Explanation Given information: EI is constant. Assumptions: Consider that the right side force as positive and left side as negative in horizontal components of forces. Consider that the upward force as positive and downward as negative in vertical components of forces. Consider that the clockwise moment as negative and anticlockwise moment as positive. Virtual-work equation for the beam as follows: 1 ⋅ θ = ∫ 0 L m θ M E I d x (1) Here, 1 is external virtual unit load acting on the beam in the direction of θ , θ is slope caused by the real loads on the beam, m θ is internal virtual moment in the beam, expressed as a function of x and caused by the external virtual unit moment, 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: Show the free-body diagram of the entire beam as in Figure 1. Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 w × 2 L × 2 L 2 − B y ( L ) = 0 B y ( L ) = w × 2 L × 2 L 2 B y = 2 w L Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of forces. ∑ F x = 0 A x = 0 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 − w × 2 L = 0 (2) Substitute 2 w L for B y in Equation (2). A y + 2 w L − 2 w L = 0 A y = 0 Region 0 ≤ x < L : Show the free body diagram of the beam for the region 0 ≤ x < L 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 + w × x 1 × x 1 2 = 0 M 1 = − w x 1 2 2 Region L < x ≤ 2 L : Show the free body diagram of the beam for the region L < x ≤ 2 L 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 ( x 2 ) − w × x 2 × x 2 2 = 0 (3) Substitute 0 for A y in Equation (3). − M 2 + 0 ( x 2 ) − w ( x 2 ) 2 2 = 0 M 2 = − w x 2 2 2 Internal virtual moment in the beam ( m ): Apply a unit moment at point B and calculate the moments. Determine the reactions: Show the free-body diagram of the entire beam as in Figure 4. Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 B y ( L ) − 1 = 0 B y = 1 L Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of forces

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

See similar textbooks

Videos

Question

Chapter 14.7, Problem 14.122P

To determine

The slope at point B.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.7, Problem 14.121P

Chapter 14.9, Problem 14.123P

Students have asked these similar questions

Correct Answer is written below. Detailed and complete fbd only please. I will upvote, thank you. 1: The assembly shown is composed of a rigid plank ABC, supported by hinge at A, spring at B and cable at C.The cable is attached to a frictionless pulley at D and rigidly supported at E. The cable is made of steel with E = 200,000MPa and cross-sectional area of 500 mm2. The details of pulley at D is shown. The pulley is supported by a pin, passingthough the pulley and attached to both cheeks. Note that E is directly above B.Given: H = 3 m; L1 = 2 m; L2 = 4 m; w = 12 kN/m; x:y = 3:4Spring Parameters:Wire diameter = 30 mmMean Radius = 90 mmNumber of turns = 12Modulus of Rigidity = 80 GPaAllowable stresses:Allowable shear stress of Pin at D = 85 MPaAllowable normal stress of cheek at D = 90MPaAllowable bearing stress of cheek at D = 110MPa1. Calculate the reaction of spring Band tension in cable at C.2. Calculate the vertical displacementat C and the required diameter ofpin at D.3.…

Correct answer and complete fbd only. I will upvote. The compound shaft, composed of steel,aluminum, and bronze segments, carries the two torquesshown in the figure. If TC = 250 lb-ft, determine the maximumshear stress developed in each material (in ksi). The moduliof rigidity for steel, aluminum, and bronze are 12 x 106 psi, 4x 106 psi, and 6 x 106 psi, respectively

Can you explain the algebra steps that aren't shown but stated to be there, on how to get this equation

Chapter 14 Solutions

Mechanics of Materials

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 point B .

Videos

The slope at point B.

Want to see the full answer?

Chapter 14 Solutions