Problem Statement:
Determine the vertical deflection at joint 2 and the force in member 4 of the truss in prob. 14-3. Take A = 0.0015 m2 and E = 200 GPa for each member.
Solution:
The Structural Stiffness matrix was derived in the solution of problem 14-3, please click here to see.
Structure Stiffness matrix, K = AE*
[[ 0.4 0. 0. 0. -0.2 0. -0.2 0. ]
[ 0. 0.25 0. -0.25 0. 0. 0. 0. ]
[ 0. 0. 0.1905 0. -0.0952 -0.0762 -0.0952 0.0762]
[ 0. -0.25 0. 0.3719 -0.0762 -0.0609 0.0762 -0.0609]
[-0.2 0. -0.0952 -0.0762 0.2952 0.0762 0. 0. ]
[ 0. 0. -0.0762 -0.0609 0.0762 0.0609 0. 0. ]
[-0.2 0. -0.0952 0.0762 0. 0. 0.2952 -0.0762]
[ 0. 0. 0.0762 -0.0609 0. 0. -0.0762 0.0609]]
Check the complete Book Solution Manual
for  Structural Analysis by R C Hibbeler (Eighth Edition)
K11 = AE*
[[ 0.4 0. 0. 0. -0.2 ]
[ 0. 0.25 0. -0.25 0. ]
[ 0. 0. 0.19045581 0. -0.0952279 ]
[ 0. -0.25 0. 0.37189172 -0.07618232]
[-0.2 0. -0.0952279 -0.07618232 0.2952279 ]]
K12 = AE*
[[ 0. 0. -0.07618232 -0.06094586 0.07618232]
[-0.2 0. -0.0952279 0.07618232 0. ]
[ 0. 0. 0.07618232 -0.06094586 0. ]]
Inverse of K11, i.e. K11^(-1) = [[ 5. 3.125 2.5 3.125 5. ]
[ 3.125 16.11 3.125 12.11 6.25 ]
[ 2.5 3.125 7.751 3.125 5. ]
[ 3.125 12.11 3.125 12.11 6.25 ]
[ 5. 6.25 5. 6.25 10. ]]
As per the compatibility, Qk = K11Du hence, [[ 0.] [-15.] [ 0.] [ 0.] [ 0.]] = AE[[ 0.4 0. 0. 0. -0.2 ]
[ 0. 0.25 0. -0.25 0. ]
[ 0. 0. 0.19 0. -0.095]
[ 0. -0.25 0. 0.372 -0.076]
[-0.2 0. -0.095 -0.076 0.295]]* Du
Or Du = K11^(-1) * Qk / AE
or Du =
[[ 5. 3.125 2.5 3.125 5. ]
[ 3.125 16.11 3.125 12.11 6.25 ]
[ 2.5 3.125 7.751 3.125 5. ]
[ 3.125 12.11 3.125 12.11 6.25 ]
[ 5. 6.25 5. 6.25 10. ]]*[[ 0.]
[-15.]
[ 0.]
[ 0.]
[ 0.]] / AE
Solving the matrix algebraically,
D1 = -46.875/AE
or D1 = -0.00016 m
D2 = -241.654/AE
or D2 = -0.00081 m
D3 = -46.875/AE
or D3 = -0.00016 m
D4 = -181.654/AE
or D4 = -0.00061 m, vertical deflection at joint 2 is -ve, i.e. it is downwards.
D5 = -93.75/AE
or D5 = -0.00031 m
Similarly, Qu = K21*Du or Qu = AE*
[[ 0. 0. -0.07618232 -0.06 0.07618232] [-0.2 0. -0.0952279 0.07618232 0. ] [ 0. 0. 0.07618232 -0.06094586 0. ]]1/AE*[[ -46.875 ]
[-241.65379394]
[ -46.875 ]
[-181.65379394]
[ -93.75 ]]
or Qu =
[[7.5]
[0. ]
[7.5]]
Q6 = 7.5 kN,
Q7 = 0.0 kN,
Q8 = 7.5 kN,
for member 1 :
k’T = AE/5.0 * [ 1. -0. -1. 0.],
and D =
[[ -46.875]
[-241.654]
[ 0. ]
[ 0. ]] m
Hence, q1 = k’T*D = [-9.375] kN
for member 2 :
k’T = AE/5.0 * [-1. -0. 1. 0.],
and D =
[[ -46.875]
[-241.654]
[ -93.75 ]
[ 0. ]] m
Hence, q2 = k’T*D = [-9.375] kN
for member 3 :
k’T = AE/6.403 * [ 0.78086881 -0.62469505 -0.78086881 0.62469505],
and D =
[[ -46.875]
[-181.654]
[ 0. ]
[ 0. ]] m
Hence, q3 = k’T*D = [12.006] kN
for member 4 :
k’T = AE/6.403 * [-0.78086881 -0.62469505 0.78086881 0.62469505],
and D =
[[ -46.875]
[-181.654]
[ -93.75 ]
[ 0. ]] m
Hence, q4 = k’T*D,
or q4 = 12.006 kN, force in member 4 is tensile in nature.
for member 5 :
k’T = AE/4.0 * [-0. -1. 0. 1.],
and D =
[[ -46.875]
[-181.654]
[ -46.875]
[-241.654]] m
Hence, q5 = k’T*D = [-15.] kN
