ME 323: Mechanics of Materials Homework Set 8 solutions
Fall 2019 Due: Wednesday, October 23
Problem 8.1 (10 points)
A steel (πΈ = 30,000ππ π) square beam with a side lengthπ = 2" is subject to loading as shown in Fig. 7.4. Determine the deflection curve of the beam using superposition. Use the table of beam deflections from the R. R. Craig textbook distributed via email. Compare this with the result found using second order integration in Homework 7.
Fig. 8.1
Solution:
2β
6β
2β
Po = 100 lb/ft
x D
C B
A
(1)
(2)
(3)
=
_
+
π0 = 2ππ‘
4π0 = 8ft
5π0 = 10ft
The following parameters are given
π0 = 100lb/ft
πΈ = 3 Γ 10;ksi
πΌ =112(2)@(2)in; = 1.33in; = 6.4 β 10EFft;
Using the tables and solving for deflections for each of the sections we get the following:
(1)
The deflection for section AB is as follows π£H(π₯) =
JKLM
N;OP(6π0N β 4π0π₯ + π₯N) , 0 β€ π₯ β€ π0
The deflection for the section CD is as follows
π£H(π₯) =(JK)TUV
N;OP(4π₯ β π0),π0 β€ π₯ β€ 5π0
(2)
The deflection for section AC is as follows
π£N(π₯) = βJKLM
N;OP(6(4π0)N β 4(4π0)π₯ + π₯N)
= β JKLM
N;OP(96π0N β 16π0π₯ + π₯N) 0 β€ π₯ β€ 4π0
The deflection for section CD is as follows
π£N(π₯) = βJK(;TU)V
N;OP(4π₯ β 4π0), 4π0 β€ π₯ β€ 5π0
(3)
The deflection for the entire beam AD is given by
π£@(π₯) =XYLM
ZOP(15π0 β π₯), 0 β€ π₯ β€ 5π0
Thus, by principle of superposition we can say that the deflections will be the sum of the deflections from the sections (1), (2) and (3) for the respective ranges of lengths:
JKLM
N;OP(β90π0N + 12π0π₯) +
XYLM
ZOP(15π0 β π₯),0 β€ π₯ β€ π0
π(π₯) = JKN;OP
(4π0@π₯ β π0; β 96π0Nπ₯N + 16π0π₯@ β π₯;) + XYLM
ZOP(15π0 β π₯), π0 β€ π₯ β€ 4π0
JKTKV
N;OP(β252π₯ + 255π0) +
XYLM
ZOP(15π0 β π₯), 4π0 β€ π₯ β€ 5π0
Using the boundary condition that, the deflection at the roller D is 0,π(5π0) = 0 we solve for D]
π0π0@
24πΈπΌ(β252(5π0) + 255π0) +
25π·_π0N(15π0 β 5π0)6πΈπΌ
= 0
π0π`@
24(β1005π0) +
250π·_π0@
6= 0
β1005π π0; + 1000π·_π0@ = 0
π«π = π.ππππ·πππ = πππππ
The final deflection can be described by the following equations:
JKLM
N;OP(β90π0N + 12π0π₯) +
XYLM
ZOP(15π0 β π₯),0 β€ π₯ β€ π0
π(π₯) = JKN;OP
(4π0@π₯ β π0; β 96π0Nπ₯N + 16π0π₯@ β π₯;) + XYLM
ZOP(15π0 β π₯),π0 β€ π₯ β€ 4π0
JKTKV
N;OP(β252π₯ + 255π0) +
XYLM
ZOP(15π0 β π₯), 4π0 β€ π₯ β€ 5π0
4.17π₯N(24π₯ β 360) + 33.5π₯N(30 β π₯), 0 β€ π₯ β€ 2ft
EIπ(π₯) = 4.17(β16 + 32π₯ β 384π₯N + 32π₯@ β π₯;) + 33.5π₯N(30 β π₯), 2ft β€ π₯ β€8ft
33.34(β252π₯ + 510) + 33.5π₯N(30 β π₯) β, 8ft β€ π₯ β€ 10ft
β495.12π₯N + 66.52π₯@,0 β€ π₯ β€ 2ft
EIπ(π₯) = β4.17π₯; + 99.84π₯@ β 595.13π₯N + 133.34π₯ β 66.67,2ft β€ π₯ β€ 8ft
β8401.68π₯ + 17003.4 + 1005π₯N β 33.5π₯@ β ,8ft β€ π₯ β€ 10ft
EIπH(π₯) = β495. 12π₯N + 66.52π₯@ , 0 β€ π₯ β€ 2ft
EIπN(π₯) = β4.17π₯; + 99.84π₯@ β 595.13π₯N + 133.34π₯ β 66.67 ,2ft β€ π₯ β€ 8ft
EIπ@(π₯) = β8401.68π₯ + 17003.4 + 1005π₯N β 33.5π₯@ ,8ft β€ π₯ β€ 10ft
From HW problem 7.4 we have the following deflection equations:
EIπH(π₯) =πoπ₯N
2+πoπ₯@
6 ,0 β€ π₯ β€ 2ft
EIπ£N(π₯) =qrLM
N+ srLV
Zβ NF
Z(π₯ β 2); + πΆ@π₯ + πΆ; ,2ft β€ π₯ β€ 8ft
EIπ£@(π₯) =qrLM
