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I'm supposed to be finding the reaction forces first but from the FBD of the entire truss where am I supposed to start. If I take the moment at A then I get 3 unknowns at Bx Dy and Dx. If I get the sum of forces in x direction equals to 0 then I get 3 unknowns again. Same thing with the Y direction, I get two unknowns.

  • $\begingroup$ On your digital sketch E has a different connection to the rest. Is it also pinned? $\endgroup$
    – SlydeRule
    Oct 10, 2015 at 8:23
  • 2
    $\begingroup$ From the way it is drawn I would suspect that the connection at E is a roller. In which case the reactions at D can be obtained by considering a FBD of the 7.5m element connected to D. $\endgroup$
    – atom44
    Oct 10, 2015 at 9:43
  • $\begingroup$ it says The contact at point E between members AFCE and DE is smooth $\endgroup$
    – Bsoo1996
    Oct 10, 2015 at 14:02
  • 1
    $\begingroup$ @Bsoo1996 which means it is treated as a roller. Try drawing a FBD of just the 7.5m element connected to D. You could even imagine that if you rotated that element horizontal you could treat it as a beam with a pin at D, a roller at E and a cantilever load at the end. $\endgroup$
    – atom44
    Oct 10, 2015 at 18:52
  • 1
    $\begingroup$ "The contact at point E between members AFCE and DE is smooth" - so there is no friction force acting on DE, and therefore you know the direction of the force at E. You can find the magnitude of the force at E by considering the rod DE on its own. $\endgroup$
    – alephzero
    Oct 10, 2015 at 21:16

1 Answer 1


You seem to be treating all the connections between members as fully fixed but rather they are all hinges with the exception of joint E. The contact at E can be treated like a roller, thus there's a force normal to member DE as shown below which I have called $F_E$:

Member DE

Consider the sum of moments about D:

$\sum M_D = 0: F_E(5) = 650(7.5)$

$\therefore F_E = 975 \text{ N}$

Consider the sum of forces in the direction normal to the member:

$\sum F_n = 0: F_E = 650 + F_D$

$F_D = F_E - 650$

$F_D = 975 - 650 $

$\therefore F_D = 325 \text{ N}$

Now we can split $F_E$ into its x and y components:

$F_{E,x} = F_E\text{cos}(30) = 975\text{cos}(30) = 844.4 \text{ N}$

$F_{E,y} = F_E\text{sin}(30) = 975\text{sin}(30) = 487.5 \text{ N}$

Let's look at the rest of the system including the resultant force of $F_E$ at E - remember to switch the directions of $F_E$ as this is how it acts on member AE:

The rest of the system

I won't do the rest but you should be able to do some force balances in the x and y-axes and then a moment balance about A to get the reactions: $A_x, A_y \text{ & } B_x $

You could use SkyCiv Structural 3D to check your solution for this as shown below but you need to be careful so that you model the connections properly:

SkyCiv 3D Structural Analysis Software


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