ME 497M Examination # 1 Name
Transcript of ME 497M Examination # 1 Name
Page 1 of 18
Name (Print)__________________________________________________________________
(Last) (First)
ME 323 - Mechanics of Materials
Exam # 1
Date: October 5, 2016 Time: 8:00 β 10:00 PM
Instructions:
Circle your lecturerβs name and your class meeting time.
Gonzalez Krousgrill Zhao Sadeghi
4:30-5:20PM 11:30-12:20PM 12:30-1:20PM 2:30-3:20PM
Begin each problem in the space provided on the examination sheets. If additional space is required,
use the paper provided.
Work on one side of each sheet only, with only one problem on a sheet.
Please remember that for you to obtain maximum credit for a problem, you must present your solution
clearly.
Accordingly,
coordinate systems must be clearly identified,
free body diagrams must be shown,
units must be stated,
write down clarifying remarks,
state your assumptions, etc.
If your solution cannot be followed, it will be assumed that it is in error.
When handing in the test, make sure that ALL SHEETS are in the correct sequential order.
Remove the staple and restaple, if necessary.
Prob. 1 ______________________
Prob. 2 ______________________
Prob. 3 ______________________
Prob. 4 ______________________
Total ________________________
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Useful Equations
ex
=1
Es
x-n s
y+s
z( )éë
ΓΉΓ»+aDT
ey
=1
Es
y-n s
x+s
z( )éë
ΓΉΓ» +aDT
ez=
1
Es
z-n s
x+s
y( )éë
ΓΉΓ»+aDT
gxy
=1
Gt
xyg
xz=
1
Gt
xzg
yz=
1
Gt
yz
G =E
2(1+ v)
0
4
4 4
( ) dx
( ) ( )
polar area moment for solid circular cross section32
( ) polar a32
p
L
p p
p
op i
Tr
I
T x TL
G x I x GI
dI
I d d
3
4
rea moment for a circular hollow cross section
1 second area moment for a rectangular cross section
12
second area moment for a circular cross section64
I bh
dI
0 0
cos sin
L LF FL
e dx Tdx e L TEA EA
e u v
Yield Strength failure StressFactor of Safety or
Allowable Stress Stress member
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ME 323 Examination # 1 Name ___________________________________
October 5, 2016 (Print) (Last) (First)
Instructor _______________________________
Problem 1 (25 points): A rigid bar HD is pinned to ground at H, and to two elastic rods (1) and (2) at
B and D, respectively. Rods (1) and (2) are pinned to a wall at M and N, respectively. The length,
Youngβs modulus, and cross sectional area of each rod is shown in the following figure. The material
for each rod possesses a coefficient of thermal expansion of πΌ. A point load π acts at point C on the
beam in the negative π¦ direction. The temperature of rod (2) is increased by a temperature Ξπ, while
the temperature of rod (1) is kept constant. Assuming small rotations:
1. Write down the equilibrium conditions for the rigid bar.
2. Write down the compatibility conditions that relate the elongations of rods (1) and (2).
3. Determine the axial stresses in rods (1) and (2).
Express your answers in terms of π, πΏ, π΄, πΈ, πΌ, and Ξπ.
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Solution: 1. Let πΉ1 and πΉ2 represent the tensile axial forces carried by rods (1) and (2) respectively.
From the FBD of beam HD,
Ξ£ππ» = πΉ1(2πΏ) β π(3πΏ) β πΉ2(4πΏ) = 0
β πΉ1 β 2πΉ2 =3
2π (1.1)
2. From the (exaggerated) figure showing the displaced points Bβ, Cβ and Dβ, where π’π΅ and π’π·
represent the downward displacements of B and D respectively,
π’π΅
2πΏ=
π’π·
4πΏβ 2π’π΅ = π’π· (1.2)
The rotation indicated represents and
elongation of rod (1) and a shortening of
rod (2), i.e.
