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Modern University For Information and Technology

Electrical Engineering Department

Lectures Notes of Strength of Materials

BIO 207

Prepared By Dr: Shaimaa Mostafa

(First Edition 2021)

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Vision

The vision of the Faculty of Engineering at MTI university is to be

a center of excellence in engineering education and scientific

research in national and global regions. The Faculty of Engineering

aims to prepare graduates meet the needs of society and contribute

to sustainable development.

Mission

The Faculty of Engineering MTI university aims to develop

distinguished graduates that can enhance in the scientific and

professional status, through the various programs which fulfill the

needs of local and regional markets. The Faculty of Engineering

hopes to provide the graduates a highly academic level to keep up

the global developments.

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COURSE OUTLINE

Chapter Title Page

Chapter (1): Introduction – Concept of Stress …………………...3

Chapter (2): Axial Loading: Shearing Stress - Bearing Stress …..10

Chapter (3): Stress and Strain – Axial Loading…………………..21

Chapter (4): Composite/Compound Bars ………………………...28

Chapter (5): Torsion………………………………….……..........37

Chapter (6): Beams under concentrated or distributed loads……..48

Chapter (7): Strain Energy and Resilience………...……………..59

Sheets…………………………………………………………….78

References:

[1] Mechanics of Materials, Ferdinand P. Beer, E. Russell Johnston, Jr., John

T. DeWolf, David F. Mazurek, Seventh Edition, 2015.

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Chapter (1)

Introduction – Concept of Stress

Contents

1.1. Concept of Stress

1.2. Review of Statics

1.3. Component Free-Body Diagram

1.4. Method of Joints

1.5. Stress Analysis

1.6. Stresses in the Members of a Structure

1.6.1. Axial Loading: Normal Stress

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1.1. Concept of Stress

The main objective of the study of mechanics of materials is to

provide the future engineer with the means of analyzing and

designing various machines and load bearing structures.

Both the analysis and design of a given structure involve the

determination of stresses and deformations. This chapter is devoted

to the concept of stress.

1.2. Review of Statics

The structure is designed to support

a 30 kN load. Perform a static

analysis to determine the internal

force in each structural member

and the reaction forces at the

supports.

1.3. Component Free-Body Diagram

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Conditions for static equilibrium

Consider a free-body diagram for the boom

substitute into the structure equilibrium

equation:

Cy = 30kN

Results:

A = 40kN → Cx = 40kN ← Cy = 30kN ↑

1.4. Method of Joints

The members are subjected to only two forces which are applied at

member ends. Joints must satisfy the conditions for static

equilibrium which may be expressed in the form of a force triangle:

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1.5. Stress Analysis

Can the structure safely support the 30 kN load?

From a statics analysis

FAB = 40 kN (compression)

FBC = 50 kN (tension)

At any section through member BC

(dBC = 20 mm), the internal force is 50

kN with a force intensity or stress of:

From the material properties for steel, the allowable stress is: σ all

= 165MPa

Conclusion: the strength of member BC is adequate.

1.6. Stresses in the Members of A Structure

1.6.1. Axial Loading: Normal Stress

The first and necessary step in the analysis of a structure is to find

forces in individual members.

The normal stress in a member of cross-sectional area A subjected

to an axial load P is obtained by:

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Axial force represents the resultant force and uniform stress distribution

Solved Example (1):

Two solid cylindrical rods AB and BC are welded

together at B and loaded as shown. Knowing that the

average normal stress must not exceed 175 MPa in rod

AB and 150 MPa in rod BC, determine the smallest

allowable values of d1 and d2.

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Solution:

Solved Example (2):

Two steel plates are to be held together by means of 16-mm diameter

high-strength steel bolts fitting snugly inside cylindrical brass

spacers. Knowing that the average normal stress must not exceed

200 MPa in the bolts and 130 MPa in the spacers, determine the

outer diameter of the spacers that yields the most economical and

safe design.

