Mechanics of solids 1 lecture-1

Khalid Yousaf BSc Civil Engineering 03217182339 Senior Resident Engineer at Engineering Services Consultants (Pvt.) Ltd.
MECHANICS OF SOLIDS-I
1. Stress, Strain and Mechanical Properties of Materials
 Uniaxial state of stress and strain
 Relationships between elastic Constants
 Response of materials under different sets of monotonic
loading (including impact)
 Normal and shearing stress and strains
 Distribution of direct stresses on uniform and non-
uniform members
 Thermal stresses and strains
2. Bending Theory
 Shear Force and Bending Moment Diagrams
 Relationship between load, shear force and bending
moment
 Theory of bending
 Moment of resistance and section modulus
 Bending and shearing stress distribution in beams
 Stresses in composite sections
3. Deflections of Beams
 Curvature, slope and deflection of beams using
integration methods
4. Theory of Torsion
 Theory of torsion of solids and hollow circular shafts
 Shearing stress distribution, angle of twist, strength and
stiffness of shaft
5. Stress and Strain Transformations
 Biaxial state of stresses
 Resolution of stresses
 Principal plane, principal stresses and strains,
 Graphical representation of stress and strains, Mohr’s
circle of stresses and strains
Recommended Books:
1. Pytel, A. & F. L. Singer, Strength of Materials, 4th Edition, Harper & Row Publishers, New York.
2. Hibbler, R. C., Mechanics of Materials, Prentice Hall, 10th Edition, 2016.
3. Warnock, F. V., Benham, P. P., Mechanics of Solids and Strength of Materials, Pitman Publishing, 1970.
4. James M. Gere & Barry. J. Gonodo, Mechanics of Materials, 9th Edition, 2008, CL Engineering
5. Beer F.P , Johnston E.R & Stephen P. Timoshenko, Mechanics of Materials, 7th Edition, 2015, PWS Pub Co.

Normal Stresses: Stress is defined as the strength of a material per unit area or unit strength. It is the force on a
member divided by area, which carries the force, formerly express in psi, now in N/mm2 or MPa. Mathematically
𝛔 = 𝑃
𝐴
where P is the applied normal load in Newton and A is the area in mm2. The maximum stress in tension or compression occurs
over a section normal to the load.
Normal stress is either tensile stress or compressive stress.
Members subject to pure tension (or tensile force) is under
tensile stress, while compression members (members
subject to compressive force) are under compressive stress.
Compressive force will tend to shorten the member.
Tension force on the other hand will tend to lengthen the member.

Problem#101: The bar ABCD in Fig. (a) consists of three cylindrical steel segments with different lengths and cross-sectional
areas. Axial loads are applied as shown. Calculate the normal stress in each segment.
Solution: We begin by using equilibrium analysis to
compute the axial force in each segment of the bar.
Step#1:Considering section 1 and summing up the forces
acting on it.
𝑃𝐴𝐵= 4000 lb
𝐴 𝐴𝐵=1.2 in2
𝛔 = 𝑃
𝐴
𝛔 𝐴𝐵 =
𝑃 𝐴𝐵
𝐴
𝛔 𝐴𝐵 =
𝑃 𝐴𝐵
𝐴
𝛔 𝐴𝐵 =
4000
1.2
𝛔 𝐴𝐵 = 3333.3 𝑃𝑠𝑖 (𝑇)
Step#2: Adding up forces in section 2, we get
𝑃𝐵𝐶= - 5000 lb
𝐴 𝐵𝐶=1.8 in2
𝛔 𝐵𝐶 =
𝑃 𝐵𝐶
𝐴
𝛔 𝐵𝐶 = −
5000
1.8
𝛔 𝐵𝐶 = −2777.8 𝑃𝑠𝑖 =2777.7 Psi (C)

Step#3: Taking section 3 under consideration.
𝑃𝐶𝐷= 7000 lb
𝐴 𝐶𝐷=1.6 in2
𝛔 𝐶𝐷 =
𝑃 𝐶𝐷
𝐴
𝛔 𝐶𝐷 =
7000
1.6
𝛔 𝐶𝐷 = 4375 𝑃𝑠𝑖 (𝑇)