π’π΅ = π1; π’π· = βπ2 (1.3π)
β 2π1 = βπ2 (1.3π)
The elongations can be written in terms of the tensile loads carried, and temperature rise as:
π1 =πΉ1πΏ1
πΈ1π΄1=
πΉ1πΏ
(2πΈ)(2π΄)= 0.25
πΉ1πΏ
πΈπ΄(1.4π)
π2 =πΉ2πΏ2
πΈ2π΄2+ πΏ2πΌΞπ =
2πΉ2πΏ
πΈπ΄+ 2πΏπΌΞπ (1.4π)
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3. From (4.3b), (4.4a), and (4.4b),
2 Γ 0.25πΉ1πΏ
πΈπ΄= β
2πΉ2πΏ
πΈπ΄β 2πΏπΌΞπ
β πΉ1 + 4πΉ2 = β4πΈπ΄πΌΞπ (1.5) Solving (4.1) and (4.5),
πΉ1 = β4
3πΈπ΄πΌΞπ + π ; πΉ2 = β
2
3πΈπ΄πΌΞπ β
1
4π (1.6)
Therefore, the axial stresses,
π1 =πΉ1
2π΄= β
2
3πΈπΌΞπ +
π
2π΄
π2 =πΉ1
π΄= β
2
3πΈπΌΞπ β
π
4π΄
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ME 323 Examination # 1 Name ___________________________________
October 5, 2016 (Print) (Last) (First)
Instructor _______________________________
Problem 2 (30 points): A shaft is made up of three components: a solid circular shaft (1) of length 2L
and diameter d; solid circular shaft (2) of length L and diameter 2d; and circular-cross section tubular
shaft (3) of length 3L, outer diameter of 4d and inner diameter of 3d. All three components are made of
the same material having a shear modulus of G. Shafts (1) and (2) are connected by a thin, rigid
connector C, whereas shafts (2) and (3) are connected by a rigid connector D. Shaft 1 is rigidly
attached to ground at B. Torques T and 2T are applied to C and D, respectively, as shown in the figure
below. It is desired to know the torque carried by each of the shaft components. To this end:
a) Draw free body diagrams (FBDs) of connectors C and D below. Write down the appropriate
equilibrium equations from these FBDs.
b) Write down the torque/rotation equations for the three shaft components.
c) Write down the appropriate compatibility equations.
d) Solve for the torques in the three shaft components.
Express your final answers in terms of T, L, d and G.
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Solution
Equilibrium
πͺππππππππ πͺ: Ξ£ππΆ = βπ1 β π + π2 = 0 β π1 = π2 β π (2.1) πͺππππππππ π«: Ξ£ππ· = βπ2 β π3 + 2π = 0 β π3 = 2π β π2 (2.2)
Torque/rotation equations
πΌπ1 =π
32π4; πΌπ2 =
π
32(2π)4 =
16π
32π4; πΌπ3 =
π
32{(4π)4 β (3π)4} =
175π
32π4
Ξπ1 = (ππΏ
πΊπΌπ)
1
=π1(2πΏ)(32)
ππΊπ4=
64 π1πΏ
ππΊπ4 (2.3)
Ξπ2 = (ππΏ
πΊπΌπ)
2
=π2(πΏ)(32)
16ππΊπ4=
2 π2πΏ
ππΊπ4 (2.4)
Ξπ3 = (ππΏ
πΊπΌπ)
3
=π3(3πΏ)(32)
175ππΊπ4=
96 π3πΏ
175 ππΊπ4 (2.5)
Compatibility equations ππ‘πππ π: ππΆ = ππ΅ + Ξπ1 = Ξπ1
ππ‘πππ π: ππ· = ππΆ + Ξπ2 = Ξπ1 + Ξπ2 (2.6)
ππ‘πππ (π): ππ· = ππ΅ + Ξπ3 = Ξπ3 (2.7)
Equating (2.6) and (2.7):
Ξπ1 + Ξπ2 = Ξπ3 (2.8) Solve
Using equations (2.3)-(2.5) and (2.8): 64 π1πΏ
ππΊπ4+
2 π2πΏ
ππΊπ4=
96 π3πΏ
175 ππΊπ4 β 64π1 + 2π2 =
96
175π3 (2.9)
Using equations (2.1), (2.2) and (2.9):
64(π2 β π) + 2π2 =96
175(2π β π2) β π2 = 0.978 π
Using equations (2.1) and (2.2):
π1 = π2 β π = β0.022 π π3 = 2π β π2 = 1.022 π
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ME 323 Examination # 1 Name ___________________________________
October 5, 2016 (Print) (Last) (First)
Instructor _______________________________
Problem 3 (25 Points): A simply supported beam shown in Fig 3.1, is 10 ππ‘ long and is supported at B
and M. It is subjected to the following loads as shown. The cross section of the beam is uniform
rectangular of height 4 ππ and width 2 ππ as shown in Fig 3.2. For the state of loading shown:
1. Construct shear force and bending moment diagrams for the beam AM. Mark numerical values for
points A through M along with max/min values on these diagrams for the shear, and moment
diagrams respectively. Provide justification for your diagrams in terms of calculations.