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Solution:

The upper plate is pulled down by the tensile force Pb of the bolt

The spacer pushes that plate upward with a compressive force Ps

To satisfy equilibrium: Pb = Ps

Solved Example (3):

A cast iron link, as shown in the figure is required to transmit a

steady tensile load of 45 kN. Find the tensile stress induced in the

link material at sections A-A and B-B.

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Solution:

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Chapter (2)

Axial Loading

Contents

2.1. Shearing Stress

2.2. Bearing Stress

2.3. Factor of Safety

Solved Examples

Problems

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2.1. Shearing Stress

Shearing stress is obtained when an equal and transverse forces P

and P’ are applied to a member AB. Normal stress caused by forces

perpendicular to the area on which it act, while shear stress caused

by forces parallel to the area resisting the forces.

Bolt subject to single shear

average shearing stress in case of single shear:

average shearing stress in case of double shear:

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Solved Example (1)

Compute the shearing stress in the pin at

B for the member supported as shown.

The pin diameter is 20 mm.

Solution:

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Solved Example (2)

Three steel bolts are to be used to attach the steel

plate shown to a wooden beam. Knowing that the

plate will support a 110-kN load, that the shearing

stress for the steel used is 107 MPa, determine the

required diameter of the bolts.

Solution:

Shear stress τ = 107 MPa

F

3A=

P

3π

4d2

d = √4p

3πτ

d = √4×110×1000

3π×107 = 20.8 mm

Solved Example (3)

A pull of 80 kN is transmitted from a bar X to the bar Y through a

pin. If the maximum permissible tensile stress in the bars is 100

N/mm2 and the permissible shear stress in the pin is 80 N/ mm2, find

the diameter of bars and of the pin.

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Solution:

Given: P = 80 KN = 80×103 N

σt = 100 MPa

τ = 80 MPa

Diameter of the bar

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Diameter of the pin

Problem (1)

When the force P reached 8 kN, the

wooden specimen shown failed in shear

along the surface indicated by the dashed

line. Determine the average shearing

stress along that surface at the time of failure.

2.2. Bearing Stress

Bolts, rivets, and pins create stresses on the points of contact

(bearing surfaces).

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Bearing stress obtained by:

Solved Example (4)

A 6-mm diameter pin is used at the connection c of the pedal shown.

knowing that P = 500 N, determine: (a) the average shearing stress

in the pin, (b) the nominal bearing stress in the pedal C. (c) the

nominal bearing stress in each support bracket at C.

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Solution:

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Solved Example (5)

An axial load P is supported by a short

W 8 × 40 column of cross-sectional

area A = 11.7 in.2 and is distributed to

a concrete foundation by a square

plate as shown. Knowing that the

average normal stress in the column

must not exceed 30 ksi and that the

bearing stress on the concrete

foundation must not exceed 3 ksi, determine the side a of the plate

that will provide the most economical and safe design.

Solution:

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2.3. Factor of Safety

The maximum load that a structural member or a machine

component will be allowed to carry is smaller than the ultimate

load. This smaller load is the allowable load.

The ratio of the ultimate load to the allowable load is used to

define the factor of safety:

Solved Example (6)

Link AB is to be made of a steel for

which the ultimate normal stress is

450 MPa. Determine the cross-

sectional area for AB for which the

factor of safety will be 3.5. Assume

that the link will be adequately

reinforced around the pins at A and B.

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Solution:

Problem (1)

Two loads are applied to the bracket BCD

as shown:

(a) Knowing that the control rod AB is to

be made of a steel having an ultimate

normal stress of 600 MPa, determine the

diameter of the rod for which the factor of

safety is 3.3.

(b) The pin at C is to be made of a steel

having an ultimate shearing stress of 350

MPa. Determine the diameter of the pin C

for which the factor of safety is also 3.3.

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(c) Determine the required thickness of the bracket supports at C,

knowing that the allowable bearing stress of the steel used is 300

MPa.

Problem (2)

Each of the steel links AB and CD is connected to a support and to

member BCE by 25-mm diameter steel pins acting in single shear.

