 − A y ( x 1 ) = 0 (4) Substitute 8 kN for A y Equation (4). M 1 − 8 ( x 1 ) = 0 M 1 = 8 x 1 Region 1 m < x < 1.5 m : Show the free body diagram of the shaft for the region 1 m < x < 1.5 m as in Figure 3. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 2 = 0 M 2 − A y ( 1 + x 2 ) + 8 ( x 2 ) = 0 (5) Substitute8 kN for A y in Equation (5). M 2 − 8 ( 1 + x 2 ) + 8 ( x 2 ) = 0 M 2 − 8 − 8 x 2 + 8 x 2 = 0 M 2 = 8 kN-m Internal virtual moment in the beam ( m ): Apply a unit moment at point A and calculate the moments. Determine the reactions: Entire shaft: Show the free-body diagram of the entire shaft as in Figure 4. Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 − 1 ( 1.5 ) + B y ( 3 ) = 0 (6) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y + B y − 1 = 0 (7) Show the calculation of values as follows: Conclusion: Solve Equation (6). 3 B y = 1.5 B y = 0.5 Substitute 0.5 for B y in Equation (7). A + 0.5 − 1 = 0 A y = 0.5 Region 0 ≤ x < 1 m : Show the free body diagram of the shaft for the region 0 ≤ x < 1 m as in Figure 5. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 1 = 0 m 1 − A y ( x 1 ) = 0 (8) Substitute 0.5for A y in Equation (8). m 1 − 0

9th Edition

ISBN: 2810014920922

Author: HIBBELER

Publisher: PEARSON

See similar textbooks

Concept explainers

Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and af…

Videos

Question

Chapter 14.7, Problem 14.104P

To determine

The slope at A of the 2014-T6 aluminum shaft.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.7, Problem 14.103P

Chapter 14.7, Problem 14.105P

Students have asked these similar questions

This is an old practice exam. The answers are OAB = 19.10 ksi OBC = 2.228 ksi OCD = −2.865 ksi v = 0.2792delta Ltot = 0.01585 in (increase) but why

A random poly(styrene-butadiene) copoly- mer has a number-average molecular weight of 350,000 g/mol and a degree of polymerization of 5000. Compute the fraction of styrene and buta- diene repeat units in this copolymer. H H | | -C-C- 방 H

Design and assemble on the fluidsim (or a draft) the Hydraulic Drive Circuit, with the following characteristics: (a) Sequential operation, pressure, for the advance and return of the cylinders (according to the proper operation for the device) controlled by a directional 4x3 way, closed center; (b) Speed control for the cylinders, according to the load signal; (c) Pressure counterbalance for cylinder A, in order to compensate for the weight of the assembly.

Chapter 14 Solutions

MECHANICS OF MATERIALS-TEXT

Ch. 14.2 - A material is subjected to a general state of...Ch. 14.2 - The strain-energy density for plane stress must be...Ch. 14.2 - 14–3. Determine the strain energy in the stepped...Ch. 14.2 - Prob. 14.4P Ch. 14.2 - Using bolts of the same material and...Ch. 14.2 - If P = 60 kN, determine the total strain energy...Ch. 14.2 - Determine the maximum force P and the...Ch. 14.2 - *14–8. Determine the torsional strain energy in...Ch. 14.2 - Determine the torsional strain energy in the A-36...Ch. 14.2 - The shaft assembly is fixed at C. The hollow...

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.

The slope at A of the 2014-T6 aluminum shaft.

Solution Summary: The author shows the free-body diagram of the 2014-T6 aluminum shaft as in Figure 1. Determine the vertical reaction at point A by resolving vertical component of forces.

Concept explainers

Videos

The slope at A of the 2014-T6 aluminum shaft.

Want to see the full answer?

Chapter 14 Solutions