1
Problem 102: For the truss shown in the
shown in the figure. Determine the stress in
the member AC and BD. The cross sectional
area of the each member of the truss is 900 mm2.
Solution: The three assumptions taken in the
elementary analysis of the truss are ,
1- Weights of the members are neglected.
2- All connections are smooth pins.
3- All external loads
are applied directly to the pins.
Step#1: Calculate the reactions at each support
∑ Fy =0
Ay + Hy = 50 kN
∑ MA =0
30 × 4 + 70 × 12 − Hy × 16 = 0
Hy = 60 kN
And
Ay = 40 kN
Step#2: Considering Section 1
∑ Fy =0
Ay + 𝟑
𝟓
FAB = 0
FAB = − 𝟓
𝟑
Ay
FAB = − 𝟓
𝟑
× 40 = − 66.7 kN

∑ Fy =0
FAC + 𝟒
𝟓
FAB = 0
FAC = − 𝟒
𝟓
FAB
FAC = − 𝟒
𝟓
× (−66.7) C
FAC = 53.4 𝑘𝑁
Step#3:
∑ ME =0
Ay × 8 + FBD × 3 − 30 × 4 = 0
40 × 8 + FBD × 3 − 30 × 4 = 0
3FBD = −200
FBD = −66.7 kN
Step#4:
The area of each member 𝐴 = 900 mm2 =900 × 10−6 m2
The stresses in members AC and BD are,
𝛔 = 𝑃
𝐴
𝛔 𝐴𝐶 =
53.4×103
900×10−6
𝛔 𝐴𝐶 = 59.3 × 106
𝛔 𝐴𝐶 = 59.3MPa (T)
𝛔 𝐵𝐷 =
−66.7×103
900×10−6
𝛔 𝐵𝐷 = −74.1 × 106
𝛔 𝐵𝐷 = −74.1MPa =74.1MPa (C)

Problem 103: Figure (a) shows a two-member truss supporting a block of weight W. The cross-sectional areas of the members are 800 mm2 for
AB and 400 mm for AC . Determine the maximum safe value of W if the working stresses are 110 MPa for AB and 120 MPa for AC.
Solution. Being members of a truss, AB and AC can
be considered to be axially loaded bars . The forces
in the bars can be obtained by analyzing the FBD of
pin A in Fig. (b). The equilibrium equations are,
∑ Fx =0
PAC 𝐶𝑜𝑠60 𝑜
- PAB 𝐶𝑜𝑠40 𝑜
= 0
PAC 𝑆𝑖𝑛60 𝑜
+ PAB 𝑆𝑖𝑛40 𝑜
- W= 0
Solving above equations simultaneously,
PAC = 0.7779𝑊
PAB = 0.5077𝑊
Area of each member A = 800 mm2 = 800 × 10−6 m2
Stress in each member:
𝛔= 𝑃
𝐴
𝜎𝐴𝐶 = 𝑃 𝐴𝐶
𝐴
120 × 106 = 0.7779W
800× 10−6
W = 123409 N = 123 kN
𝜎𝐴𝐵 = 𝑃 𝐵𝐶
𝐴
110 × 106 = 0.5077W
800× 10−6
W = 173330 N = 173 kN For safe loading, use W= 123 kN

Alternatively: Make a polygon with the force vectors as shown in the figure.
Applying Law of Sine to this polygon , we have
For Member AC:
𝑃 𝐴𝐶
sin 50o = W
sin 100o
𝑃𝐴𝐶=0.7779W
𝛔 𝐴𝐶 𝐴=0.7779W
120(800 × 10−6)=0.7779W
W = 123409 N =123 kN
For wire AB:
𝑃 𝐴𝐵
sin 30o = W
sin 100o
𝑃𝐴𝐵=0.5077W
𝛔 𝐴𝐵 𝐴=0.5077W
110 × 106(800 × 10−6)=0.5077W
W = 173330 N = 173 kN
For safe load W use W = 123 kN