Knowing that the ultimate shearing stress is 210 MPa for the steel

used in the pins and that the ultimate normal stress is 490 MPa for

the steel used in the links. Determine the allowable load P if the

factor of safety is 3.

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Chapter (3)

Stress and Strain – Axial Loading

Contents

3.1. Stress & Strain: Axial Loading

3.2. Normal Strain

3.3. Stress-Strain Test

3.4. Stress-Strain Diagram: Ductile and Brittle Materials

3.5. Hooke’s Law: Modulus of Elasticity

3.6. Deformations Under Axial Loading

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3.1. Stress & Strain: Axial Loading

Suitability of a structure or machine may depend on the

deformations in the structure as well as the stresses induced under

loading. This chapter concerned with deformation of a structural

member under axial loading.

3.2. Normal Strain

Consider a homogeneous rod BC of length L and

uniform cross section of area A subjected to a centric

axial load P, the normal stress relation become:

The normal strain is:

Where: σ: normal stress

P: applied load

A: cross-sectional area of the rod

ε: normal strain

δ: deformation of the rod (increase in length)

L: length of the rod

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3.3. Stress-Strain Test

Universal test machine for tensile test Tensile test specimen with tensile

load and

the fracture

3.4. Stress-Strain Diagram: Ductile and Brittle Materials

Stress-Strain Diagram: Ductile Materials

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Stress-Strain Diagram: Brittle Materials

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3.5. Hooke’s Law: Modulus of Elasticity

The deformations of a structural member such as a

rod, bar, or plate is due to axial loading applied to

the structure.

Consider a homogeneous rod BC of length L and

uniform cross section of area A subjected to a

centric axial load P, from Hooke’s Law the

Modulus of Elasticity or Youngs Modulus equal to:

E = 𝜎

= 𝑃𝐿

𝐴𝛿 (

𝑁.𝑚𝑚

𝑚𝑚2.𝑚𝑚 = MPa)

3.6. Deformations Under Axial Loading

From Hooke’s Law the deformation become:

δ = 𝑃𝐿

𝐴𝐸 (mm)

Where:

δ: is the deformation of the member (increase in length)

P: is the applied load

E: is the modulus of elasticity

With variations in loading, cross-section or material properties:

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Solved Example (1)

A 25 mm diameter bar is subjected to an axial

tensile load of 100 kN. Under the action of this

load a 200 mm gauge length is found to extend

0.19×10-3 mm. Determine the modulus of

elasticity for the bar material.

Solution:

Solved Example (2)

The assembly consists of three titanium (Ti-6A1-4V) rods and a

rigid bar AC. The cross-sectional area of each rod is given in the

figure. If a force of 6 kip is applied to the ring F, determine the

horizontal displacement of point F if E = 17.4×103 psi.

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Solution:

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Problem (1)

The A-36 steel rod is subjected to the loading shown. If the diameter

of the rod is 8 mm, determine the displacement of its end D.

Knowing that the modulus of elasticity E= 200 Gpa.

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Chapter (4)

Composite and Compound Bars

Contents

4.1. Non-uniform Bars

4.2. Composite/Compound Bars

4.3. Thermal Stresses

4.4. Types Of Installations and Strength

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4.1. Non-uniform Bars

Bars with cross-sections varying in steps are called non-uniform

bars. A typical bar with cross-sections varying in steps and subjected

to axial load is as shown in the figure. Let the length of three portions

be L1, L2 and L3 and the respective cross-sectional areas of the

portion be A1, A2, A3 and E be the Young’s modulus of the material

and P be the applied axial load. The lower figure shows the forces

acting on the cross-sections of the three portions.

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Hence total change in length of the bar:

δt = δ1+ δ2+ δ3 = 𝑃𝐿1

𝐴1𝐸+

𝑃𝐿2

𝐴2𝐸 +

𝑃𝐿3

𝐴3𝐸

Solved Example (1)

The bar shown in the Figure is tested in universal testing machine.

