Fixed-end moments for Trapezoidal load on part of span?

[TarikHKJ]

Hi,

Trying to solve an indeterminate beam using moment distribution displacement method, I used the general fixed-end moment equations, but one span has a trapezoidal load on part of the span which does not have a fixed-end moment equation,

here is the problem:

https://goo.gl/photos/dnY2sz6vZ4BxESA8A

How to calculate the fixed-end moments for that span?


Many Thanks!

 

[BAretired]

Knowing the fixed end moments of a concentrated load P, namely Pab²/L² and Pa²b/L² respectively, the trapezoidal loading may be represented by a few point loads. The fixed end moments for the trapezoidal loading will be approximately equal to the sum of the fixed end moments for the point loads.

If more precision is required, M₁ = ∫_a^a+b wx(L-x)²dx/L² where:
a = start of trapezoidal loading
b = length of load
w = load per unit length
For trapezoidal loading, w = w₁ + (w₂-w₁)(x-a)/b

Similarly,
M₂ = ∫_a^a+b wx²(L-x)dx/L²

BA

 

[TarikHKJ]

Thanks BA,

just to make sure, if I want to calculate M1 with the integration, are the ( a , b , w1 , w2 ) in this image correct

or do I need to switch w1 and w2?

https://goo.gl/photos/BJhcfstsicLBqoUWA

I entered the values of a=1.5 b=4 w1=50 w2=70 L=7 like this :

http://s19.postimg.org/3o3cljuv7/screenshot_46.jpg

and it gave incorrect answer. what am I doing wrong in the integration?




Kindest regards!

 

[BAretired]

TarikHKJ,
Your diagram is the reverse of mine (I move from left to right), but it should produce the correct answer as your numbering system is consistent with a right to left system. Your labeling appears correct; do not switch w1 and w2.

Your limits of integration are (0, L) which is incorrect. The limits of integration are (a, a+b) or (1.5, 5.5). You should be integrating only the loaded portion of span, not the whole span because integration is a summation of the effects of the load.

I cannot follow your expression to be integrated. There should be some terms with higher powers of x.

For example, in the expression:
M1 = ∫_a^a+b wx(L-x)²dx/L²
there are terms with x, x² and x³ even when w is constant.

When w varies as a function of x, the expression for M1 contains a term with x⁴ as well as the lower powers of x.

In the case of trapezoidal loading,
w = w1 + (w2-w1)(x-a)/b
which can be written w = w1 - (w2-w1)a/b + (w2-w1)x/b
In the present case, this would be w = 42.5 + 5x
and when you do the multiplication, you should arrive at the correct expression to be integrated.

Edit: shown in red.

BA

 

[TarikHKJ]

Thanks BA!

is the expression correct now?

http://s19.postimg.org/424ol5eyr/screenshot_47.jpg

 

[TarikHKJ]

After edit

http://s19.postimg.org/ltgazludf/screenshot_48.jpg

 

[BAretired]

Getting warmer!
It should be:
M1 = ∫_1.5^5.5 (42.5+5x)x(7-x)²dx/49

BA

 

[TarikHKJ]

http://s19.postimg.org/whk1yg4cj/screenshot_49.jpg

 

[TarikHKJ]

Solved for limits 1.5, 5.5 and the answer was 1690.6 which is far off the true value of M1 = 181.8 kN.m

 

[TarikHKJ]

@BA here is the result

http://s19.postimg.org/t0hzvh5ab/screenshot_50.jpg

 

[BAretired]

Why do you persist in including the expression 1.5^5.5? That term makes no sense at all. The fact that you are including it suggests that you are not following the reasoning at all. The expression ∫_1.5^5.5 simply means you are evaluating the integral between those limits.

If you divide your result (1690.64938) by (1.5)^5.5, the corrected result is 181.78.

BA

 

[TarikHKJ]

Sorry about that.


BA you were very helpful and I hope someone else benefits from this..



Thanks!