Problem 104
A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside
diameter of the tube if the stress is limited to 120 MN/m2.
Solution: Given that
Inside diameter d = 100 mm
Tensile load P = 400 kN = 400000 N
Stress 𝛔 = 120 MN/m2
Outer diameter D = ?
Since
𝛔 = 𝑃
𝐴
or 𝑃 = 𝛔A
Area 𝐴 = π
4
(D2 − d2)
400000 = 120 π
4
(D2 − d2)
400000 = 30π (D2 − 1002)
𝐷2 =
400000+300000π
30π
𝐷 = 119.35 𝑚𝑚

Problem 105: A homogeneous 800 kg bar AB is supported at either end by a cable as shown in Fig. P-105. Calculate the smallest area of
each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel.
Solutions: Given that
Mass of the bar M = 800 kg
W = 7848 N
Stress in bronze σbr = 90 Mpa
Stress in steel σst = 90 Mpa
By symmetry:
Pbr =Pst =12(7848)
Pbr=3924 N
Pst=3924 N
For bronze cable:
Pbr = σbrAbr
3924 = 90Abr
Abr = 43.6 mm2
For steel cable:
Pst = σstAst
3924 = 120Ast
Ast = 32.7 mm2

Problem106: The homogeneous bar shown in Fig. P-106 is supported by a smooth pin at C and a cable that runs
from A to B around the smooth peg at D. Find the stress in the cable if its diameter is 0.6 inch and the bar weighs
6000 lb.
Diameter of the cable d = 0.6 in
Weight of the bar w = 600 lb
ΣMC = 0
5T+10( 𝟑
𝟑𝟒
T) = 5(6000)
T=2957.13 lb
T = σA
2957.13 = σ[
1
4π
(0. 62)]
𝛔 = 10458.72 psi

Problem 107: A rod is composed of an aluminum section rigidly attached between steel and bronze sections, as
shown in Fig. P-107. Axial loads are applied at the positions indicated. If P = 3000 lb and the cross sectional area of
the rod is 0.5 in2, determine the stress in each section.
Applied load P = 3000 lb = 3 kips
cross-Sectional Area A = 0.5 in2
For steel:
σstAst = Pst
σst(0.5) = 4P
σst(0.5) =4 ⨉ 3
σst = 24 ksi
For aluminum:
σalAal = Pal
σal(0.5) = 4P
σal(0.5) = 4 ⨉ 3
σal = 24 ksi
For bronze:
σbrAbr = Pbr
σbr(0.5) = 9
σbr = 18 ksi

Problem 108: An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in Fig. P-108.
Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in steel of
140 MPa, in aluminum of 90 MPa, or in bronze of 100 MPa.
Solutions: Given that
Stress in steel 𝛔 𝑠𝑡 = 140 Mpa
Steel area 𝐴 𝑠𝑡= 500 mm2
Stress in aluminum 𝛔 𝑎𝑙 = 90 Mpa
Aluminum area 𝐴 𝑎𝑙 = 400 mm2
Stress in bronze 𝛔 𝑏𝑟 = 100 Mpa
Bronze area 𝐴 𝑏𝑟 = 200 mm2
Max. applicable load P = ?
For bronze:
σbrAbr = Pbr
1000(200) = 2P
P = 10000 N
For aluminum:
σalAal = Pal
90(400) = P
P = 36000 N
For steel:
σstAst = Pst
100(500) = 5P
P = 10000 N For safe value of P , use P = 10000 N = 10 kN