N+ srLV
Zβ 100π₯@ + 1500π₯N + πΆFπ₯ + πΆZ ,8ft β€ π₯ β€ 10ft
Where,
πo = β990lb. ft
πo = 399lb
πΆ@ = 0
πΆ; = 0
πΆF = β8400
πΆZ = 17000
πX = 201lb
The reaction force at D, πXis exactly equal to the reaction force calculated from the superposition method, π«π = π½π« = ππππ₯π
The final deflection of the beam from HW 7.4 is given by:
EIπH(π₯) = β495π₯N + 66.5π₯@ , 0 β€ π₯ β€ 2ft
EIπN(π₯) = β4.17π₯; + 99.84π₯@ β 595π₯N + 133.34π₯ β 66.67,2ft β€ π₯ β€ 8ft
EIπ@(π₯) = β8400π₯ + 17000 + 1005π₯N β 33.5π₯@ , 8ft β€ π₯ β€ 10ft
We can see from the numerical values that the deflection of the three sections of the beams calculated using method of superposition match very closely with the deflections calculated from the method of integration.
Further, if we compare and plot we get the following results.
Problem 8.2 (10 points)
The solid circular rod shown below has an axial force P0 acting at H and an axial force F0 acting at C. All components of the rod are made of steel (E = 200 GPa). Using P0 = 2 kN, F0 = 5 kN, d = 10 mm, and L = 200 mm, determine:
1) The strain energy stored in the rod. 2) The magnitude and direction of the horizontal displacement at H using Castiglianoβs
theorem. 3) The magnitude and direction of the horizontal displacement at C using Castiglianoβs
theorem.
Fig. 8.2
Solution:
Given:
πΈ = 200,000MPa
π0 = 2000N, πΉ0 = 5000N
π = 10 = 1 Γ 10ENm
πΏ = 200mm = 0.2m
π΄H =ππN
4= 7.85 Γ 10EFmN
π΄N = ππN = 3.14 Γ 10E;mN
π΄@ =9ππN
4= 7.065 Γ 10E;mN
Equilibrium in the shafts gives us following equations:
πΉH = π0
πΉH = πΉN = π0
πΉ@ = π0 β πΉ0
To find the strain energy we use the following formulation:
π =12οΏ½
πΉN
πΈπ΄ππ₯
οΏ½
`
Strain energy in shaft 1:
πH = οΏ½π0N
2πΈπ΄Hππ₯
;οΏ½
NοΏ½=π0N2πΏ2πΈπ΄H
= 50.9N.mm
Strain energy in shaft 2:
πN = οΏ½π0N
2πΈπ΄Nππ₯
NοΏ½
οΏ½=
π0NπΏ2πΈπ΄N
= 6.4N.mm
Strain energy in shaft 3:
π@ = οΏ½(π0 β πΉ0)N
2πΈπ΄@ππ₯
οΏ½
`=(π0 β πΉ )NπΏ2πΈπ΄@
= 6.4N.mm
Total strain energy can then be calculated to be,
ποΏ½0οΏ½οΏ½T = πH + πN + π@ = 63.7N.mm = 0.0637J
To find the displacement at point H we take the partial derivative with respect to the applied force ( π0 ) at that point and get:
π’οΏ½ =πΏποΏ½0οΏ½οΏ½TπΏπ0
π’οΏ½ =πΏπΏπ0
οΏ½π0N2πΏ2πΈπ΄H
+π0NπΏ2πΈπ΄N
+(π0 β πΉ )NπΏ2πΈπ΄@
οΏ½
π’οΏ½ =2π0πΏπΈπ΄H
+π0πΏπΈπ΄N
+(π0 β πΉ0)πΏπΈπ΄@
π’οΏ½ = 0.053ππ to the left
To find the displacement at point C we take the partial derivative with respect to the applied force at that point ( πΉ0) and get:
π’οΏ½ =οΏ½οΏ½οΏ½KοΏ½οΏ½οΏ½οΏ½οΏ½K
π’οΏ½ =πΏπΏπΉ0
οΏ½(π0 β πΉ0)NπΏ2πΈπ΄@
οΏ½
π’οΏ½ = β(π0 β πΉ0)πΏπΈπ΄@
π’οΏ½ = 0.00424ππ to the right
Problem 8.3 (10 points)
The beam shown below has elastic modulus E, second area moment I, and cross-sectional area A. It is loaded by a point force P at H. Neglect the shear strain energy due to bending. Using Castiglianoβs theorem, determine:
1) The magnitude and direction of the vertical displacement at D. 2) The magnitude and direction of the in-plane rotation angle at D. 3) The magnitude and direction of the horizontal displacement at H.