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2. Determine the cross section (π₯), and locations (π¦) of the max bending normal (flexural) stress in
the beam. Also determine the magnitude of the corresponding flexural stress. State whether the
stress is compressive, or tensile. Represent the stress element at this location, and label the non-
zero stress components in Fig 3.3.
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Solution: 1. Net intensity of the uniform load: 500 Γ 3 = 1500 ππ acting at π₯ = 1.5 ππ‘. Net intensity
of the triangular load: 0.5 Γ 800 Γ 2 = 800 ππ, acting at π₯ = 10 β 0.67 = 9.33 ππ‘
From the FBD of beam AD,
Ξ£πΉπ₯ = π΅π₯ = 0 (3.1)
Ξ£πΉπ¦ = β1500 + π΅π¦ β 1000 β 800 + πΉπ = 0 β πΉπ = 3300 β π΅π¦ (3.2π)
Ξ£ππ = 1500 Γ 8.5 β π΅π¦ Γ 8 + 1000 Γ 5 + 200 + 800 Γ 0.67 = 0 (3.3π)
β π΅π¦ = 2310.75 ππ (3.3π)
Substituting the value of π΅π¦ into (3.2π),
πΉπ = 3300 β 2310.75 = 989.25 ππ (3.2π)
Inspection approach for the shear force diagram:
Since there are no point loads at A, π(0) = 0. Between A and B, shear force drops uniformly with a
slope that is equal to the intensity of the uniformly distributed load.
ππ΅β = π(0) β 500 Γ 2 = β1000 ππ
Due to the reaction force π΅π¦, the shear force jumps up by an equivalent amount, i.e.
ππ΅+ = ππ΅
β + π΅π¦ = β1000 + 2310.75 = 1310.75 ππ
Between B and C, shear force drops uniformly with a slope that is equal to the intensity of the
uniformly distributed load.
ππΆ = ππ΅+ β 500 Γ 1 = 1310.75 β 500 = 810.75 ππ
Since there are no applied loads between C and D, shear force remains constant, i.e.
ππ·β = ππΆπ· = ππΆ = 810.75 ππ
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Due to the applied downward load of 1000 ππ at D, shear force drops by an equivalent amount, i.e.
ππ·+ = ππ·
β β 1000 = β189.25 ππ
Since there are no applied forces between D and K, shear force remains constant, i.e.
ππΎ = ππ·πΎ = ππ· = β189.25 ππ
Due to the triangular downward load between K and M, shear force drops parabolically by an amount
that is equal to the net intensity of the applied load, i.e.
ππ = ππ β 800 = β989.25 ππ = βπΉπ (check)
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Inspection approach for the bending moment diagram:
Since there are no applied moments at A, π(0) = 0. Between A and B, bending moment drops
parabolically by an amount equal to the triangular area (Ξππ΄π΅) in the shear force diagram in this range.
ππ΅ = π(0) β 0.5 Γ 1000 Γ 2 = β1000 ππ β ππ‘
Between B and C, bending moment increases parabolically by an amount equal to the area of the
trapezium (Ξππ΅πΆ) in the shear force diagram in this range. However, at B, the bending moment would
have a higher positive slope, than at C, due to the corresponding shear forces.
