Problem Statement:
Determine the Vertical Deflection of joint 1 and the force in member 2 of the truss in problem 14-1.
Reference: Problem 14-2, chapter – 14, Truss Analysis using the Stiffness Method, Structural Analysis by R C Hibbeler.
Solution:
Structure Stiffness matrix (as derived in solution of problem 14.1)
Check the complete Book Solution Manual
for Structural Analysis by R C Hibbeler (Eighth Edition)
K = AE*
[[ 0.3659 0.0055 -0.0707 0.0707 -0.2 0. -0.0952 -0.0762]
[ 0.0055 0.1317 0.0707 -0.0707 0. 0. -0.0762 -0.0609]
[-0.0707 0.0707 0.0707 -0.0707 0. 0. 0. 0. ]
[ 0.0707 -0.0707 -0.0707 0.0707 0. 0. 0. 0. ]
[-0.2 0. 0. 0. 0.2 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[-0.0952 -0.0762 0. 0. 0. 0. 0.0952 0.0762]
[-0.0762 -0.0609 0. 0. 0. 0. 0.0762 0.0609]]
Now, let’s write out K11 and K21 separately.
K11 = AE*
[[0.3659 0.0054]
[0.0054 0.1316]]
K12 = AE*
[[-0.0707 0.0707]
[ 0.0707. -0.07071]
[-0.2 0.0 ]
[ 0.0 0.0 ]
[-0.095 -0.0761]
[-0.076 -0.0609]]
Inverse of K11, i.e. K11^(-1) =
[[ 2.734 -0.114]
[-0.114 7.6 ]]
As per the compatibility, Qk = K11*Du hence,
[[ 0.]
[-40.]] = AE[[0.366 0.005]
[0.005 0.132]]* Du
Or Du = K11^(-1) * Qk / AE
or Du =
[[ 2.734 -0.114]
[-0.114 7.6 ]]*[[ 0.]
[-40.]] / AE
or Solving the matrix algebraically,
D1 = 4.546/AE
or D1 = 4.54 *10-5 m
D2 = -304.01/AE
or D2 = -0.003 m, -ve sign shows downward deflection.
Similarly, Qu = K21Du
or Qu = AE[[-0.07071068 0.07071068] [ 0.07071068 -0.07071068] [-0.2 0. ] [ 0. 0. ] [-0.0952279 -0.07618232] [-0.07618232 -0.06094586]]*
1/AE*[[ 4.545]
[-304.009]]
or Qu =
[[-21.818]
[ 21.818]
[ -0.909]
[ 0. ]
[ 22.727]
[ 18.182]]
Hence, Q3 = -21.818 kN,
Q4 = 21.818 kN,
Q5 = -0.909 kN,
Q6 = 0.0 kN,
Q7 = 22.727 kN,
Q8 = 18.182 kN,
for member 1 :
k’T = AE/7.071 * [ 0.7071 -0.7071 -0.7071 0.7071],
and D =
[[ 4.546]
[-304.01 ]
[ 0. ]
[ 0. ]] m
Hence, q1 = k’T*D = [30.856] kN
for member 2 :
k’T = AE/5.0 * [ 1.0 -0.0 -1.0 0.0],
and D =
[[ 4.546]
[-304.01 ]
[ 0. ]
[ 0. ]] m
Hence, q2 = k’T*D or
q2 = 0.909 kN (+ve sign shows tensile force.)
Please note that q2 = Q5, which was needed as per the compatibility of the given truss joint 3, which shows that the solution is correct.
Check the complete Book Solution Manual
for Structural Analysis by R C Hibbeler (Eighth Edition)
for member 3 :
k’T = AE/6.403 * [ 0.78086881 0.62469505 -0.78086881 -0.62469505],
and D =
[[ 4.546]
[-304.01 ]
[ 0. ]
[ 0. ]] m
Hence, q3 = k’T*D = -29.105 kN (-ve sign shows compressive force.)
Please note that the final answer given in the book seems to be wrong.