It is observed that at a load of 40 kN the total extension of the bar is

0.280 mm. Determine the Young’s modulus of the material.

Solution:

Extension of portion 1: δ1 =

Extension of portion 2: δ2 =

Extension of portion 3: δ3 =

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Problem (1)

Determine the deformation of

deflections. the steel rod shown under

the given loads.

Problem (2)

The specimen shown has been cut

from a 0.25 in. thick sheet of vinyl (E

= 0.45×106 psi) and is subjected to a

350-lb tensile load. Determine (a)

deformation of its central portion BC.

(b) the total deformation of the specimen

4.2. Composite/Compound Bars

Bars made up of two or more materials are called

composite/compound bars. They may have same length or different

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lengths as shown in the figure. The ends of different materials of the

bar are held together under loaded conditions.

Consider a member with two materials. Let the load shared by

material 1 be P1 and that by material 2 be P2. Then:

i. From equation of equilibrium of the forces, we get:

P = P1 + P2

ii. Since the ends are held securely, we get δ1 = δ2 where δ1 and

δ2 are the extension of the bars of material 1 and 2 respectively

i.e.

Using this equation, P1 and P2 can be found uniquely.

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Solved Example (2)

A compound bar of length 600 mm consists of a

strip of aluminium 40 mm wide and 20 mm thick

and a strip of steel 60 mm wide × 15 mm thick

rigidly joined at the ends. If elastic modulus of

aluminium and steel are 1 × 105 MPa and 2 × 105

MPa, determine the stresses developed in each

material and the extension of the compound bar when axial tensile

force of 60 kN acts. Take Ea = 1 × 105 MPa and Es = 2 × 105 MPa

Solution:

L = 600 mm

P = 60 kN = 60 × 1000 N

Aa = 40 × 20 = 800 mm2

As= 60 × 15 = 900 mm2

Ea = 1 × 105 MPa

Es = 2 × 105 MPa

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Solved Example (3)

A compound bar consists of a circular rod of steel

of 25 mm diameter rigidly fixed into a copper tube

of internal diameter 25 mm and external diameter

40 mm. If the compound bar is subjected to a load

of 120 kN, find the stresses developed in the two materials.

Solution:

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4.3. Thermal Stresses

Every material expands when temperature rises and

contracts when temperature falls. It is established

experimentally that the change in length δ is directly

proportional to the length of the member L and

change in temperature (ΔT). Thus

δ ∝ (ΔT) L

δ = α (ΔT) L

The constant of proportionality α is called

coefficient of thermal expansion and is defined

as change in unit length of material due to unit

change in temperature.

the support force P develops to keep it at the

original position. Magnitude of this force is such that contraction is

equal to free expansion:

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𝑃𝐿1

𝐴1𝐸 = α (ΔT) L

σ = E α (ΔT) where, σ is the stress in this case.

Solved Example (4)

The composite bar shown is rigidly fixed

at the ends A and B. Determine the

reaction developed at ends when the

temperature is raised by 18°C. Knowing:

Ea = 70 kN/mm2

Es = 200 kN/mm2

αa = 11 × 10–6/°C

αs = 12 × 10–6/°C

Solution:

Free expansion = α (ΔT) La + α (ΔT) Ls =

= (11 × 10–6 × 18 × 1500) + (12 × 10–6 × 18 × 3000) =

0.945 mm

δ = (𝑃𝐿

𝐴𝐸)𝑎 + (

𝑃𝐿

𝐴𝐸 ) 𝑠

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Chapter (5)

Torsion

Contents

5.1. Torsional Loads on Circular Shafts

5.2. The Shear Stress Formula

5.3. Angle of Twist in Elastic Range

5.4. Power Transmitted by a Shaft

5.5. Stress in Rotating Arms (Fatigue)

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5.1. Torsional Loads on Circular Shafts

Torsion interested in stresses and strains of circular shafts subjected

to twisting couples or torques. Turbine exerts torque T on the shaft,

then the shaft transmits the torque to the generator. Generator

creates an equal and opposite torque T’. So, shaft subjected to equal

and opposite torque, producing shear stress τ.