Ξππ΅πΆ = 0.5(1310.75 + 810.75) Γ 1 = 1060.75 ππ β ππ‘
ππΆ = ππ΅ + Ξππ΅πΆ = β1000 + 1060.75 = 60.75 ππ β ππ‘
Between C and D, bending moment increases linearly with a slope that is equal to the intensity of
uniform shear force 810.75 ππ in this range.
ππ· = ππΆ + 810.75 Γ 2 = 1682.25 ππ β ππ‘
Between D and H, bending moment decreases linearly with a slope that is equal to the intensity of
uniform shear force β189.25 ππ in this range.
ππ»β = ππ· β 189.25 Γ 2 = 1303.75 ππ β ππ‘
Due to the counter-clockwise moment at H, moment drops by 200 ππ β ππ‘.
ππ»+ = ππ»
β β 200 = 1103.75 ππ β ππ‘
Between H and K, bending moment decreases linearly with a slope that is equal to the intensity of
uniform shear force β189.25 ππ in this range.
ππΎ = ππ»+ β 189.25 Γ 1 = 914.5 ππ β ππ‘
Between K and M, bending moment drops cubically (order 3) by an amount equal to the parabolic area
(ΞππΎπ) in the shear force diagram in this range. Since end M is pinned, ππ = 0.
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2. The magnitude of maximum bending moment occurs at D (π₯ = 5ππ‘). Therefore, maximum flexural
stress also occurs at this cross section. Since the bending moment is positive and the cross section
symmtrical about the neutral plane, maximum tensile stress is experienced at the bottom plane (π¦ =β2 ππ), and maximum compressive stress at the top plane (π¦ = 2 ππ).
ππππ₯ = 1682.25 ππ β ππ‘ = 1682.25 Γ 12 ππ β ππ
For a rectangular cross section,
πΌπ§ =πβ3
12=
(2)(4)3
12= 10.67 ππ4
Therefore, at = 5ππ‘, π¦ = β2 ππ,
ππππ‘π‘ππ = βππππ₯(β2)
πΌπ§=
1682.25 Γ 12 Γ 2
10.67= 3783.88 ππ π = 3.78 ππ π
At = 5ππ‘, π¦ = +2 ππ,
ππππ‘π‘ππ = βππππ₯(2)
πΌπ§= β
1682.25 Γ 12 Γ 2
10.67= β3783.88 ππ π = β3.78 ππ π
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ME 323 Examination # 1 Name ___________________________________
October 5, 2016 (Print) (Last) (First)
Instructor _______________________________
Problem #4 (20 Points): PART A β 8 points
A T-beam of length 3a is supported at the two ends and loaded by forces PB and PC. The line of action
of the forces is indicated (dashed lines) but the direction is to be determined. The correct moment
diagram is properly shown below.
(a) Indicate the cross section(s) where the maximum tensile stress is attained:
(i) x = 0
(ii) x = a
(iii) x = 2a
(iv) x = 3a
(b) Indicate the cross section(s) where the maximum compressive stress is attained:
(i) x = 0
(ii) x = a
(iii) x = 2a
(iv) x = 3a
(c) Indicate the value of the reaction at A, that is of Ay:
(i) Ay = M0/a
(ii) Ay = - M0/a
(iii) Ay = 2M0/a
(iv) Ay = - 2M0/a
(v) Ay = 3M0/a
(vi) Ay = - 3M0/a
(d) Indicate the value of the load at B, that is of PB:
(i) PB = M0/a
(ii) PB = - M0/a
(iii) PB = 2M0/a
(iv) PB = - 2M0/a
(v) PB = 3M0/a
(vi) PB = - 3M0/a
a a a
PB
BA C D
PCPositiveforceconvention
Cross section
of the T-beam
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Solution The distance from the neutral plane is maximum at the bottom of the cross section. Letβs assume that
the bottom of the cross section is at (π₯, βπ¦π).
Flexural stress:
ππππ₯ = βπ(βπ¦π)
πΌπ§
(a) Ans: (ii) π₯ = π. Since maximum positive bending moment occurs at π₯ = π, the maximum tensile
stress occurs here, i.e.
