Finite element method, HW-1 Solution
1. A concrete table column-support with the profile shown in the accompanying figure is to carry a load of 500
lb. Using the direct method, determine the deflection and average normal stress along the column. Divide the
column into five elements. 𝐸 = 3.27 × 103 ksi.
Element
Nodes
Average cross-sectional
area (in2)
1
2
3
4
5
1-2
2-3
3-4
4-5
5-6
33.75
29.25
27.00
29.25
33.75
𝑘𝑖 =
(𝐴𝑖 + 𝐴𝑖+1 )𝐸
𝑘
, [𝐊](𝑖) = [ 𝑖
−𝑘
2𝑙𝑖
𝑖
Length (in)
6
6
4
6
6
𝑘1
−𝑘1
−𝑘𝑖
0
] → [𝐊](𝐺) =
0
𝑘𝑖
0
[ 0
18.39 −18.39
−18.39 34.35
0
−15.94
= 106 ×
0
0
0
0
[ 0
0
Modulus of
elasticity (psi)
Element’s stiffness
coefficient (lb/in)
3.27×106
18.39×106
15.94×106
22.07×106
15.94×106
18.39×106
−𝑘1
𝑘1 + 𝑘2
−𝑘2
0
0
0
0
0
0
−15.94
0
0
38.01 −22.07
0
−22.07 38.01 −15.94
0
−15.94 34.35
0
0
−18.39
0
−𝑘2
𝑘2 + 𝑘3
−𝑘3
0
0
0
0
−𝑘3
𝑘3 + 𝑘4
−𝑘4
0
0
0
0
lb/in
0
−18.39
18.39 ]
0
0
0
−𝑘4
𝑘4 + 𝑘5
−𝑘5
0
0
0
0
−𝑘5
𝑘5 ]
{𝐑} = [𝐊](𝐺) {𝐮} − {𝐅}
𝑅1
18.39 −18.39
0
−18.39 34.35 −15.94
0
0
−15.94 38.01
0
→
= 106 ×
0
0
−22.07
0
0
0
0
0
{0}
[ 0
0
0
𝑢1
0
0
0
0
𝑢
0
0
0
2
0
𝑢3
−22.07
0
0
0
−
lb
𝑢
38.01 −15.94
0
4
0
−15.94 34.35 −18.39 𝑢5
0
{
]
{
𝑢
}
0
−18.39 18.39
6
−500}
Boundary condition: 𝑢1 = 0
1
0
−18.39 34.35
0
−15.94
→ 106 ×
0
0
0
0
[ 0
0
𝑢1
0
0
0
0
0
𝑢
−15.94
0
0
0
2
0
𝑢3
38.01 −22.07
0
0
0
=
lb
𝑢
−22.07 38.01 −15.94
0
4
0
0
−15.94 34.35 −18.39 𝑢5
0
{
]
{
𝑢
}
0
0
−18.39 18.39
6
−500}
0
−0.272
−0.585
→ {𝐮} =
× 10−4 in
−0.812
−1.126
{−1.398}
−14.81
14.81
−17.09
17.09
𝑘𝑖 (𝑢𝑖+1 − 𝑢𝑖 )
𝑢𝑖+1 − 𝑢𝑖
{𝛔}
𝜎𝑖 =
= 𝐸(
)→
= −18.52 psi, or 18.52 psi (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒)
𝐴𝑎𝑣𝑔
𝑙
−17.09
17.09
{−14.81}
{14.81}
2. A shaft is made of three parts, as shown in the accompanying figure, Parts AB and CD are made of the same
material with a modulus of rigidity of 𝐺 = 9.8 × 103 ksi, and each has a diameter of 1.5 in. Segment BC is
made of a material with a modulus of rigidity of 𝐺 = 11.2 × 103 ksi and has a diameter of 1 in. The shaft
is fixed at both ends. A torque of 2400 lb∙in is applied at C. Using three elements, determine the angle of twist
at B and C and the torsional reactions at the boundaries.
