Moment equation for a beam with multiple loads

[chetzread]

Homework Statement

refer to the circled part , i don't 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 ) ...

 

[chetzread]

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

 

[SteamKing]

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

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 the net distributed load is zero.

 

[chetzread]

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 the net distributed load is zero.

why ? i still don't understand , i can only understand the 450x , why should be -0.5(300)(x^2) +0.5(300)[(x-2)^2 ]

 

[chetzread]

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 the net distributed load is zero.

why shouldn't it be 450x - 150(4-x) -300(0.5)(2)(x-1) ??

 

[SteamKing]

why shouldn't it be 450x - 150(4-x) -300(0.5)(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.

 

[chetzread]

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.

i don't understand why it is -0.5(300)(x^2) +0.5(300)[(x-2)^2 ] , can you explain ?

 

[chetzread]

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 the net distributed load is zero.

why it's 0.5(300)(x^2) ? the 300N/m only 'defined ' from R1 to 2m from R1, am i right ? why shouldn't it be 0.5(300)(2)(x-2) instead? as you said , after 2m , net distributed load is zero

 

[chetzread]

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.

the location of 300n/m is at R1 and extend to 2m from R1, so ,the EIy" should be = 450x-300(2)(x-1) ?
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?

 

[chetzread]

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