CIV3505 Structural Analysis: Truss and Beam Analysis using Strand7

This report details the structural analysis of trusses and beams using the stiffness method, validated with Strand7 software. It includes the determination of the global stiffness matrix, calculation of nodal displacements and member forces for a given truss structure. The assignment further involves a comparative analysis of member forces obtained through manual calculations and Strand7, accompanied by a discussion on the comparison of values between the two methods. Additionally, the report presents the beam analysis using stiffness matrix method. The deformed configuration of the truss is visualized using a displacement scale of 10%. The report also includes the shear force and bending moment diagrams.

Running head: STRUCTURAL ANALYSIS 1
Structural Analysis
Name
Institutional Affiliation

STRUCTURAL ANALYSIS 2
1. Question 1
a) Stiffness matrix K for the truss
The origin of the global coordinate system will be set at joint 1
λx=cos θx= X f − XN
L = X f − XN
√ ( X f − XN )2− ( Y f −Y N ) 2
λ y=cos θ y= Y f −Y N
L = Y f −Y N
√ ( X f − XN )2− ( Y f −Y N )2
Member 1
L=2 m λx= 0−2
2 =−1 λ y= 0−0
2 =0
k1 =0.0015 [200 ( 109 ) ]
2 [ 1 0 −1 0
0 0 0 0
−1 0 1 0
0 0 0 0 ] ¿
[ 15(107) 0 −15(107 ) 0
0 0 0 0
−15 (107) 0 15 (107 ) 0
0 0 0 0 ]
Member 2
L=2 m λx= 2−4
2 =−1 λ y= 0−0
2 =0
k 2=0.0015 [200 ( 109 ) ]
2 [ 1 0 −1 0
0 0 0 0
−1 0 1 0
0 0 0 0 ] ¿
[ 15( 107) 0 −15(107 ) 0
0 0 0 0
−15 (107) 0 15 (107 ) 0
0 0 0 0 ]
For
Member 3
L=2 √ 2 m λx= 2−4
2 √ 2 =− √ 2
2 λ y= 2−0
2 √ 2 =− √ 2
2

STRUCTURAL ANALYSIS 3
k3 = 0.0015[200 ( 109 ) ]
√ 2 [ 1 −1 −1 1
−1 1 1 −1
−1 1 1 −1
1 −1 −1 1 ]
¿
[ 5.303(107) −5.303(107 ) −5.303(107) 5.303(107 )
−5.303( 107) 5.303(107 ) 5.303(107) −5.303(107 )
−5.303( 107) 5.303(107 ) 5.303(107) −5.303(107 )
5.303(107) −5.303(107 ) −5.303(107) 5.303(107 ) ]
Member 4
L=2 m λx= 0−0
2 =0 λ y= 0−2
2 =−1
k 4= 0.0015[200 ( 109 ) ]
2 [ 0 0 0 0
0 1 0 −1
0 0 0 0
0 −1 0 1 ]=
[ 0 0 0 0
0 15 (107 ) 0 −15(107 )
0 0 0 0
0 −15 (107 ) 0 15(107 ) ]
Member 5
L=2 √ 2 m λx= 2−0
2 √ 2 =− √ 2
2 λ y= 2−4
2 √ 2 =− √ 2
2
k5 = 0.0015[200 ( 109 ) ]
√ 2 [ 1 1 1 −1
1 1 −1 −1
−1 −1 1 1
−1 −1 1 1 ]
Type equation here .
¿
[ 5.303(107) −5.303(107 ) −5.303(107) 5.303(107 )
−5.303( 107) 5.303(107 ) 5.303(107) −5.303(107 )
−5.303( 107) 5.303(107 ) 5.303(107) −5.303(107 )
5.303(107) −5.303(107 ) −5.303(107) 5.303(107 ) ]

STRUCTURAL ANALYSIS 4
Member 6
L=2 m λx= 0−2
2 =−1 λ y= 0−0
2 =0
k 6= 0.0015[200 ( 109 ) ]
2 [ 1 0 −1 0
0 0 0 0
−1 0 1 0
0 0 0 0 ] ¿
[ 15( 107) 0 −15(107 ) 0
0 0 0 0
−15 (107) 0 15 (107 ) 0
0 0 0 0 ]
The stiffness matrix of the truss is a 10X10 matrix which is
1 2 3 4 5 6 7 8 9 10
¿
[ 203.033 −53.033 −53.033 53.033 −150 0 0 0 0 0
−53.033 53.033 53.033 −53.033 0 0 0 0 0 0
−53.033 53.033 256.066 0 0 0 −53.033 −53.033 −150 0
53.033 −53.033 0 256.066 0 −150 −53.033 −53.033 0 0
−150 0 0 0 300 0 −150 0 0 0
0 0 0 −150 0 150 0 0 0 0
0 0 −53.033 −53.033 −150 0 203.033 53.033 0 0
0 0 −53.033 −53.033 0 0 53.033 53.033 0 0
0 0 150 0 0 0 0 0 150 0
0 0 0 0 0 0 0 0 0 0
] 1
2
3
4
5
6
7
8
9
10
(106 )
b) Vertical displacement at joint 2
The force applied downward in member 5
The global force of members, Qk=
[ 0
−30 ( 102 )
0
0
0
0
] 1
2
3
4
5
6

