Problem doing force analysis using simultaneous equations

1. Aug 16, 2011

liamh05

hi guys

ive got a problem whils solving a force analysis on a crane truss, when i get to a joint where simultaneous equations are needed. have a look at what ive got and itll make more sense. when i equate the two equations they dont balance.

from my previous working i get F2=9810, F3=15473.81, F5=5663.81

i get

RESOLVING FORCES IN X-DIRECTION=0 --->= +IVE

-F2 + F5COS30 + F7COS15 + F8COS60 = 0
9810 = 5663.81COS30 + F7COS15 + F8COS60
F7COS15 = (4905 - F8COS60)/COS15

RESOLVING FORCES UPWARDS=0 UPWARDS= +IVE

F3 + F5COS60 - F7SIN15 - F8SIN60 = 0
15473.81 + 5663.81COS60 - F7SIN15 - F8SIN60 = 0
18305.72 - F7SIN15 - F8SIN60 = 0
F7SIN15 = -F8SIN60 + 18305.72
F7 = (-F8SIN60 + 18305.72)/SIN15

the problem is if i substitute say 10 in for F8 the equations dont balance.... i must have done something wrong musn't i ?

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2. Aug 16, 2011

Spinnor

You have two equations in two unknowns. On the last line you have an expression for F7, substitute that in the second to last line and solve for F8.

It seems set up right, just complete it. Did you get values for F7 and F8?

3. Aug 17, 2011

liamh05

the problem is that the when i equate the two equations to F7 they dont balance ie the value of F7 comes out different for each one. so when i work out F8, which i got as F8= 6219.3 in compression, but then depending on which equation i put the value of F8 into, i could get either, 1858.68 in tension or 49917.69 in tension

4. Aug 17, 2011

Spinnor

When you resolved forces in the upwards direction you have F5cos30 and it should be F5sin30, missed that first time around.

Good luck.

5. Aug 17, 2011

vela

Staff Emeritus
Of course it doesn't work. Unless F8=10 happens to be the solution, the two equations won't be satisfied simultaneously.

\begin{align*}
F_7 \cos 15^\circ + F_8 \cos 60^\circ &= F_2 - F_5\cos 30^\circ \\
F_7 \sin 15^\circ + F_8 \sin 60^\circ &= F_3 + F_5\sin 30^\circ
\end{align*}
after you fix the error Spinnor found. It's two linear equations and two unknowns, the kind of thing you've been solving in algebra class for years.

How did you solve for F8 above?

6. Aug 18, 2011

liamh05

spinnor- there isnt F5cos30 when resolving upwards, im not sure if ive misunderstood, ive looked back over the FBD and i think my method is correct. could you explain what you meant mate?

vela- thanks for your reply first of all, i think that you have misunderstood what i meant. i rearranged your equations for F7 and made them equal to each other, then no matter what value of F8 the left side should equal the right side? right?

7. Aug 18, 2011

Spinnor

Right, you have F5cos60 which is the same as F5sin30.

No, there should be only one value of F8 that solves the problem. Re-draw your diagram with F7, F8 and the resultant of F2 + F3 + F5 = FR say. There is only one solution such that F7 + F8 = - FR.

I think you must be making a math error?

8. Aug 18, 2011

liamh05

i think ive got it cracked, worked out F7 as 7752.48 and F8 as -15608.7

thanks for your help guys, its much appreciated! :D

9. Aug 18, 2011

vela

Staff Emeritus
Those don't match the answers I got. With your results, you get
\begin{align*}
F_7 \cos 15^\circ + F_8 \cos 60^\circ &= -316 \\
F_7 \sin 15^\circ + F_8 \sin 60^\circ &= 11511
\end{align*}
However, the other three forces sum to
\begin{align*}
F_x = F_2 - F_5\cos 30^\circ &= 4905 \\
F_y = F_3 + F_5\sin 30^\circ &= 18306
\end{align*}
Clearly, the five forces are not going to sum to 0.