ππππ₯,π‘πππ πππ = βπ(βπ¦π)
πΌπ§=
π0π¦π
πΌπ§
(b) Ans: (iii) π₯ = 2π. Since maximum negative bending moment occurs at π₯ = 2π, the maximum
compressive stress occurs here, i.e.
ππππ₯,ππππ = β(βπ0)(βπ¦π)
πΌπ§= β
π0π¦π
πΌπ§
(c) Ans: (i) π΄π¦ = π0/π.
The slope of the linear bending moment is the shear force π(0) = π΄π¦.
(d) Ans: (vi) ππ΅ = β3π0/π.
Since bending moment is linear in π β€ π₯ β€ 2π,
ππΆ = ππ΅ + ππ΅+ Γ π
β ππ΅+ =
ππΆ β ππ΅
π= β
2π0
π
ππ΅+ = ππ΅
β + ππ΅ , πππ ππ΅β = π(0) = π΄π¦ =
π0
π
β ππ΅ = ππ΅+ β
π0
π= β
2π0
πβ
π0
π= β
3π0
π
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PART B β 6 points
Use only the compatibility condition for the truss
structure shown in the figure to find the value of the
elongation of member 2 (π2) in terms of the
elongation of member 1 (π1), and the elongation of
member 3 (π3). Specifically:
(a) Determine an expression for the compatibility
condition at A:
π2 = π π1 + π π3 where π and π are numbers.
Please notice that you are not required to solve for the
internal axial forces at equilibrium.
Please show your work and thought process.
Solution:
The angular orientation of the three joints are shown below:
π1 = 315Β°, ππ β 45Β°
π2 = tanβ13
4β cos π2 =
3
5, sin π2 =
4
5
π3 = 180Β°
Compatibility:
π1 = π’π΄ cos π1 + π£π΄ sin π1 =1
β2(π’π΄ β π£π΄) (1)
π2 = π’π΄ cos π2 + π£π΄ sin π2 =3
5 π’π΄ +
4
5 π£π΄ (2)
π3 = π’π΄ cos π3 + π£π΄ sin π3 = βπ’π΄ (3)
The following algebraic operation with equations (1) - (3):
4β2 π1 = 4 π’π΄ β 4 π£π΄ [4β2 Γ (1)]
+5π2 = 3 π’π΄ + 4 π£π΄ [5 Γ (2)] +7π3 = β7π’π΄ [7 Γ (3)]
4β2 π1 + 5π2 + 7π3= 0
β π2 = β4β2
5 π1 β
7
5 π3
(1)
(2)
(3)
3 ft.
4 ft.
3 ft.
3 ft.
Page 17 of 18
PART C β 6 points
For each state of plane stress shown below, i.e., for configurations (a) and (b), indicate whether each
component of the state of strain is:
= 0 (equal to zero)
> 0 (greater than zero)
< 0 (less than zero)
The material is linear elastic with Poissonβs ratio π (0 < π < 0.5), and the deformations are small.
(a) (b)
ππ₯
ππ¦
ππ§
πΎπ₯π¦
πΎπ₯π§
πΎπ¦π§
Fill in with β= 0β, β> 0β, or β< 0β.
(a) (b)
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Solution:
(a) ππ₯ = ππ¦ = βπ; ππ§ = 0; ππ₯π¦ = βπ/2; ππ¦π§ = ππ₯π§ = 0
(b) ππ₯ = π; ππ¦ = βπ; ππ§ = 0; ππ₯π¦ = ππ¦π§ = ππ₯π§ = 0
(a) (b)
ππ₯ 1
πΈ(βπ + ππ) = β
π
πΈ(1 β π) (< π)
1
πΈ(π + ππ) =
π
πΈ(1 + π) (> π)
ππ¦ 1
πΈ(βπ + ππ) = β
π
πΈ(1 β π) (< π)
1
πΈ(βπ β ππ) = β
π
πΈ(1 + π) (< π)
ππ§ 1
πΈ(βπ(βπ β π)) =
2ππ
πΈ (> π)
1
πΈ(βπ(βπ + π)) = π
πΎπ₯π¦ ππ₯π¦
πΊ= β
π
2πΊ (< π) π
πΎπ₯π§ π π
πΎπ¦π§ π π