Problem 109: Determine the largest weight W that can be supported by two wires shown in Fig. P-109. The stress in
either wire is not to exceed 30 ksi. The cross-sectional areas of wires AB and AC are 0.4 in2 and 0.5 in2, respectively.
Maximum applicable stress 𝛔 = 30 ksi
Cross-sectional area of AB 𝐴 𝐴𝐵 = 0.4 in2
Cross-sectional area of AB 𝐴 𝐵𝐶 = 0.5 in2
Largest supportable weight W = ?
The free body diagram of the joint A
is given by
For wire AB: By sine law (from the force polygon):
𝑇 𝐴𝐵
sin 40o = W
sin 80o
𝑇𝐴𝐵=0.6527W
𝛔 𝐴𝐵 𝐴 𝐴𝐵=0.6527W
30(0.4)=0.6527W
W = 18.4 kips
For wire AC:
𝑇 𝐴𝐶
sin 60o = W
sin 80o
𝑇𝐴𝐶=0.8794W
𝛔 𝐴𝐶 𝐴 𝐴𝐶=0.8794W
30(0.5)=0.6527W
W = 17.1 kips For safe load W use W = 17.1 kips

Problem 110: A 12-inches square steel bearing plate lies between an 8-inches diameter wooden post and a concrete
footing as shown in Fig. P-110. Determine the maximum value of the load P if the stress in wood is limited to 1800
psi and that in concrete to 650 psi.
Area of steel bearing plate 𝐴 𝑠𝑡 = 12 in2
Diameter of wooden post 𝑑 𝑤𝑑 = 8 in
Stress in wood 𝛔 𝑤𝑑 = 1800 Psi
Stress in concrete 𝛔 𝑐𝑜𝑛 = 650 Psi
For Wood:
Pw = σwAw
Pw = σw
π
4
d2
Pw = 1800[π
4
82]
Pw = 90477.9 lb
For concrete:
PC = 𝛔CAC
PC = 650 ⨉ 122
PC = 93600 lb.
For safe load ‘P’ use P = Pw = 90477.9 lb

Problem 111: For the truss shown in Fig. P-111, calculate the stresses in members CE, DE, and DF. The cross
sectional area of each member is 1.8 in2. Indicate tension (T) or compression (C).
Cross Sectional Area of each member A = 1.8 in2
σCE = ?
σDE = ?
σDF = ?
Σ 𝐹𝑦 = 0
RA + RF =30 (1)
Σ 𝑀𝐴 = 0
24 RF =16(30)
RF =20k put in (1)
RA = 10k
At joint F:
Σ 𝐹𝑦 = 0
- 3
5
DF = 20
DF = -33.33k = 33.33k (C)
At joint D:
By symmetry

BD = DF = 33.33k (C)
Σ 𝐹𝑦 = 0
𝐷𝐸 = 3
5
𝐵𝐷 + 3
5
DF
𝐷𝐸 = 3
5
33.33 + 3
5
33.33
𝐷𝐸 = 40k (T)
At joint E:
Σ 𝐹𝑦 = 0
3
5
𝐶𝐸 + 30 = 40
CE = 10.67k (T)
Stresses: (Stress = Force/ Area)
σCE = 10.67
1.8
𝜎 𝐶𝐸 = 5.93 𝑘𝑠𝑖 (T)
𝜎 𝐷𝐸 = 40
1.8
𝜎 𝐷𝐸 = 22.22 𝑘𝑠𝑖 (𝑇)
𝜎 𝐷𝐹 = 33.33
1.8
𝜎 𝐷𝐹 = 18.52 𝑘𝑠𝑖 (𝐶)

Problem#112: Determine the cross-sectional areas of members AG, BC, and CE for the truss shown in Fig. P-112.
The stresses are not to exceed 20 ksi in tension and 14 ksi in compression. A reduced stress in compression is
specified to reduce the danger of buckling.
Soln.
∑𝐹𝑦 = 0
𝐴 𝑦 = 40 + 25 = 65 𝑘
∑𝑀𝐴 = 0
18 × 𝐷 𝑥 = 40 × 4 + 25 × 8
𝐷 𝑥 = 20𝑘
∑𝐹𝑥 = 0
𝐴 𝑥 = 𝐷 𝑥 = 20 𝑘
Check:
∑𝑀 𝐷 = 0
12 × 𝐴 𝑦 = 18 × 𝐴 𝑥 + 40 × 8 + 25 × 4
12 × 65 = 18 × 20 + 40 × 8 + 25 × 4
780 𝑘𝑖𝑝. 𝑓𝑡 = 780 𝑘𝑖𝑝. 𝑓𝑡 (ok!)
For member AG (At joint A):
∑𝐹𝑦 = 0
3
13
𝐴𝐵 = 65
AB = 78.12 k

