URGENT: Truss Question (1 Viewer)

[Husayn]

B is ~ 81 kN. Cannot work it out.

EDIT: DAMN TRUSS IS NOW RIGHT

 

[gordo]

if the diagram is correct,
i.e the 50Kn foce causes no moment at B (sliding vecotr on a line through B)
and u have just a normal external force at B and normal and tangent external force at A,
then i got the force at B in the x-direction as ~125 kN

(just using a free body diagram, no trusses needed, let the forces in the x-dir = 0 and the moment at B = 0, solve for Ax and then Bx)

 

[Husayn]

The answer at the back of the book says B is ~ 81.

EDIT: I fixed up the Truss, the 50 kN was in the wrong place.

 

[Husayn]

Come oonn people, surely there must be someone out there who can do this in no time at all?

B is 81, thats all I want to know how to do.

 

[fakingtheday]

are you sure you didn't miss a distance for the 55 KN? otherwise it's getting pretty tricky.

 

[Husayn]

Argghhh, the distance for the 50 kN force is 6 m.

 

[fakingtheday]

Ok well given the diagram i got 131Kn going horizontally left. Here is my working

Moments About A
clock = anti
(50x6)+(7x75)+(14x33) = (B x (Tan30 x 17))
B = 131.3Kn Going horizontally left

To Find A
Sum the forces in vertical direction = 0
A(v) = 55 x sin30 + 75 33
A(v) = 135.5

A(h) = 131 - 55 x cos 30
A(h) = 83.4

Therefore resultant at A=
square root of (135.5(2) + 83.4(2))
=159.1 Kn

Angle =
Tan (*) = 135.5/83.4
(*) = 58 degrees

Maybe you've copied it out wrong again? check those botom distances

Unless i've missed some
thing that's the answer. Are you sure you've copied all the distances right?

 

[Husayn]

You didn't include the 50kN force as being inclined - it's at 30 degrees incase you didnt see.

 

[fakingtheday]

I included it, the 6m is the perpendicular distance, so that's the distance you use.

 

[Husayn]

How can it be perpendicular when it's clearly on an angle? The Force that is.

 

[Husayn]

Oh wait I see you included it when working A. OK.

But why not with moments around A?

 

[fakingtheday]

As a general rule, take moments about the fixed support, especially in this situation, there is no need to take moments about A because you sum the forces in vertical and horizontal direction, they should all be equal.

And that 6m is it's perdicular distance from A, with moments it's always force times the perpendicular distance.

 

[TheBearThtWasnt]

bear not understand this

he not like

 

[Husayn]

fakingtheday - You must be doing something wrong considering your answer was way off.

Have you overlooked it again?

 

[fakingtheday]

Please recheck your diagram. My working is right. That 6m is the perpendicular. Those bottom distances must be wrong, 17m seems awefully long for a truss.

 

[Husayn]

The measurements are correct, I assure you. That's what's in the book.

Now, whether the answer in the back are wrong is probable.

What are others' opinions on the answer?

 

[fakingtheday]

The measurements are correct, I assure you. That's what's in the book.

Now, whether the answer in the back are wrong is probable.

What are others' opinions on the answer?

What book is it? looking over it again, it's a moments problem. Clock = Anti

 

[Husayn]

Paul Copeland's 'Engineering Studies Volume 2'.

 

[fakingtheday]

i found it in the book, but i gotta work. my guess is he's just made a printing error, because these things are nearly always to scale.

My conclusion: Copeland sucks, get Rochfore

 

[Husayn]

Well when you get back, I'd like to know why that 50 kN force doesn't have components when taken about A, because I haven't figured out the reason yet.