Book Solution Manual – Problem 14-3, Structural Analysis by R C Hibbeler, – Determine the Stiffness Matrix.

Problem Statement:

Determine the Stiffness Matrix K for the truss. Take A = 0.0015 m² and E = 200 GPa for each member.

Solution:

Known Loads, Qk =
[[ 0.0]
[-15.0]
[ 0.0]
[ 0.0]
[ 0.0]]
Unknown Deflections, Du =
[‘D1’, ‘D2’, ‘D3’, ‘D4’, ‘D5’]

Check the complete Book Solution Manual

for  Structural Analysis by R C Hibbeler (Eighth Edition)

Nodal dofs: [node1 (7, 8), node2(3, 4), node3(5, 6), node1(1, 2)]

Global Stiffness Matrix is given as
KG7 =
[[ λx^2 λxλy -λx^2 -λxλx ]
(AE/L)[ λxλy λy^2 -λxλx -λy^2 ]
[-λx^2 -λxλx λx^2 λxλy ]
[-λxλx -λy^2 λxλy λy^2 ]]

The nodes are (x,y) :

Node 1 = (0.0, 0.0) m
Node 2 = (5.0, -4.0) m
Node 3 = (10.0, 0.0) m
Node 4 = (5.0, 0.0) m

For Member 1:

starting node = 4
end node = 1
x = 0.0 – 5.0 = -5.0 m, and y = 0.0 – 0.0 = 0.0 m
length L = √(x^2 + y^2) = 5.0 m
λx = Cosθ = x/L = -1.0,
λy = = Sinθ y/L = 0.0

Node Columns and Rows IDs =
[1. 2. 7. 8.]

Global Stiffness matrix,
KG1 = AE*
[[ 0.2 -0. -0.2 0. ]
[-0. 0. 0. -0. ]
[-0.2 0. 0.2 -0. ]
[ 0. -0. -0. 0. ]]

or KG1 : AE*
[[ 0.2 -0. 0. 0. 0. 0. -0.2 0. ]
[-0. 0. 0. 0. 0. 0. 0. -0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[-0.2 0. 0. 0. 0. 0. 0.2 -0. ]
[ 0. -0. 0. 0. 0. 0. -0. 0. ]]

For Member 2:

starting node = 4
end node = 3
x = 10.0 – 5.0 = 5.0 m, and y = 0.0 – 0.0 = 0.0 m
length L = √(x^2 + y^2) = 5.0 m
λx = Cosθ = x/L = 1.0,
λy = = Sinθ y/L = 0.0

Node Columns and Rows IDs =
[1. 2. 5. 6.]

Global Stiffness matrix,
KG2 = AE*
[[ 0.2 0. -0.2 -0. ]
[ 0. 0. -0. -0. ]
[-0.2 -0. 0.2 0. ]
[-0. -0. 0. 0. ]]

or KG2 : AE*
[[ 0.2 0. 0. 0. -0.2 -0. 0. 0. ]
[ 0. 0. 0. 0. -0. -0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[-0.2 -0. 0. 0. 0.2 0. 0. 0. ]
[-0. -0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]]

For Member 3:

starting node = 2
end node = 1
x = 0.0 – 5.0 = -5.0 m, and y = 0.0 – -4.0 = 4.0 m
length L = √(x^2 + y^2) = 6.403 m
λx = Cosθ = x/L = -0.781,
λy = = Sinθ y/L = 0.625

Node Columns and Rows IDs =
[3. 4. 7. 8.]

Global Stiffness matrix,
KG3 = AE*
[[ 0.0952 -0.0762 -0.0952 0.0762]
[-0.0762 0.0609 0.0762 -0.0609]
[-0.0952 0.0762 0.0952 -0.0762]
[ 0.0762 -0.0609 -0.0762 0.0609]]

or KG3 : AE*
[[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0.0952 -0.0762 0. 0. -0.0952 0.0762]
[ 0. 0. -0.0762 0.0609 0. 0. 0.0762 -0.0609]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. -0.0952 0.0762 0. 0. 0.0952 -0.0762]
[ 0. 0. 0.0762 -0.0609 0. 0. -0.0762 0.0609]]

For Member 4:

starting node = 2
end node = 3
x = 10.0 – 5.0 = 5.0 m, and y = 0.0 – -4.0 = 4.0 m
length L = √(x^2 + y^2) = 6.403 m
λx = Cosθ = x/L = 0.781,
λy = = Sinθ y/L = 0.625

Node Columns and Rows IDs =
[3. 4. 5. 6.]

Global Stiffness matrix,
KG4 = AE*
[[ 0.0952 0.0762 -0.0952 -0.0762]
[ 0.0762 0.0609 -0.0762 -0.0609]
[-0.0952 -0.0762 0.0952 0.0762]
[-0.0762 -0.0609 0.0762 0.0609]]

or KG4 : AE*
[[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0.0952 0.0762 -0.0952 -0.0762 0. 0. ]
[ 0. 0. 0.0762 0.0609 -0.0762 -0.0609 0. 0. ]
[ 0. 0. -0.0952 -0.0762 0.0952 0.0762 0. 0. ]
[ 0. 0. -0.0762 -0.0609 0.0762 0.0609 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]]

For Member 5:

starting node = 2
end node = 4
x = 5.0 – 5.0 = 0.0 m, and y = 0.0 – -4.0 = 4.0 m
length L = √(x^2 + y^2) = 4.0 m
λx = Cosθ = x/L = 0.0,
λy = = Sinθ y/L = 1.0

Node Columns and Rows IDs =
[3. 4. 1. 2.]

Global Stiffness matrix,
KG5 = AE*
[[ 0. 0. -0. -0. ]
[ 0. 0.25 -0. -0.25]
[-0. -0. 0. 0. ]
[-0. -0.25 0. 0.25]]

or KG5 : AE*
[[ 0. 0. -0. -0. 0. 0. 0. 0. ]
[ 0. 0.25 -0. -0.25 0. 0. 0. 0. ]
[-0. -0. 0. 0. 0. 0. 0. 0. ]
[-0. -0.25 0. 0.25 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. ]]

Structure Stiffness matrix = ∑KGi

Structure Stiffness matrix, K = AE*
[[ 0.4 0.0 0.0 0.0 -0.2 0.0 -0.2 0.0 ]
[ 0. 0.25 0.0 -0.25 0.0 0.0 0.0 0.0 ]
[ 0. 0. 0.1905 0. -0.0952 -0.0762 -0.0952 0.0762]
[ 0. -0.25 0.0 0.3719 -0.0762 -0.0609 0.0762 -0.0609]
[-0.2 0.0 -0.0952 -0.0762 0.2952 0.0762 0. 0. ]
[ 0. 0 -0.0762 -0.0609 0.0762 0.0609 0.0 0.0 ]
[-0.2 0.0 -0.0952 0.0762 0. 0. 0.2952 -0.0762]
[ 0. 0. 0.0762 -0.0609 0. 0. -0.0762 0.0609]]

Please note that AE = 1500*200 = 3*10⁵ kN

The Answer given at the end of the book Structural Analysis by R C Hibbeler (9th edition) is incorrect, the correct stiffness matrix is as shown above. It is noted that the 8th edition of this book had different problems, and hence it is likely that the answer given is not properly checked before publication.