∑𝐹𝑥 = 0
𝐴𝐺 + 20 =
2
13
𝐴𝐵
𝐴𝐺 + 20 =
2
13
× 78.12
𝐴𝐺 = 20.33 𝑘 (𝑇)
since
𝜎 =
𝑃
𝐴
𝜎𝑡𝑒𝑛𝑠𝑖𝑜𝑛 =
𝑃
𝐴
20=
20.33
𝐴 𝐴𝐺
𝐴 𝐴𝐺 = 1.17 𝑖𝑛2
For member BC (At section through MN):
∑𝑀 𝐹 = 0
12 × 𝐷 𝑥 = −6 ×
2
13
𝐵𝐶
12 × 20 = −6 ×
2
13
𝐵𝐶
B𝐶 = −72.11 𝑘 = 72.11 𝑘 (𝐶)
𝜎𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖𝑜𝑛 =
𝑃
𝐴 𝐵𝐶
14 =
72.11
𝐴 𝐵𝐶
𝐴 𝐵𝐶 = 5.15 𝑖𝑛2

For member CE (At joint D):
∑𝐹𝑥 = 0
20 =
2
13
𝐶𝐷
CD = 36.06 k
∑𝐹𝑦 = 0
𝐷𝐸 =
3
13
𝐶𝐷
𝐷𝐸 =
3
13
× 36.06
𝐷𝐸 = 30 𝑘
At joint E:
∑𝐹𝑦 = 0
30 =
3
13
𝐸𝐹
𝐸𝐹 = 36.06 𝑘
∑𝐹𝑥 = 0
C𝐸 = −
2
13
𝐸𝐹
C𝐸 = −
2
13
× 36.06
𝐶𝐸 = −20 𝑘 = 20 𝑘 (𝐶)
𝜎𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖𝑜𝑛 =
𝑃
𝐴
14 =
20
𝐴 𝐴𝐺
𝐴 𝐴𝐺 = 1.43 𝑖𝑛2

Problem#113:Find the stresses in members BC, BD, and CF for the truss shown in Fig. P-113. Indicate the tension or
compression. The cross sectional area of each member is 1600 mm2.
Soln.
For member BD: (See FBD 01)
∑𝑀 𝐶 = 0
3 ×
4
5
𝐵𝐷 = 3 × 60
𝐵𝐷 = 75 𝑘 (T)
𝜎 =
𝑃
𝐴
𝜎 𝐵𝐷 =
75(1000)
1600
𝜎 𝐵𝐷 = 46.875 Mpa (T)
For member CF: (See FBD 01)
∑𝑀 𝐷 = 0
4 ×
1
2
𝐶𝐹 = 4 × 900 + 7 × 60
𝐶𝐹 = 275.77 𝑘𝑁 (𝐶)
𝜎 =
𝑃
𝐴
𝜎 𝐶𝐹 =
275.77(1000)
1600
𝜎 𝐶𝐹 = 172.375𝑀𝑃𝑎 (𝐶)

For member BC: (See FBD 02)
∑𝑀 𝐷 = 0
4 × 𝐵𝐶 = 7 × 60
𝐵𝐶 = 105 𝑘𝑁 (𝐶)
𝜎 =
𝑃
𝐴
𝜎 𝐵𝐶 =
105(1000)
1600
𝜎 𝐵𝐶 = 62.625𝑀𝑃𝑎 (𝐶)

