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STATICS - EXAMPLE

    Example


Complex Truss with Loading
 

Determine the force in members BE, BJ, DJ, and IJ. (ADJ and HDB are not continues, each consist of two elements.)

   
    Solution


Reaction Forces at End Supports
 

The first step is to find the support reaction forces. Vertical reaction force at F can be found by taking the moment about L.

     ΣML= 0
            = - 6Fy + 5Gy + 4Hy + 3Iy + 2Jy
            = - 6Fy + 5(2) + 4(3) + 3(5) + 2(2)
            = - 6Fy + 41 = 0

     Fy = 6.833 kN

Summation of forces in the y direction should be zero.

     ΣF= 0
     Fy + Ly - (Gy + Hy + Iy + Jy) = 0
     6.833 + Ly - (2 + 3 + 5 + 2) = 0
     Ly = 5.167 kN

Another method to find Ly is to take the moment about F. It is a good way to check to see if the previous value for Ly was found correctly.

     ΣMF = 0
             = 6Ly - ( Gy + 2Hy + 3Iy + 4Jy )
             = 6Ly - ( 2 + 2(3) + 3(5) + 4(2) )
             = 6Ly - 31 = 0

     Ly = 5.167 kN

     

Method of Section (Cut 1)
to Determine FBE
 

To find the value of the forces it is easier to use method of sections. Since the problem asks for four member forces, BE, BJ, DJ, and IJ, the problem needs to be solved taking two cuts. The first cut , shown at the left, will be used to solve for member force in member BE. Then another cut will be used to find the other three member forces.

Moments about J can be taken to find the value of the FBE (notation EJ represents the length of the member EJ).

      ΣMJ = 0
              = 2L + FBE(EJ)

JBE is a right triangle and BE is perpendicular to EJ. The length of KE and JK is 1. Based on Pythagorean theorem, EJ is the square root of 2. (EJ = 1.414).

     2L + FBE(1.414) = 0
     2(5.167) + FBE(1.414) = 0
     FEB = -7.308 kN

     


Method of Section (Cut 2)
to Determine FJB, FJD, and FIJ

 

Now that FBE is known, another cut can be made to find FJB, FJD, and FIJ. To find FJD and FJB, moment about L can be used with the sum of the vertical forces (notation JL mean length of JL)

      ΣML = 0
               = JL (2kN) - FJD(BD) - FJB(JL)
               = 2(2) - FJD(1.414) - FJB(2)

     ΣFy = 0
            = FJD + FJB + 5.167 - 2
                       -7.308(sin45)

FJD and FJB can be found by solving the above system of equations to give

      FJD = -0.8327(10-3) kN    (very small)

      FJB = 2.000 kN

FJD is found to be 0.8 newton which is negligible and is because of the rounding errors.

Summation of forces in x direction can be written in order to find the value of FIJ

      ΣFx = 0
             = FEB(cos45) + FJD(cos45) + FIJ
             = -7.308 (0.707) + 0 (0.707) + FIJ

      FIJ = 5.167 kN

FIJ can also be found by taking the moment about joint B

     ΣMB = 0
               = L(JL) - FJD(BD) - FIJ(BJ)
               = 5.167(2) + 0 (1.414) - FIJ(2)

     FIJ = 5.167 kN

     
   
 
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