Complex scissors truss question

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Homework Help Overview

The discussion revolves around a symmetrical scissors truss supported at points A and D, with specific angles and loads applied at various joints. Participants are tasked with determining the reaction force at D and the axial forces in certain members under different loading conditions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of moment equations about different joints to find reaction forces and member forces. Some question the application of the sum of moments and the implications of the distances used in calculations.

Discussion Status

There is an ongoing exchange of ideas regarding the calculation of reaction forces and member forces. Some participants have provided guidance on how to set up equations based on the sum of moments, while others are clarifying their understanding of the problem setup and assumptions.

Contextual Notes

Participants note that the problem involves multiple variables and that assumptions about the lengths and forces must be carefully considered. There is also mention of the need for additional equations to solve for unknowns in the context of changing load conditions.

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Homework Statement


e6y3ya.png


The symmetrical scissors truss is supported at point A and D. The support at D is a rocker bearing and does not produce a horizontal reaction. A, B, C, D, E, & F are pin joints. The members of the truss are not joined to each other where they cross. At A, the angle between AB and AE is 45 degrees and the angle between AC and AE is 26.57 degrees. The magnitude of the two loads, E and F are given as 192.6 kN and 135.8 kN respectively. L is equal to 3m.

What is the magnitude of the reaction force at D?

If the load at F has changed so that the vertical reaction at A is 125.4 kN:

what is the magnitude of the axial force in members AB, CD and AE?


Homework Equations



Sum of all moments = 0
Sum of all forces = 0
Sum Fx = 0
Sum Fy = 0
M=Fd


The Attempt at a Solution



I worked out the three moment equations about B, C and E, however I have got too many variables to solve the equations.
 
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haydez98 said:

Homework Statement


e6y3ya.png


The symmetrical scissors truss is supported at point A and D. The support at D is a rocker bearing and does not produce a horizontal reaction. A, B, C, D, E, & F are pin joints. The members of the truss are not joined to each other where they cross. At A, the angle between AB and AE is 45 degrees and the angle between AC and AE is 26.57 degrees. The magnitude of the two loads, E and F are given as 192.6 kN and 135.8 kN respectively. L is equal to 3m.

What is the magnitude of the reaction force at D?

If the load at F has changed so that the vertical reaction at A is 125.4 kN:

what is the magnitude of the axial force in members AB, CD and AE?


Homework Equations



Sum of all moments = 0
Sum of all forces = 0
Sum Fx = 0
Sum Fy = 0
M=Fd


The Attempt at a Solution



I worked out the three moment equations about B, C and E, however I have got too many variables to solve the equations.
The first part of the question asks you to determine the reaction at D. Have you done this part yet by summing moments about A? For part 2, under a different loading case, you are asked to find member forces. It's usually best, after determining support reactions, to look at the easier joints first, which in this case are E and F. You can solve then for the diagonal member forces emanating from those joints, and the rest of the forces can then be calculated.
 
Would the sum of the moments about A just be:
(192.6 x 1) + (135.8 x 2)
?
And so the reaction at D is the sum of the moments at A/3?
 
haydez98 said:
Would the sum of the moments about A just be:
(192.6 x 1) + (135.8 x 2)
?
And so the reaction at D is the sum of the moments at A/3?
I don't know what you mean by A/3. You know that the sum of the moments about any point must be zero. If you choose joint A as the point about which you sum moments, then the reaction at A will not enter into the equation, because it produces no moment about point A. So representing the vertical reaction at D as an upward force D_y, your equation becomes
\Sigma M_A = 0
192.6(L/3) + 135.8(2L/3) - D_y(?) = 0 (you fill in what to use for (?)).
Edit: I re-read your post and it looks like we are saying the same thing, D_y = (sum of moments about A from the applied loads at E and F)/3
 
Last edited:
ah cool, thank you very much, that explains lots

what I meant by A/3 was that:
because sum of moments at A = 0, and L =3

Dy x 3 = (192.6 x 1) + (135.8 x 2)
therefore:

Dy = [(192.6 x 1) + (135.8 x 2)] /3With the vertical reaction at A has different for Part 2, do I sum the y-components to find the load at F?

0 = Dy + Ay - E - F ?
 
Last edited:
haydez98 said:
ah cool, thank you very much, that explains lots

what I meant by A/3 was that:
because sum of moments at A = 0, and L =3

Dy x 3 = (192.6 x 1) + (135.8 x 2)
therefore:

Dy = [(192.6 x 1) + (135.8 x 2)] /3
yes this is OK, you've assumed L=3, but actually, L can be any number, and since it cancels out of the equation, it does not affect the analysis in this problem.
With the vertical reaction at A has different for Part 2, do I sum the y-components to find the load at F?

0 = Dy + Ay - E - F ?
This equation is correct, but since both F and Dy are unknown, you need another equation before you can solve for F (Dy will not be the same as before). This 2nd equation comes from summing moments about D. Then you can get F right away, and then find Dy from your first equation.
 
ok thanks ^_^
 
Hey I'm new to this forum and am doing the exact same queastion but with different values, how would you find the members, using method of sections, i know you cannot cut through more than 3 members or cut through more than 2 braching from the same joint. How would I approach the rest of the question
 
hey guys this helped heaps..

LAgreek - here's how to find member CD

CD = E / sqrt(4.5)
 

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