Problem#114:The homogeneous bar ABCD shown in Fig. P-114 is supported by a cable that runs from A to B around
the smooth peg at E, a vertical cable at C, and a smooth inclined surface at D. Determine the mass of the heaviest bar
that can be supported if the stress in each cable is limited to 100 MPa. The area of the cable AB is 250 mm2 and that
of the cable at C is 300 mm2.
Soln.
∑𝐹𝑥 = 0
𝑇𝐴𝐵 𝐶𝑜𝑠300 = 𝑅 𝐷Sin300
𝑅 𝐷 = 1.1305𝑇𝐴𝐵
∑𝐹𝑦 = 0
𝑇𝐴𝐵 + 𝑇𝐶 + 6 × 𝑇𝐴𝐵Sin300 + 𝑅 𝐷Cos500 = 𝑊
𝑇𝐴𝐵 + 𝑇𝐶 + 𝑇𝐴𝐵Sin300 + 1.1305𝑇𝐴𝐵Cos500 = 𝑊
𝑇𝐶 + 2.226𝑇𝐴𝐵 = 𝑊
𝑇𝐶 = 𝑊 − 2.226𝑇𝐴𝐵
∑𝑀 𝐷 = 0
4 × 𝑇𝐴𝐵 + 2 × 𝑇𝐶 + 6 × 𝑇𝐴𝐵Sin300 = 3𝑊
7 × 𝑇𝐴𝐵 + 2 × (𝑊 − 2.226𝑇𝐴𝐵) = 3𝑊
2.5466𝑇𝐴𝐵 = 𝑊
𝑇𝐴𝐵 = 0.3927𝑊
𝑇𝐶 = 𝑊 − 2.226𝑇𝐴𝐵
𝑇𝐶 = 𝑊 − 2.2267(0.3927𝑊)
𝑇𝐶 = 0.1256𝑊

Based on cable AB:
𝜎𝐴𝐵 =
𝑇 𝐴𝐵
𝐴 𝐴𝐵
100 =
0.3967𝑊
250
𝑊 = 63661.83 𝑁
Based on cable at C:
𝜎 𝐶 =
𝑇 𝐶
𝐴 𝐶
100 =
0.1256𝑊
300
𝑊 = 238853.5 𝑁
Safe value of W
𝑊 = 63661.83 𝑁
𝑊 = 𝑚𝑔
63661.83 = 𝑚 (9.81)
𝑚 = 6490 𝑘𝑔
𝑚 = 6.49 𝑀𝑔

Assignment.
Q#1: Axial loads are applied to the compound rod that is
composed of an aluminum segment rigidly connected
between steel and bronze segments. What is the stress in
each material given that P = 10 kN?
Q#2: Axial loads are applied to the compound rod that is
composed of an aluminum segment rigidly connected
between steel and bronze segments. Find the largest safe
value of P if the working stresses are 120 MPa for steel,
68MPa for aluminum, and110 MPa for bronze.
Q#3: The wood pole is supported by two cables of 0.25 in.
diameter. The turnbuckles in the cables are tightened until
the stress in the cables reaches 60000 psi. If the working
compressive stress for wood is 200 psi, determine the
smallest permissible diameter of the pole.

Q#4:Find the maximum allowable value of P for the
column. The cross-sectional areas and working stresses
(𝜎 𝑤) are shown in the figure.
Q#5:Determine the smallest safe cross-sectional areas of
members CD, GD, and GF for the truss shown. The
working stresses are 140 MPa in tension and 100 MPa in
compression.(The working stress in compression is smaller
to reduce the danger of buckling.)
Q#6.Find the stresses in members BC, BD, and CF for the
truss shown.Indicate tension or compression.The cross-
sectional area of each member is 1400 mm2.
Q#7:Determine the smallest allowable cross-sectional
areas of members BD, BE, and CE of the truss shown. The
working stresses are 20 000 psi in tension and 12000 psi in
compression. (A reduced stress in compression is specified

Q#8: The uniform 300-lb bar AB carries a 500-lb vertical
force at A. The bar is supported by a pin at B and the 0.5 in.
diameter cable CD. Find the stress in the cable.
Q#9: Determine the smallest allowable cross-sectional areas
of members CE, BE, and EF for the truss shown. The
working stresses are 20 ksi in tension and 14 ksi in
compression. (The working stress in compression is smaller
Q#10:For the Pratt bridge truss and loading shown, determine
the average normal stress in member BE, knowing that the
cross-sectional area of that member is 5.87 in2.
Q#11:A solid brass rod AB and a solid aluminum rod BC
are connected together by a coupler at B, as shown in
Figure Q#11.The diameters of the two segments are 𝑑1 =
60𝑚𝑚 and 𝑑2 = 50𝑚𝑚, respectively. Determine the axial
stresses 𝜎1 (in rod AB) and 𝜎2 (in rod BC).