5.2. The Shear Stress Formula

Torque applied to shaft produces shearing stresses

on the faces perpendicular to the axis. The

shearing stress results are known as:

τ = 𝑇⋅𝐶

𝐼𝑝

Where,

τ: shearing stress

T: twisting moment (torque)

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C: shaft radius

Ip: polar moment of area (second moment)

Ip = 𝜋

32𝑑4 =

𝜋

2𝐶4 (solid shaft)

Ip = 𝜋

32(𝑑0

4 − dⅈ4) (Tubular shaft)

➢ Therefore, total torque transmitted by a circular solid

shaft:

τmax = 𝑇⋅𝐶

𝐼𝑝 =

𝑇⋅(𝑑

2)

𝜋

32𝑑4

= 16 𝑇

𝜋𝑑3 solid shaft

τmax = 16 𝑇𝐷

𝜋(𝐷4− 𝑑4

) hollow shaft

Solved Example (1)

The solid rod AB has a diameter dAB 60

mm and is made of a steel for which the

allowable shearing stress is 85 MPa. The

pipe CD, which has an outer diameter of 90

mm and a wall thickness of 6 mm, is made

of an aluminum for which the allowable

shearing stress is 54 MPa. Determine the

largest torque T that can be applied at A.

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Solution:

Rod AB:

τ = 85 MPa C = 30 mm

Ip = 𝜋

2𝐶4 =

𝜋

2(30)4 = 1.3×106 mm4

T = τ⋅𝐼𝑝

𝐶 =

85 ×1.3 × 106

30 = 3.6×106 N.mm

Pipe CD:

τ = 54 MPa Co = 45 mm

Ci = Co - t = 45 – 6 = 39 mm

Ip = 𝜋

2(𝐶0

4 − Cⅈ4) = 𝜋

2(454 − 394) =2.8×106 mm4

T = τ⋅𝐼𝑝

𝐶 =

54 ∗1.3 * 106

45 = 3.3×106 N.mm

Allowable torque is the smaller value: T = 3.3×106 N.mm

Problem (1)

Determine the torque T that causes a

maximum shearing stress of 70 MPa in the

cylindrical steel shaft shown.

Problem (2)

A solid steel shaft is to transmit a torque of 10 KN.m. If the shearing

stress is not exceed 45 MPa, find the minimum diameter of the shaft.

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5.3. Angle of Twist in Elastic Range

the angle of twist of the shaft is

proportional

to the applied torque and to the shaft

length.

φ ∝ T φ ∝ L

𝜙 =𝑇 ⋅ 𝐿

𝐼𝑝 ⋅ 𝐺 (𝑟𝑎𝑑)

Where,

𝜙 : angle of twist

T: twisting moment (torque)

L: shaft length

Ip: polar moment of area

G: modulus of rigidity of the

shaft material

Solved Example (2)

The torques shown are exerted on pulleys B, C, and D. Knowing

that the shaft is made of steel (G = 27 GPa), determine the angle of

twist between (a) B and C, (b) B and D.

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Solution:

Shaft BC:

C = 15 mm

Ip = 𝜋

2C4 =

𝜋

2 (15)4 = 79.5×103 mm4

LBC = 800 mm

T BC = + 400×103 N.mm

G = 27×103 MPa

φBC =𝑇𝐵𝐶⋅𝐿𝐵𝐶

𝐼𝑝⋅𝐺

φBC = 400×103 ×800

79.5×103×27×103 = 0.14 rad×180/π = 8.5

o

(b) Shaft CD:

C = 18 mm

Ip = 𝜋

2𝐶4 =

𝜋

2(18)4 = 164.8×103 mm4

LCD = 1000 mm

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T CD = - 500×103 N.mm

G = 27×103 MPa

φCD =𝑇𝐶𝐷⋅𝐿𝐶𝐷

𝐼𝑝⋅𝐺

φCD =− 500×103×1000

164.8×103×27×103 = - 0.11 rad

φBD = φBC + φCD = 0.14 - 0.11 = 0.03 rad

φBD = 0.03×180/ π = 1.7o

Problem (3)

