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anonymous

  • one year ago

Help guys , truss :)))

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @mathmate

  3. mathmate
    • one year ago
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    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?

  4. anonymous
    • one year ago
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    i dont know how to solve :(( @mathmate

  5. rajat97
    • one year ago
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    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!

  6. rajat97
    • one year ago
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    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

  7. mathmate
    • one year ago
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    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.

  8. mathmate
    • one year ago
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    @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.

  9. anonymous
    • one year ago
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    we have to find the angle, with is beta

  10. anonymous
    • one year ago
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    @mathmate

  11. anonymous
    • one year ago
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    @rajat97

  12. mathmate
    • one year ago
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    |dw:1439640442912:dw| Is angle \(\beta\) given as equal to angle DBC, or is it given a different value?

  13. mathmate
    • one year ago
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    Correction: Is angle β given as equal to angle DBE, or is it given a different value?

  14. anonymous
    • one year ago
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    diffrent value

  15. anonymous
    • one year ago
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    i got the β = 30.96

  16. mathmate
    • one year ago
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    @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.

  17. anonymous
    • one year ago
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    vertical

  18. anonymous
    • one year ago
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  19. anonymous
    • one year ago
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    @mathmate

  20. anonymous
    • one year ago
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    @nincompoop

  21. anonymous
    • one year ago
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    @goformit100

  22. mathmate
    • one year ago
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    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.

  23. mathmate
    • one year ago
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    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

  24. anonymous
    • one year ago
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    my fx and fy is diffrent from you

  25. anonymous
    • one year ago
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  26. anonymous
    • one year ago
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    @abb0t

  27. rajat97
    • one year ago
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    i'm sorry for the wrong statements

  28. anonymous
    • one year ago
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    i got the all internal answers :)) thank for your help :))

  29. mathmate
    • one year ago
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    "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.

  30. mathmate
    • one year ago
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    If you wish, I will check your internal forces and get back to you. Please let me know.

  31. mathmate
    • one year ago
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    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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