- #1

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## Homework Statement

refer to the circled part , i dont understand the author's working .

## Homework Equations

## The Attempt at a Solution

I think it should be 450x - 150(4-x) -30092)(x-1) , correct me if i'm wrong . (taking moment about x ) ....

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- Thread starter chetzread
- Start date

- #1

- 801

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refer to the circled part , i dont understand the author's working .

I think it should be 450x - 150(4-x) -30092)(x-1) , correct me if i'm wrong . (taking moment about x ) ....

- #2

- 801

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is the author wrong , coz the distributed load start from R1 to 2m from R1 ...

- #3

SteamKing

Staff Emeritus

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Homework Helper

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No, the author is not wrong.is the author wrong , coz the distributed load start from R1 to 2m from R1 ...

The bending moment function EIy" is carefully constructed to show that the distributed load starts at x = 0 and runs to x = 2, after which an equal and opposite distributed load is applied to the beam at x > 2, so that the

- #4

- 801

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why ? i still dont understand , i can only understand the 450x , why should be -0.5(300)(x^2) +0.5(300)[(x-2)^2 ]No, the author is not wrong.

The bending moment function EIy" is carefully constructed to show that the distributed load starts at x = 0 and runs to x = 2, after which an equal and opposite distributed load is applied to the beam at x > 2, so that thenet distributed load is zero.

- #5

- 801

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why shouldnt it be 450x - 150(4-x) -300(0.5)(2)(x-1) ??No, the author is not wrong.

The bending moment function EIy" is carefully constructed to show that the distributed load starts at x = 0 and runs to x = 2, after which an equal and opposite distributed load is applied to the beam at x > 2, so that thenet distributed load is zero.

- #6

SteamKing

Staff Emeritus

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I'm not going to speculate on that.why shouldnt it be 450x - 150(4-x) -300(0.5)(2)(x-1) ??

Look, the author has a method of analysis of this beam which he is trying to teach. If you want to develop your own methods, write your own textbook.

- #7

- 801

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i dont understand why it is -0.5(300)(x^2) +0.5(300)[(x-2)^2 ] , can you explain ?I'm not going to speculate on that.

Look, the author has a method of analysis of this beam which he is trying to teach. If you want to develop your own methods, write your own textbook.

- #8

- 801

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why it's 0.5(300)(x^2) ? the 300N/m only 'defined ' from R1 to 2m from R1, am i right ? why shouldnt it be 0.5(300)(2)(x-2) instead? as you said , after 2m ,No, the author is not wrong.

The bending moment function EIy" is carefully constructed to show that the distributed load starts at x = 0 and runs to x = 2, after which an equal and opposite distributed load is applied to the beam at x > 2, so that thenet distributed load is zero.

- #9

- 801

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the location of 300n/m is at R1 and extend to 2m from R1, so ,the EIy" should be = 450x-300(2)(x-1) ???I'm not going to speculate on that.

Look, the author has a method of analysis of this beam which he is trying to teach. If you want to develop your own methods, write your own textbook.

why there's -0.5(300)(x^2) +0.5(300)( (x-2)^2 )????

I'm confused

Or do you mean the 300(X) force still exist at X >2?

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- #10

- 801

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