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Question:

1) Two vertical forces are applied to a beam of the cross-section shown above. Determine the maximum tensile and compressive stresses in portion BC of the beam.

2) Draw the free-body diagram

3) Draw the shear diagram.

4) Draw the moment diagram.

5) Determine the equation for the shear force (V) as a function of x using singularity functions.

6) Determine the equation for the internal (M) as a function of x using singularity functions.

 

 

Shear Force:

It is the algebraic summation of all the forces acting on a beam either at left or right of the section of the beam.

Bending Moment:

It is the algebraic summation of all the moments caused by the forces acting on a beam either at left or right of the section o

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Manoj Pandey
Manoj PandeyLv10
15 Sep 2020

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

A uniform beam resting on two pivots has a length L=6.00 m and mass M=90.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot located at a distance l=4.00 m from the left end exerts a normal force n2. A woman of mass m=55.0 kg steps onto the left end of the beam and begins walking to the right as in the figure. The goal is to find the woman’s position when the beam begins to tip.

(a) What is the appropriate analysis model for the beam before it begins to tip?

(b) Sketch a force diagram for the beam, labeling the gravitational and normal forces acting on the beam and placing the woman a distance x to the right of the first pivot, which is the origin.

(c) Where is the woman when the normal force n1 is the greatest?

(d) What is n1 when the beam is about to tip?

(e) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip.

(f) Using the result of part (d) and the force equation of equilibrium, with torques computed around the second pivot, find the woman’s position x when the beam is about to tip.

(g) Check the answer to part (e) by computing around the first pivot point.

<p>A Simple Atwood Machine consists of two masses m1 and m1 that are connected by a string wound over a pulley, as seen in the figure below. m2 is larger than m1.<br />On a piece of paper, draw two free body diagrams; one for each of the masses, showing all forces acting on each mass. Then answer the following questions.<br /><br /><br />(a) Based on your free body diagram, identify the forces acting on the mass m1. Select all that apply.<br />(m1+m2)a<br />m1g<br />FN<br />m1a<br />(m1+m2)g<br />m2a<br />f<br />m2g<br />T<br /><br /><br /><br />(b) Using the direction rosette indicate the direction for each of the forces acting on m1 that you identified in part (a). If any of the forces listed below is not applicable, enter NA.<br />m1a= <br />m2a= <br />m1g= <br />m2g= <br />(m1+m2)a= <br />(m1+m2)g= <br />f= <br />T= <br />FN= <br /><br />(c) Based on your free body diagram, identify the forces acting on the mass m2. Select all that apply.<br />T<br />(m1+m2)g<br />m2g<br />m1g<br />(m1+m2)a<br />m1a<br />m2a<br />f<br />FN<br /><br /><br /><br />(d) Using the direction rosette indicate the direction for each of the forces acting on m2 that you identified in part (c). If any of the forces listed below is not applicable, enter NA.<br />m1a= <br />m2a= <br />m1g= <br />m2g= <br />(m1+m2)a= <br />(m1+m2)g= <br />f= <br />T= <br />FN= <br /><br />(e) Write Newton's Second Law equation (in the form Fnet = ma) for m1 in the y-direction.<br />Use the first answer box for the left hand side (LHS) of the equation and the second answer box for the right hand side (RHS) of the equation.<br />It is important to use the correct signs for each of the forces. Motion in the upward direction is positive and motion in the downward direction is negative. (Use m1, g, a, and T as appropriate.)<br />LHS = <br />RHS = <br /><br />(f) Write Newton's Second Law equation (in the form Fnet = ma) for m2 in the y-direction.<br />Use the first answer box for the left hand side (LHS) of the equation and the second answer box for the right hand side (RHS) of the equation.<br />It is important to use the correct signs for each of the forces. Motion in the upward direction is positive and motion in the downward direction is negative.(Use m2, g, a, and T as appropriate.)<br />LHS = <br />RHS = <br /><br />(g) Use your answers from parts (e) and (f) to get an expression for the magnitude of the tension T. (Use m1, m2, and g as appropriate.)<br />magnitude of T = <br /><br />(h) Use your answers from parts (e), (f), and (g) to find the magnitude of a, the acceleration of the system. (Use m1, m2, and g as appropriate.)<br />magnitude of a = <br /><br />(i) In what direction will each mass accelerate.<br />For mass m1: <br />For mass m2: <br /><br />(j) Suppose m1 = 7 kg and m2 = 11 kg. What is the tension in the string? <br />T = <br /><br />(k) Suppose m1 = 7 kg and m2 = 11 kg. What is the value of the acceleration?<br />a = <br /><br />(l) Suppose that m2 starts from rest at a height of 5 m. Use the Kinematic Equations to determine how long it takes for m2 to hit the ground.</p>

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