# Mechanics of Materials Explanation

I was reading through my mechanics of materials book (Mechanics of Materials 8th ed. R.C. Hibbler) learning about stress and came across example 1.6 pg. 28. while most of the example makes sense the last bit of part b) confused me. I am unsure as to why for Pcd he only showed the 22 kN force acting in both directions but neglected the two 4 kN forces acting in the positive x direction..

If I need to I could further explain the whole problem but if someone happens to have the same book and can explain this to me that would be great. If I need to I can further explain his example I just figured I would try to save time if possible.

I hope this is the right place to post this question, it's not technically a homework question since its just an example in the book.

well thanks for any help ahead of time.

Also.. I don't know if I am allowed to do this but I can scan and upload an image of the example problem if needed. Let me know if this is possible as it might help.

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I happen to have the seventh edition of Hibbeler's Mechanics of Materials; it sounds like the example hasn't changed.

One issue I have always had with Hibbeler (I've used three of his books) is that he tends to assume you're comfortable enough with previous material to not need it repeated. Sure it's fair for him to think that you understood everything from prerequisite classes, but it would truly help out students if he reminded them of details. Hopping off my soapbox, you might want to review the method of sections (Hibbeler's Statics, section 6.4 in eleventh edition).

For the bar to be in equilibrium and not accelerating, the forces in all directions must be equal. For this example, there are only forces in one direction which I will call the x-direction with negative to the left and positive to the right (adherence to standard protocol). What are the forces in the negative x-direction? (12+9+9)kN = 30 kN. What about in the positive x-direction? (4+4+22)kN = 30 kN. Summing the forces in the x-direction:
$$\Sigma$$=-30 kN + 30 kN = 0 kN.
So, the forces are equal in all directions which means that there is no acceleration.

Now, look at each load point. -12 kN at point A. -18 kN at point B. +8 kN at point C. +22 kN at point D. Hibbeler shows you the loads and resultant forces on each section in figure 1-16 b. For example, the -12 kN load at point A must be counteracted by +12 kN on the other side of point A (Newton's third law of motion). If we keep moving along the bar to the right (positive x-direction) we encounter two -9 kN forces at point B. So what happens to the right of point B? Well, there must be a force counteracting the forces at A and B for a total of 30 kN (again, Newton's third law). If we keeping moving to the right nothing happens until we get to the next load at point C.

So, what's going on between point B and C? Well, we know that at point B there are a total of 30 kN acting in the negative x-direction. We also have to assume that there are 30 kN acting in the positive x-direction to counteract the load and keep the bar from accelerating (equilibrium).

Wrapping this all up, Hibbeler assumes you realize that to the left of point C you'll see a 30 kN force acting in the negative x-direction. You weren't given a drawing of that because it's assumed to be obvious from your course in Statics. If he had drawn it, it would look a lot like you'd expect it to and something like the attached image.

#### Attachments

• Example1_6.png
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If any one knows my question ans. please ans. it... Q: If a coin having uniform mass is rolled around the ground it will fall after few seconds....My question is how many flips will it take before falling?need a general equation for it... If any one knows ans. please post

If any one knows my question ans. please ans. it... Q: If a coin having uniform mass is rolled around the ground it will fall after few seconds....My question is how many flips will it take before falling?need a general equation for it... If any one knows ans. please post

Asking an unrelated question in someone else's post is just silly. As is expecting a rolling coin already in contact with the ground to "flip".

If any one knows my question ans. please ans. it... Q: If a coin having uniform mass is rolled around the ground it will fall after few seconds....My question is how many flips will it take before falling?need a general equation for it... If any one knows ans. please post

Hello porav05 and welcome to Physics Forums.

Generally, new questions are asked by creating a new thread/topic. To ask a new question, go to the appropriate subject (for Mechanical Engineering questions use https://www.physicsforums.com/forumdisplay.php?f=101") and click on the "New Topic" button located just to above and to the left of the list of threads.

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Thank you for the detailed explanation. I figured out my mistake I wasn't putting together that the 30kN force to the left would have the two 4kN forces to the right subtracted from it giving you the 22kN I don't know how I missed that. Thanks so much for the help