Help guys , truss :)))

- anonymous

Help guys , truss :)))

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- anonymous

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- anonymous

@mathmate

- mathmate

Do you have the angle of the 10KN?
As a start, can you work out the reactions by taking moments of external forces about each support?

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## More answers

- anonymous

i dont know how to solve :(( @mathmate

- rajat97

what actually do we have to find???
i mean that i havent come accross these type of questions
would feel great to learn something new!

- rajat97

and the angle for the 10KN can be found easily as the triangles drawn seem to be right angled triangles and they are made of the pythagorial triplet 3-4-5 and the angles are 37deg, 53deg and 90deg

- mathmate

1. nothing on the diagram says that the 10 kN is lined up with the diagonal member,
2. and the triangles are not 3-4-5, they are 3, 5, sqrt(34).
The objective of the problem is to solve for the forces of each and every member, with
which member sizes can be designed (i.e. the shape of the member, as shown in the handbooks) using the lengths and stresses.

- mathmate

@jacalneaila
Do you have the angle of the 10KN?
It "appears" to line up with the diagonal, but in science and especially engineering, we need explicit information or else we may be wasting time solving the wrong problem, or even worse, giving the wrong answers.

- anonymous

we have to find the angle, with is beta

- anonymous

@mathmate

- anonymous

@rajat97

- mathmate

|dw:1439640442912:dw|
Is angle \(\beta\) given as equal to angle DBC,
or is it given a different value?

- mathmate

Correction:
Is angle β given as equal to angle DBE,
or is it given a different value?

- anonymous

diffrent value

- anonymous

i got the β = 30.96

- mathmate

@jacalneaila
1. is 30.96 with the horizontal or with the vertical?
2. What you are basically saying is that the force is collinear with member BE, is that correct? If the force is collinear with BE, then we don't need the angle, we just need the sine and cosine, which we have.

- anonymous

vertical

- anonymous

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- anonymous

@mathmate

- anonymous

@nincompoop

- anonymous

@goformit100

- mathmate

You said \(\beta\)=30.96 is the angle with the vertical (i.e. collinear with BE).
But in your calculation sheet, you had the 10kN force at 30.96 with the horizontal.
Please confirm if the angle \(\beta\)=30.96 (=atan(3/5)) is with the horizontal or the vertical.
If you would like help, you need to define the problem clearly.

- mathmate

IF \(\beta\) is atan(3/5) with respect to the horizontal, then your calculations are all correct except for the last decimal place.
1. Calculation of reactions (and check)
My values are:
Ly=17.1283
Ay=13.0167
Ax=13.5749
Bx=8.5749 (to the right)
By=5.1450 (down)
\(\sum F_x \)=Ax-(5+Bx)=13.5749-13.5749=0
\(\sum F_y \)=Ay+Ly-(5*5+By+0)=31.1450-31.1450=0

- anonymous

my fx and fy is diffrent from you

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- anonymous

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- anonymous

@abb0t

- rajat97

i'm sorry for the wrong statements

- anonymous

i got the all internal answers :)) thank for your help :))

- mathmate

"my fx and fy is diffrent from you "
Sum of Fx and Fy are both zero, I don't see hw they can be different.

- mathmate

If you wish, I will check your internal forces and get back to you. Please let me know.

- mathmate

Congrats, our results all match, except for the 4th digits after the decimal and up.
beta = angle ABC…. = angle of 10 kN load with horizontal
sb=sin(beta)=3/sqrt(34)= 0.514495755
cb=cos(beta)=5/sqrt(34)= 0.857492926
External forces
Ax=13.5749
Ay=13.0167
Lx=0
Ly=17.1283
Bx=8.57493
By=5.14496
Internal forces
AB=15.1799 (C)
BC=5 (T)
AC=CE=21.3849 (T)
BE= 3.34894 (T)
BD= 18.1079 (C)
DE= 2.1283 (T)
EG= 23.1079 (T)
DG= 2.4820 (C)
DF= 16.8310 (C)
FG= 7.1283 (T)
GI= 21.8310 (T)
FI= 8.3130 (C)
FH= 12.5540 (C)
HI= 12.1283 (T)
IK= 17.5540 (T)
HK= 14.1439 (C)
HJ= 5.2770 (C)
JK= 17.1283 (T)
KL= 10.2770 (T)
JL= 19.9749 (C)

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