Express all answers in terms of P, E, I, A, and a.
Fig. 8.3
Solution:
In order to find the displacement at D, we apply a dummy moment ποΏ½and a dummy force ποΏ½ at point D
βποΏ½ = πΆ_π β ποΏ½(2π) β ππ + ποΏ½
πΆ_ = 2ποΏ½ + π β1πποΏ½
Making a cut through section DH,
βποΏ½οΏ½οΏ½ = 0
ππ¦ +πH = 0
π΄π(π) = π·π
Making a cut through section CD,
ππ = βπ·
βποΏ½οΏ½οΏ½ = 0
π΄π(ππ) = βπ·π ππ +π΄π β π·π
Making a cut through section BC,
βποΏ½οΏ½οΏ½ = 0
βπ@(π₯@) + πΆ_π₯@ β ποΏ½(π₯@ + π) + ποΏ½ β ππ = 0
π@(π₯@) = Β‘2ποΏ½ β1ππJΒ’ + πΒ£π₯@ β ποΏ½π₯@ β ποΏ½π + ποΏ½ β ππ
π΄π(ππ) = π·π (ππ β π) +π΄π Β₯π βπππΒ¦ + π·(ππ β π)
ππ = βπ·
π = οΏ½πHN
2πΈπΌππ¦
οΏ½
`+ οΏ½
πNN
2πΈπΌππ₯N
οΏ½
`+οΏ½
πΉNN
2πΈπ΄ππ₯N
οΏ½
`+ οΏ½
π@N
2πΈπΌππ₯@
οΏ½
`+οΏ½
πΉ@N
2πΈπ΄ππ₯@
οΏ½
`
πΏπH
πΏποΏ½= 0,
πΏπN
πΏποΏ½= 0,
πΏπΉNπΏποΏ½
= 0,πΏπ@
πΏποΏ½= π₯@ β π,
πΏπΉ@πΏποΏ½
= 0
πΏπH
πΏποΏ½= 0,
πΏπN
πΏποΏ½= 1,
πΏπΉNπΏποΏ½
= 0,πΏπ@
πΏποΏ½= 1 β
π₯@π,
πΏπΉ@πΏποΏ½
= 0
πΏπH
πΏπ= π¦,
πΏπN
πΏπ= βπ,
πΏπΉNπΏπ
= β1,πΏπ@
πΏπ= π₯@ β π,
πΏπΉ@πΏποΏ½
= β1
πΏππΏποΏ½
= π’X = 0 +οΏ½πN
πΈπΌ(βπ₯N)ππ₯N
οΏ½
`+ 0 +οΏ½
π@
πΈπΌ(π₯@ β π)ππ₯@
οΏ½
`+ 0
π’X =1πΈπΌοΏ½ πππ₯Nππ₯NοΏ½
`+1πΈπΌοΏ½ π(π₯@ β π)Nππ₯@οΏ½
`
π’X =ππ@
2πΈπΌ+ππΈπΌοΏ½ π₯@N β 2π₯@π + πNππ₯@οΏ½
`
π’X =ππ@
2πΈπΌ+ππΈπΌΒ‘13π@ β π@ + π@Β£
π’X =5ππ@
6πΈπΌ(πππ€ππ€πππ)
πΏππΏποΏ½
= πX = 0 +οΏ½πN
πΈπΌ(1)ππ₯N
οΏ½
`+ 0 + οΏ½
π@
πΈπΌΒ₯1 β
π₯@πΒ¦ ππ₯@
οΏ½
`+ 0
πX =1πΈπΌοΏ½ βππππ₯NοΏ½
`+1πΈπΌοΏ½ π(π₯@ β π) Β₯1 β
π₯@πΒ¦ ππ₯@
οΏ½
`
πX =βππN
πΈπΌ+ππΈπΌοΏ½ π₯@ β
1ππ₯@N β π + π₯@ππ₯@
οΏ½
`
πX =βππN
πΈπΌ+ππΈπΌΒ‘πN β
13ππ@ β πNΒ£
πX = β4ππN