STRUCTURAL ANALYSIS 5
The displacement, Dk=
[0
0
0
0 ] 7
8
9
10
Since the displacement at local members 7, 8, 9 and 10 is zero
When Q=KD
Therefore the partial matrix of Qk=K 11 Du + K12 Dk
[ 0
−30 ( 102 )
0
0
0
0
]=
[ 203.033 −53.033 −53.033 53.033 −150 0
−53.033 53.033 53.033 −53.033 0 0
−53.033 53.033 256.066 0 0 0
53.033 −53.033 0 256.066 0 −150
−150 0 0 0 300 0
0 0 0 −150 0 150 ] (106 )
[ D1
D2
D3
D4
D5
D6
] +
[ 0
0
0
0
0
0 ]Expansion of the matrix
0=203.033 D1+−53.033 D2±53.033 D3 +−150 D5 (1)
−30 ( 102 ) =−53.033 D1 +53.033 D2 +53.033 D3 −53.033 D4 (2)
0=−53.033 D1+53.033 D2 +256.066 D3 (3)
0=53.033 D1−53.033 D2 +256.066 D4 −150 D6 (4)
0=−150 D1+300 D5 (5)
0=−150 D4 +150 D6 (6)
Solution
D1=−0.4 mm D2=−2.33mm D3=0.4 mm D4 =D6=−0.966 mm
D5=−0.1997 mm
Forces in members
qF = AE
L [−λx −λ y λx λ y ]
[DN x
DNy
DFx
DFy
]

STRUCTURAL ANALYSIS 6
Member 1
L=2 m λx=−1 λ y=0
qF = 0.0015[200 ( 109 ) ]
2 [ 1 0 −1 0 ]
[ −0.0002
−0.00097
0
0 ] 5
6
7
8
¿−30 kN
Member 2
L=2 m λx=−1 λ y=0
qF = 0.0015[200 ( 109 ) ]
2 [ 1 0 −1 0 ]
[ −0.0004
−00023
−0.0002
−0.00097 ] 1
2
5
6
¿−30 kN
Member 3
L=2 √ 2 m λx=− √2
2 λ y=− √2
2
qF = 0.0015 [ 200 ( 109 ) ]
2 √2 [ √2
2
√2
2
− √2
2
− √2
2 ] [ −0.0004
−00023
−0.0004
−0.00097 ] 1
2
3
4
¿ 42.43 kN
Member 4
L=2 m λx=0 λ y=−1
qF = 0.0015[200 ( 109 ) ]
2 [0 1 0 −1 ]
[ −0.0004
−00097
−0.0002
−0.00097 ]3
4
5
6

STRUCTURAL ANALYSIS 7
¿ 0
Member 5
L=2 √ 2 m λx=− √2
2 λ y=− √2
2
qF = 0.0015 [ 200 ( 109 ) ]
2 √2 [ √2
2
√2
2
− √2
2
− √2
2 ] [ −0.0004
−000097
0
0 ] 3
4
7
8
¿−42.43 kN
Member 6
L=2 m λx=−1 λ y=0
qF = 0.0015[200 ( 109 ) ]
2 [1 0 −1 0 ]
[ −0.0004
−000097
0
0 ] 3
4
9
10
¿ 60 kN
c) Strand 7 analysis

STRUCTURAL ANALYSIS 8
The internal reaction of the Truss

STRUCTURAL ANALYSIS 9
The displacement of Truss

STRUCTURAL ANALYSIS 10
d) Comparing Axial Forces of theoretical calculation and strand 7 results
Strand 7 kN ¿ manual calculations ( kN ¿
Member 1 -30.0 −30
Member 2 -30.0 -30
Member 3 42.43 42.43
Member 4 0.0 0
Member 5 -42.43 −42.43
Member 6 60.0 60
D1=−0.4 mm D2=−2.33mm D3=0.4 mm D4 =D6=−0.966 mm
D5=−0.1997 mm
Strand 7 manual calculations
dX(m) dY (m) dX(mm) dY (mm)
Node 1 0.0 0.0 0 0
Node 2 -0.0002 -0.0009657 -0.1997 -0.966
Node 3 -0.0004 -0.002331 -0.4 -2.33
Node 4 0.0004 -0.0009657 -0.4 -0.966
Node 5 0.0 0.0 0 0
e) Deformed configuration of the truss using a displacement scale of 10%

STRUCTURAL ANALYSIS 11

STRUCTURAL ANALYSIS 12
2. Question 2
a) Beam analysis by stiffness matric method
Member 1 and member2
12 EI
L3 =12 EI
23 =3 EI
6 EI
L2 = 6 EI
22 =1.5 EI
4 EI
L = 4 EI
2 =2 EI
2 EI
L =2 EI
2 =1 EI
k1 =EI
[ 1.5 1.5 −1.5 1.5
1.5 2 −1.5 1
−1.5 −1.5 1.5 −1.5
1.5 1 −1.5 2 ]
Similarly
k 2=EI
[ 1.5 1.5 −1.5 1.5
1.5 2 −1.5 1
−1.5 −1.5 1.5 −1.5
1.5 1 −1.5 2 ]
The stiffness matric for the beam will be a 6X6 matrix
k =
[ 2 −1.5 1 0 1.5 0
−1.5 1.5 −1.5 0 −1.5 0
0 0 1 2 −1.5 1.5
1 −1.5 4 1 0 1.5
0 0 1 2 −1.5 1.5
0 0 1.5 1.5 −1.5 1.5 ] EI
a) Expanding
The known constrains
Qk= [ 0
−1 ] 1
2 D= [0
0
0 ] 4
5
6
Therefore, Q=kD