Q#12:A hollow circular nylon pipe (see Fig 1-23) supports
a load 𝑃𝐴 = 7800𝑁,which is uniformly distributed around
a cap plate at the top of the lower pipe. A second load 𝑃𝐵 is
applied at the bottom. The inner and outer diameters of the
upper and lower parts of the pipe are 𝑑1 = 51𝑚𝑚, 𝑑2 =
60𝑚𝑚,𝑑3 = 57𝑚𝑚, and 𝑑4 = 63𝑚𝑚, respectively. The
upper pipe has a length 𝐿1 = 350𝑚𝑚; the lower pipe
length is 𝐿2 = 400𝑚𝑚.Neglect the self weight of the
pipes.
(a) Find 𝑃𝐵 so that the tensile stress in upper part is 14.5
MPa. What is the resulting stress in the lower part?
(b) If 𝑃𝐴 remains unchanged, find the new value of 𝑃𝐵 so
that upper and lower parts have same tensile stress.
(c) Find the tensile strains in the upper and lower pipe
segments for the loads in part (b) if the elongation of the
upper pipe segment is known to be 3.56 mm and the
downward displacement of the bottom of the pipe is 7.63
mm.

Q#13: A 1- in. diameter solid bar (1), a square solid bar
(2), and a circular tubular member with 0.2 in. wall
thickness (3), each supports an axial tensile load of 5 kips.
(a) Determine the axial stress in bar (1).
(b) If the axial stress in each of the other bars is 6 ksi, what
is the dimension, b, of the square bar, and what is the outer
diameter, c, of the tubular member?
Q#14:The Washington Monument (Figure Q#14.) stands
555 ft. high and weighs 181,700 kips (i.e., approximately
182 million pounds). The monument was made from over
36,000 blocks of marble and granite. As shown in Fig. 1b,
the base of the monument is a square that is 665.5 in. long
on each side, and the stone walls at the base are 180 in.
thick.Determine the compressive stress that the foundation
exerts over the cross section at the base of the monument,
assuming that this normal stress is uniform.
Q#15:The pin-jointed planar truss in is subjected to a single
downward force P at joint A. All members have a cross-sectional
area of 500 mm2. The allowable stress in tension is (𝜎 𝑇)allow
300 MPa, while the allowable stress (magnitude) in compression
is (𝜎 𝐶)allow200 MPa. Determine the allowable load, 𝑃allow.

Q#16:A column in a two-story building is fabricated from
square structural tubing having the cross-sectional
dimensions shown in Fig b. Axial loads 𝑃𝐴 = 200 𝑘𝑁 and
𝑃𝐵 = 200 𝑘𝑁 are applied to the column at levels A and B,
as shown in Fig a. Determine the axial stress 𝜎1 in
segment AB of the column and the axial stress 𝜎2 in
segment BC of the column. Neglect the weight of the
column itself.
Q#17:Each member of the truss in Fig. is a solid circular
rod with diameter 𝑑 = 10 𝑚𝑚. Determine the axial stress
𝜎1 in the truss member (1) and the axial stress 𝜎2 in the
truss member (6).
Q#18:The three-part axially loaded member in Fig.
consists of a tubular segment (1) with outer diameter
(𝑑 𝑜)1 = 1.00 𝑖𝑛. and inner diameter (𝑑𝑖)1 = 0.75 𝑖𝑛. a
solid circular rod segment (2) with diameter 𝑑2 =
1.00 𝑖𝑛 ., and another solid circular rod segment (3) with
diameter 𝑑2 = 0.75 𝑖𝑛. The line of action of each of the
three applied loads is along the centroidal axis of the
member. Determine the axial stresses 𝜎1, 𝜎2, and 𝜎3 in
each of the three respective segments.

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