A steel shaft 3 ft long that has a diameter of 4 in is subjected to a

torque of 15 kip·ft. Determine the maximum shearing stress and the

angle of twist. Use G = 12 × 106 psi

5.4. Power Transmitted by a Shaft

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Let us consider that N is the R.P.M of the shaft and ω is the angular

velocity of the shaft. We have already secured here the torque

transmitted by circular solid shaft and it is mentioned above. Now

we will find here the expression for power transmitted by solid

circular shaft as mentioned here.

Power = work done in KN.m/second

Power = force × distance = T× ω = T×2πN

Solved Example (3)

A hollow shaft is to transmit 200 kW at 80 r.p.m. If the shear stress

is not to exceed 60 MPa and internal diameter is 0.6 of the external

diameter, find the diameters of the shaft.

Solution:

Given : Power (P) = 200 kW Speed of shaft (N) = 80 r.p.m.

Maximum shear stress (τ) = 60 MPa = 60 N/mm2

Inner diameter of the shaft (d) = 0.6D (D is the outer

diameter).

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Problem (4)

A steel marine propeller shaft 14 in. in diameter and 18 ft long is

used to transmit 5000 hp at 189 rpm. If G = 12 × 106 psi, determine

the maximum shearing stress.

5.5. Stress in Rotating Arms (Fatigue)

When a metal is subjected to repeated cycles of stress or strain, it

causes its structure to break down, ultimately leading to fracture.

This behavior is called fatigue, and it is usually responsible for a

large percentage of failures in connecting rods and crankshafts of

engines; steam or gas turbine blades; connections or supports for

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bridges, railroad wheels, and axles; and other parts subjected to

cyclic loading. Fatigue properties are shown on S-N diagrams.

Solved Example (4)

The fatigue data for a brass alloy are given

as follows:

(a) Make an S–N plot (stress amplitude

versus logarithm cycles to failure) using

these data.

(b) Determine the fatigue strength at 5×105

cycles.

(c) Determine the fatigue life for 200 MPa.

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Solution:

52

Chapter (6)

Beams under concentrated or distributed loads

Contents

6.1. Pure Bending

6.2. Common Load Types for Beams

6.3. Centroid and Moment of Inertia

6.4. Parallel Axes Theorem of Moment of Inertia

6.5. Stress Due to Bending

6.6. Bending Stress in Straight Beams

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6.1. Pure Bending

Pure Bending: Prismatic members subjected to equal and opposite

couples

acting in the same longitudinal plane

6.2. Common Load Types for Beams

Several common loading types for beams and frames are shown

in the figure:

54

6.3. Centroid and Moment of Inertia

Centroid (for any object in dimensional space): is the mean position

of all the points in all of the coordinates directions, it can be

determined by:

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Centroid (Center of mass): �̅� =∑ 𝑥⋅𝐴

∑ 𝐴

Centroid (Center of mass): �̅� =∑ 𝑌⋅𝐴

∑ 𝐴

Moment of inertia of the rectangular cross

section:

𝐼𝑥 =1

12𝑏ℎ3 𝐼𝑦 =

1

12ℎ𝑏3

Moment of inertia for circle: I = (π/64)d4 = (π/4)r4

6.4. Parallel Axes Theorem of Moment of

Inertia

Symmetric around y, get Iy: 𝐼𝑦 =1

12ℎ𝑏3

If symmetric around y we get Ix using

Parallel axes theory:

Ix = Σ [Ix + A(Y- �̅�)]= [Ix1 + A1(Y1- �̅�)2]+ [Ix2 + A2(Y2- �̅�)2]

Solved Example (1)

Determine the moment of inertia of the shown figure:

56

Solution:

For rectangle cross sectional area

Solved Example (2)

Calculate the location of the section centroid

and moment of inertia of the shown bar.