3πΈπΌ(πΆπ, π ππππποΏ½ππ πΆπΆπππππX < 0)
πΏππΏπ
= π’οΏ½ =1πΈπΌοΏ½ ππ¦Nππ¦οΏ½
`+1πΈπΌοΏ½ ππNππ₯NοΏ½
`+1πΈπ΄
οΏ½ πππ₯NοΏ½
`+1πΈπΌοΏ½ π(π₯@ β π)Nππ₯@οΏ½
`+1πΈπ΄
οΏ½ πππ₯@οΏ½
`
π’οΏ½ =ππ@
3πΈπΌ+ππ@
πΈπΌ+πππΈπ΄
+ππΈπΌοΏ½ π₯@N β 2π₯@π + πNππ₯@οΏ½
`+πππΈπ΄
π’οΏ½ =ππ@
3πΈπΌ+ππ@
πΈπΌ+πππΈπ΄
+ππ@
3πΈπΌ+πππΈπ΄
π’οΏ½ =5ππ@
3πΈπΌ+2πππΈπ΄
(π‘ππ‘βπππππ‘)
Problem 8.4 (10 points)
Beam BCD is loaded with a distributed load w0 between B and C and a point force P at D. The beam has elastic modulus E and has a square cross section with side length b = L / 10. Neglect the shear strain energy due to bending.
Using Castiglianoβs theorem, determine the reactions at the wall at B and the roller at C. Express your answer in terms of w0, P, L, and E.
Fig. 8.4
Solution:
Equilibrium:
βπo = πo β12π€0πΏ(πΏ) + π΅_πΏ β
32ππΏ
πo β12π€0πΏN + π΅_πΏ β
32ππΏ = 0
βπΉ_ = π΄_ + π΅_ β π β π€0πΏ = 0
Making a cut at section BC we get:
βποΏ½οΏ½οΏ½ = 0
βπH β ππ₯H = 0
πH = βππ₯H,πΏπH
πΏοΏ½Y= 0
Making a cut at section AB we get:
βποΏ½οΏ½οΏ½ = 0
βπN βHNπ€0π₯NN + π΅_π₯N β π Β₯π₯N +
�N¦ = 0
πN(π₯) =HNπ€0π₯NN + π΅_π₯N β π Β₯π₯N +
�N¦,
πΏπN
πΏπ΅_= π₯N
Total strain energy = π = β« qΒ³M
NOP
Β΄M` ππ₯H + β«
qMM
NOPοΏ½` ππ₯N
πΏππΏπ΅_
= 0 + οΏ½πN
πΈπΌπΏπN
πΏπ΅_ππ₯N
οΏ½
`= 0
1πΈπΌοΏ½ (β
12π€0π₯N@ + π΅_π₯NN β ππ₯NN β
12ππΏπ₯N)ππ₯N
οΏ½
`= 0
β18π€0πΏ +
13π΅_ β
13π β
14π = 0
13π΅_ =
18π€0πΏ +
712π
π΅_ =38π€0πΏ +
74π
π΄_ = π +π€0πΏ β π΅_ = π + π€0πΏ β38π€0πΏ β
74π
π΄_ = β34π +
58π€0πΏ
πo =12π€0πΏN β π΅_πΏ +
32ππΏ =
12π€0πΏN β
38π€0πΏN β
74ππΏ +
32ππΏ
πo =18π€0πΏN β
14ππΏ