Element
1
2
3
Nodes
1-2
2-3
3-4
Polar moment of inertia
(in4)
0.4970
0.0982
0.4970
Length (in)
24
18
24
Shear modulus of
elasticity (psi)
6
9.80×10
11.2×106
9.80×106
2.029 −2.029
0
1 4
𝐽𝑖 𝐺𝑖
−2.029 2.640 −0.611
𝐽𝑖 = 𝜋𝑟𝑖 , 𝑘𝑖 =
→ [𝐊](𝐺) = 105 × [
0
−0.611 2.640
2
𝑙𝑖
0
0
−2.029
𝑅1
2.029 −2.029
0
{𝐑} = [𝐊](𝐺) {𝛉} − {𝐓} → { } = 105 × [−2.029 2.640
0
0
−0.611
𝑅4
0
0
0
−0.611
2.640
−2.029
Boundary condition: 𝜃1 = 𝜃4 = 0
1
0
−2.029
2.640
→ 105 × [
0
−0.611
0
0
0
−0.611
2.640
0
𝜃1
0
0
𝜃2
0
0
]{ } = {
} lb
−2.029 𝜃3
2400
𝜃4
1
0
0
−450.9
−3
2.2 × 10
0
→ {𝛉} = {
} rad, {𝐑} = {
} lb ∙ in
9.6 × 10−3
0
0
−1949.1
Element’s stiffness
coefficient (lb∙in)
2.029×105
0.611×105
2.029×105
0
0
] lb ∙ in
−2.029
2.029
𝜃1
0
0
𝜃
0
0
] { 2} − {
}
−2.029 𝜃3
2400
𝜃4
2.029
0
3. Consider a plate with a thickness of 0.125 in and a variable cross section supporting a load of 1500 lb, as
shown in the accompanying figure. Using direction formulation, determine the deflection of the bar at the
locations y = 2.5 in, y = 7.5 in, and y = 10 in. The plate is made of a material with a modulus of elasticity
𝐸 = 10.6 × 103 ksi.
Element
Nodes
Average cross-sectional
area (in2)
1
2
3
4
1-2
2-3
2-3
3-4
0.4688
0.1563
0.1563
0.2813
Length (in)
2.5
5.0
5.0
2.5
𝑅1
1.988
−1.988
0
{𝐑} = [𝐊](𝐺) {𝐮} − {𝐅} → { } = 106 × [
0
0
0
0
Modulus of
elasticity (psi)
Element’s stiffness
coefficient (lb/in)
10.6×106
1.988×106
0.331×106
0.331×106
1.193×106
𝑢1
−1.988
0
0
0
𝑢2
2.650 −0.663
0
0
] {𝑢 } − {
} lb
−0.663 1.855 −1.193
0
3
𝑢4
0
−1.193 1.193
1500
Boundary condition: 𝑢1 = 0
1
0
−1.988
2.650
→ 106 × [
0
−0.663
0
0
0
−0.663
1.855
−1.193
𝑢1
0
0
0
𝑢2
0
0
0.755
]{ } = {
} lb → {𝐮} = {
} × 10−3 in
−1.193 𝑢3
0
3.019
𝑢4
1.193
1500
4.277
4. (25%) The members of the truss shown in the accompanying figure have a cross-sectional area of 15 cm2
and are made of aluminum alloy (𝐸 = 90 GPa). Using hand calculations, determine the deflection of each
joint, the stress in each member, and the reaction forces.