Solution:

location of the section centroid:

and moment of inertia of the shown bar, symmetric around x and y:

centroid �̅� and �̅� lies in the midpoint of the bar.

�̅� = 50 mm �̅� = 40 mm

Moment of inertia:

For circle I = (π/64)d4 = (π/4)r4

57

6.5. Stress Due to Bending

Formula to find the maximum tensile and compressive normal

stresses due to bending:

σm = 𝑀

𝐼⋅ 𝑦

σm: max bending stress

M: Bending moment

I: moment of inertia

y: section centroid

Solved Example (3)

Knowing that the couple shown acts in a vertical plane, determine

the: stress at (a) point A, (b) point B.

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Solution:

Solved Example (4)

Based on the given cross section geometry, calculate maximum

tensile and compressive normal stresses at point A and point B.

Solution:

Iy =1

12hb3 = (

1

12)(20)(903) + (

1

12)(40)(303)

Iy = 1305×103 mm4

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𝑌 =(𝟐𝟎×𝟗𝟎)(𝟓𝟎)+(𝟑𝟎×𝟒𝟎)(𝟐𝟎)

(𝟐𝟎×𝟗𝟎)+(𝟑𝟎×𝟒𝟎) = 38 mm

If symmetric around y we get Ix using

Parallel axes theory:

Ix = Σ [Ix + A(Y- �̅�)] = [Ix1 + A1(Y1- �̅�)2]+ [Ix2 + A2(Y2- �̅�)2]

Ix = [1

12(90×203) + (20×90)(50-38)2]+ [

1

12(30×403) + (30×40)(20-

38)2]

Ix = 868×103 mm4

σA = 𝑀

𝐼𝑥⋅ 𝑦A =

3×106

868×103 (22) = 76 MPa

σB = - 𝑀

𝐼𝑥⋅ 𝑦B = -

3×106

868×103 (38) = −131.3 MPa

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Solved Example (5)

The aluminum machine part is

subjected to a moment of 75 N.m.

Determine the maximum tensile and

compressive bending stress in the part.

Solution:

6.6. Bending Stress in Straight Beams

In engineering practice, the machine parts of structural members

may be subjected to static or dynamic loads which cause bending

61

stress in the sections. Consider a straight beam subjected to a

bending moment M as shown in the figure,

The bending equation is given by:

σm = 𝑀

𝐼⋅ 𝑦 =

𝐸

𝑅⋅ 𝑦

Solved Example (6)

A pump lever rocking shaft shown, the pump lever exerts

forces of 25 kN and 35 kN concentrated at 150 mm and 200

mm from the left and right hand bearing respectively. Find

the diameter of the central portion of the shaft if the stress is

not to exceed 100 MPa. Also find the radius of curvature of

62

the beam designed from a material having modulus of

elasticity 400 MPa.

Solution:

MA = 0

25×150 + 35×750 – RB×950 = 0 RB = 31.58 KN

Σ Fy = 0

31.58 – 35 – 25 + RA = 0 RA = 28.42 KN

MC = RA×150 = 28.42×103×150 = 4.26×106 N.mm

MD = RB×200 = 31.58×103×200 = 6.32×106 N.mm

Maximum bending moment id at point D

M = 6.32×106 N.mm

I =π

32𝑑4 y =

d

2

(a) Bending stress σm = 𝑀

𝐼⋅ 𝑦

100 MPa = 6.32×106

π

32𝑑4 ⋅

d

2

63

d3 = 643.2×103 d = 86.3 mm = 90 mm

(b) σm = 𝐸

𝑅⋅ 𝑦 100 MPa =

400

𝑅⋅

90

2

R = 180 mm

64

Chapter (7)

Strain Energy and Resilience

Contents

7.1. Toughness and Resilience

7.2. Strain Energy

7.3. Shock (impact) loading

7.4. Impact Testing Techniques

65

7.1. Toughness and Resilience

The strain energy resulting from

setting ε = εR is the modulus of

toughness.

