Explanation The Figure-(1) shows the free body diagram of the beam. Figure-(1) Here, the reaction force at A is R A and the reaction force at B is R B . Write the net moment at point B . ( R A × a ) − ( w 2 × l 2 ) = 0 (I) Here, the length of the beam is l , the distance between the point A and point B is a and uniform load intensity is w . Write the expression for the net force on the beam. R A + R B − ( w × l ) = 0 (II) Here, the net reaction force at point B is R B . Refer to Table 3-1 “Singularity (Macaulay) functions” Write the expression for load intensity acting on the beam. q = R A 〈 x 〉 − 1 − w 〈 x 〉 0 − R B 〈 x − a 〉 − 1 (III) Here, the load intensity on the beam is q , the distance from the left end is x . Write the expression for the moment. M = ∫ ∫ q d x (IV) Here, the net bending moment is M . Substitute the value from Equation (III) to Equation (IV). M = ∫ ∫ R A 〈 x 〉 − 1 − w 〈 x 〉 0 − R B 〈 x − a 〉 − 1 = R A 〈 x 〉 1 − w 2 〈 x 〉 2 − R B 〈 x − a 〉 1 (V) Write the equation for the deflection across the beam. E I d 2 y d x 2 = M (VI) Here, the modulus of elasticity is E ,the area moment of inertia is I and the deflection is y . Substitute the value from Equation (V) to Equation (VI). E I d 2 y d x 2 = R A 〈 x 〉 1 − w 2 〈 x 〉 2 − R B 〈 x − a 〉 1 (VII) Integrate Equation (VII) with respect to x . E I d y d x = R A 2 〈 x 〉 2 − w 6 〈 x 〉 3 − R B 2 〈 x − a 〉 2 + C 1 (VIII) Integrate the Equation (VIII) with respect to x . E I ∫ d y d x = ∫ [ R A 2 〈 x 〉 2 − w 6 〈 x 〉 3 − R B 2 〈 x − a 〉 2 + C 1 ] E I y = R A 6 〈 x 〉 3 − w 24 〈 x 〉 4 − R B 6 〈 x − a 〉 3 + C 1 x + C 2 (IX) Conclusion: Refer to Table A-5 “Physical Constants of Materials”, obtain the properties of modulus of elasticity for carbon steel as 30 × 10 6 psi . Substitute 100 lbf / in for w , 14 in for l and 10 in for a in Equation (I). ( R A × 10 in ) − ( ( 100 lbf / in ) 2 × ( 14 in ) 2 ) = 0 R A = 100 lbf / in × 14 in × 7 in 10 in R A = 420 lbf Thus, the reactions at the left supports is 420 lbf . Substitute 420 lbf for R A , 100 lbf / in for w and 14 in for l in Equation (II)

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

10th Edition

ISBN: 9780073398204

Author: Richard G Budynas, Keith J Nisbett

Publisher: McGraw-Hill Education

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 4, Problem 59P

To determine

The reactions at the supports.

The deflection equations of the beam.

Plot the deflection.

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Chapter 4, Problem 58P

Chapter 4, Problem 60P

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Chapter 4 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

Ch. 4 - The figure shows a torsion bar OA fixed at O,...Ch. 4 - For Prob. 41, if the simple support at point A...Ch. 4 - A torsion-bar spring consists of a prismatic bar,...Ch. 4 - An engineer is forced by geometric considerations...Ch. 4 - A bar in tension has a circular cross section and...Ch. 4 - Prob. 6P Ch. 4 - Prob. 7P Ch. 4 - Derive the equations given for beam 2 in Table A9...Ch. 4 - Derive the equations given for beam 5 in Table A9...Ch. 4 - The figure shows a cantilever consisting of steel...

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The reactions at the supports. The deflection equations of the beam. Plot the deflection.

Solution Summary: The author analyzes the reactions at the supports and the deflection equations of the beam.