Explanation Given information: The A992 steel beam has a moment of inertia of I = 125 ( 10 6 ) mm 4 Method adopted: Castigliano’s theorem. Assumptions: Young’s modulus of the A992 steel beam is E = 200 × 10 6 kN/m 2 . Explanation: Castigliano’s theorem to determine the slope for the beam is as follows: θ = ∫ 0 L M ( ∂ M ∂ M ′ ) d x E I (1) Here, θ is the slope of the tangent θ at a point on the elastic curve is to be determined, M is internal moment in the beam, expressed as a function of x and caused by both the force P and the actual loads on the beam, M ′ is the external moment, 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 ): Apply external moment M ′ at the point where the slope is to be determined. Determine the reactions: Entire beam: Show the free-body diagram of the entire beam as in Figure 1. Moment about point B : Determine the vertical reaction at point C by taking moment about point B . ∑ M B = 0 C y ( 6 ) + M ′ − 18 = 0 (2) Along the vertical direction: Determine the vertical reaction at point B by resolving the vertical component of forces. ∑ F y = 0 B y + C y = 0 (3) Along the horizontal direction: Determine the horizontal reaction at point C by resolving the horizontal component of forces. ∑ F x = 0 C x = 0 Show the calculation of values as follows: Conclusion: Solve Equation (2). C y ( 6 ) = 18 − M ′ C y = 3 − M ′ 6 Substitute ( 3 − M ′ 6 ) for C y in Equation (3). B y + 3 − M ′ 6 = 0 B y = M ′ 6 − 3 Region 0 ≤ x < 4 m from left end: Show the free body diagram of the beam for the region 0 ≤ x < 4 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 + M ′ = 0 (4) SolveEquation (4). M 1 = − M ′ Region 4 m < x < 10 m from left end: Show the free body diagram of the beam for the region 4 m < x < 10 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 + M ′ − B y ( x 2 ) = 0 (5) Substitute ( M ′ 6 − 3 ) for B y in Equation (5). M 2 + M ′ − ( M ′ 6 − 3 ) ( x 2 ) = 0 M 2 + M ′ − M ′ x 2 6 + 3 x 2 = 0 M 2 = − M ′ + M ′ x 2 6 − 3 x 2 Region 10 m < x ≤ 14 m : Show the free body diagram of the beam for the region 10 m < x ≤ 14 m as in Figure 4. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 3 = 0 − M 3 − 18 = 0 (6) Solve Equation (6). M 3 = − 18 kN-m Calculation: Tabulate the calculated values as in Table 1. Limit (m) M ∂ M ∂ M ′ M ( M ′ = 18 kN-m ) 0 ≤ x < 4 M 1 = − M ′ ∂ M 1 ∂ M ′ = − 1 M 1 = − 18 kN-m 4 < x < 10 M 2 = − M ′ + M ′ x 2 6 − 3 x 2 ∂ M 2 ∂ M ′ = − 1 + x 2 6 M 2 = − 18 kN-m 10 < x ≤ 14 M 3 = − 18 kN-m ∂ M 3 ∂ M ′ = 0 M 3 = − 18 kN-m Rearrange the Equation (1)

11th Edition

ISBN: 9780137605460

Author: Russell C. Hibbeler

Publisher: Pearson Education (US)

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 14.10, Problem 139P

To determine

The slope at A.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.10, Problem 138P

Chapter 14.10, Problem 140P

Students have asked these similar questions

a) Determine state of stress at all points (a, b and c). These points are located on the exteriorsurface of the beam. Point a is located along the centreline of the beam, point b is 15mmfrom the centreline and point c is located on the edge of the beam. Present yourresults in a table and ensure that your sign convention is clearly shownb) Construct Mohrs circle at this point andcalculate the principal stresses and maximum in‐plane shear stress (τmax) at pointc. sketch the resulting state of stress at this point clearly indicating themagnitude of the stresses and any angles associated with the state of stress (principal ormaximum in‐plane shear).

parts e,f,g

Figure 9-6 9-49 An aluminum alloy plate with dimensions 20 cm x 10 cm × 2 cm needs to be cast with a secondary dendrite arm spacing of 10-2 cm (refer to Figure 9-6). What mold constant B is required (assume n = 2 )? Secondary dendrite arm spacing (cm) - 10-1 10-2 10-3 10 41 0.1 1 Copper Zinc alloys Aluminum alloys 10 100 1,000 10,000 100,000 Solidification time (s)

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 - The A-36 steel bar consists of two segments, one...Ch. 14.2 - If P = 10 kip, determine the total strain energy...Ch. 14.2 - Determine the maximum force P and the...Ch. 14.2 - Consider the thin-walled tube of Fig.5-26 . Use...Ch. 14.2 - Determine the bending strain energy in the 2-in...Ch. 14.2 - Determine the bending strain energy in the...Ch. 14.2 - Determine the bending strain energy in the simply...Ch. 14.3 - Determine the horizontal displacement of joint A....

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 .

Solution Summary: The author explains Castigliano's theorem to determine the slope for the A992 steel beam.

Concept explainers

Videos

The slope at A.

Want to see the full answer?

Chapter 14 Solutions