Element
Nodes
1
2
3
4
1-2
2-3
3-4
2-4
Average cross-sectional
area (mm2)
1500
1803
1500
1000
1500
cos 𝜃
{𝐔} = [𝐓]{𝐮}, [𝐓] = [ sin 𝜃
0
0
Length (mm)
− sin 𝜃
cos 𝜃
0
0
0
0
cos 𝜃
sin 𝜃
Modulus of
elasticity (MPa)
Element’s stiffness
coefficient (N/mm)
9×104
9.000×104
7.489×104
9.000×104
13.50×104
0
0
]
− sin 𝜃
cos 𝜃
Take element (2) for example, cos 𝜃 = −0.832, sin 𝜃 = 0.555
−0.832
0.555
→ [𝐓](2) = [
0
0
−0.555
−0.832
0
0
0
0
−0.832 0.555
−1
0
0
−0.555 −0.832
] , [𝐓](2) = [
−0.832 −0.555
0
0
0.555 −0.832
0
0
{𝐅} = [𝐓][𝐤][𝐓]−1 {𝐔} → [𝐊] = [𝐓][𝐤][𝐓]−1
cos2 𝜃
sin 𝜃 cos 𝜃
𝜃
sin2 𝜃
= 𝑘 [ sin 𝜃 cos
2
− cos 𝜃
− sin 𝜃 cos 𝜃
− sin 𝜃 cos 𝜃
− sin2 𝜃
− cos2 𝜃
− sin 𝜃 cos 𝜃
cos 2 𝜃
sin 𝜃 cos 𝜃
− sin 𝜃 cos 𝜃
− sin2 𝜃 ]
sin 𝜃 cos 𝜃
sin2 𝜃
0
0
−0.832
−0.555
0
0
]
0.555
−0.832
0.692
−0.462
→ [𝐊](2) = 𝑘 [
−0.692
0.462
−0.462
0.308
0.462
−0.308
− 0.692
0.462
0.692
−0.462
0.462
−0.308
]
−0.462
0.308
9
0
−9
0
→ [𝐊](𝐺) = 104 ×
0
0
0
[0
0
−9
0
0
0 14.18
0 −3.456
0 −5.184
0 3.456
0
0
0
0
0
0
0
0
0
0
−3.456 −5.184 3.456
15.80
3.456 −2.304
3.456
14.18 −3.456
−2.304 −3.456 2.304
0
−9
0
−13.5
0
0
0
−9
0
0
0 14.18
0 −3.456
0 −5.184
0 3.456
0
0
0
0
0
0
−3.456
15.80
3.456
−2.304
0
−13.5
0
0
0
0
0
0
0 −13.5
N/mm
−9
0
0
0
9
0
0
13.5 ]
{𝐑} = [𝐊](𝐺) {𝐔} − {𝐅}
𝑅1𝑋
9
𝑅1𝑌
0
0
−9
0
0
→
= 104 ×
𝑅3𝑋
0
0
𝑅3𝑌
0
0
[0
{ 0 }
𝑈1𝑋
0
0
0
0
0
𝑈
0
0
0
0
0
1𝑌
𝑈2𝑋
−5.184 3.456
0
0
0
𝑈
3.456 −2.304 0 −13.5
0
2𝑌
−
N
𝑈3𝑋
14.18 −3.456 −9
0
0
𝑈3𝑌
−3.456 2.304
0
0
0
𝑈3𝑋
−9
0
9
0
0
]
{
0
0
0
13.5 {𝑈3𝑌 }
−10000}
Boundary condition: 𝑈1𝑋 = 𝑈1𝑌 = 𝑈3𝑋 = 𝑈3𝑌 = 0
1
0
−9
0
→ 104 ×
0
0
0
[0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
14.18 −3.456 −5.184
−3.456 15.80
3.456
0
0
1
0
0
0
0
0
−9
0
−13.5
0
0
0
3.456
−2.304
0
1
0
0
0
15000
0
0
−0.167
0
−0.684
0
→ {𝐔} =
mm, {𝐑} =
N
0
−15000
0
10000
0
0
{−0.758}
{ 0 }
𝑈1𝑋
0
0
0
𝑈
0
0
0
1𝑌
𝑈2𝑋
0
0
0
𝑈
0 −13.5
0
2𝑌
=
N
𝑈3𝑋
0
0
0
𝑈3𝑌
0
0
0
𝑈3𝑋
9
0
0
]
{
0 13.5 {𝑈3𝑌 }
−10000}
{𝐮} = [𝐓]−1 {𝐔}
𝑢2𝑥
−0.832
𝑢
−1
2𝑦
{𝐮}(2) = {𝑢 } = [𝐓](2) {𝐔}(2) = [−0.555
3𝑥
0
𝑢3𝑦
0
0.555
−0.832
0
0
−0.167
−0.241
0
0
−0.684
0.662
0
0
]{
}={
} mm
0
0
−0.832 0.555
0
0
−0.555 −0.832
10.00 (C)
−10.00
𝑘𝑖 (𝑢𝑖𝑥 − 𝑢𝑗𝑥 )
𝑢𝑖+1 − 𝑢𝑖
12.02 (T)
12.02
𝜎𝑖 =
= 𝐸(
) → {𝛔} = {
} MPa, or {
} MPa
0
0
𝐴
𝑙
6.67 (C)
−6.67