anonymous
  • anonymous
Help guys , truss :)))
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@mathmate
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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anonymous
  • anonymous
i dont know how to solve :(( @mathmate
rajat97
  • 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
  • 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
  • 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
  • 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
  • anonymous
we have to find the angle, with is beta
anonymous
  • anonymous
@mathmate
anonymous
  • anonymous
@rajat97
mathmate
  • mathmate
|dw:1439640442912:dw| Is angle \(\beta\) given as equal to angle DBC, or is it given a different value?
mathmate
  • mathmate
Correction: Is angle β given as equal to angle DBE, or is it given a different value?
anonymous
  • anonymous
diffrent value
anonymous
  • anonymous
i got the β = 30.96
mathmate
  • 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
  • anonymous
vertical
anonymous
  • anonymous
anonymous
  • anonymous
@mathmate
anonymous
  • anonymous
@nincompoop
anonymous
  • anonymous
@goformit100
mathmate
  • 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
  • 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
  • anonymous
my fx and fy is diffrent from you
anonymous
  • anonymous
anonymous
  • anonymous
@abb0t
rajat97
  • rajat97
i'm sorry for the wrong statements
anonymous
  • anonymous
i got the all internal answers :)) thank for your help :))
mathmate
  • 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
  • mathmate
If you wish, I will check your internal forces and get back to you. Please let me know.
mathmate
  • 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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