The strain energy resulting from setting

σ = σY is the modulus of resilience.

7.2. Strain Energy

The strain energy is defined as the energy stored in any object which

is loaded within its elastic limits. In other words, the strain energy

is the energy stored in anybody due to its

deformation. The strain energy is also

known as Resilience. The unit of strain

energy is N-m or Joules.

A uniform rod is subjected to a slowly

increasing load:

66

Work Done = Elastic strain energy

dU = Pdx = elementary work

The total work done by the load for a deformation x1,

In the case of a linear elastic deformation,

67

7.3. Shock (impact) loading

Consider a rod BD with a uniform cross

section that is hit at end B by a body of

mass m moving with a velocity v0. As the

rod deforms under the impact (shock

load), stresses develop within the rod and

reach a maximum value σm.

To determine the maximum stress σm Rod subject to impact

loading.

- Assume that the kinetic energy is

transferred entirely to the structure:

Assume that the stress-strain diagram obtained from a static test is

also valid under impact loading.

For a rod of uniform cross-section

68

Solved Example (1)

One of the two bolts need to support a sudden tensile loading. To choose it is

necessary to determine the greatest amount of strain energy each bolt can

absorb. Bolt A has a diameter of 20 mm for 50 mm length and a root

diameter of 18 mm for 6 mm length. Bolt B has 18 mm diameter for 56-mm

length. Take E = 210 GPa and σy = 310 GPa

69

Solution:

Solved Example (2)

Body of mass m with velocity v0 hits the end of the nonuniform rod

BCD. Knowing that the diameter of the portion BC is twice the

diameter of portion CD, determine the maximum value of the

normal stress in

the rod.

70

Solution:

Due to the change in diameter, the normal stress distribution is

nonuniform:

Find the static load Pm which produces the same strain energy as

the impact

Evaluate the maximum stress resulting from the static load Pm

71

7.4. Impact Testing Techniques

Impact testing techniques were established so as to ascertain the

fracture characteristics of materials at high loading rates. Two

standardized tests, the Charpy and Izod test machines were designed

and are still used to measure the impact energy as shown in the

figure. For both Charpy and Izod, the specimen is in the shape of a

bar of square cross section, into which a V-notch is machined. The

load is applied as an impact blow from a weighted pendulum

hammer that is released from a cocked position at a fixed height h.

The pendulum continues its swing, rising to a maximum height h_,

which is lower than h. The energy absorption, computed from the

difference between h and h_, is a measure of the impact energy.

72

73

Solved Example (3)

Following is tabulated data that

were gathered from a series of

Charpy impact tests on a ductile

cast iron.

(a) Plot the data as impact energy

versus temperature.

(b) Determine a ductile-to-brittle

transition

temperature as that temperature corresponding to the average of the

maximum and minimum impact energies.

(c) Determine a ductile-to-brittle transition temperature as that

temperature at which the impact energy is 80 J.

74

Solution:

75

SHEETS

76

Sheet (1)

1. A hollow steel tube with an inside diameter of 80 mm must carry an

axial tensile load of 330 kN. Determine the smallest allowable outside

diameter of the tube if the working stress is 110 MN/m2.

2. A hollow circular post ABC supports a load P1 = 7.5 KN acting at the

top. A second load P2 is uniformly distributed around the cap plate at

B. The diameters and thicknesses of the upper and lower parts of the

post are dAB= 30 mm, tAB= 5 mm, dBC= 60 mm and tBC= 10 mm.

(a) Calculate the normal stress in the upper part of the post AB.

(b) If it is desired that the lower part of the post have the same

compressive stress as the upper part, what should be the magnitude

of the load P2?

(c) If P1 remains at 7.5 KN and P2 is now set at 10 KN, what new

thickness of BC will result in the same compressive stress in both

parts?

77

3. Find the maximum allowable value of P for the column. The cross-

sectional areas and working stresses (σw) are shown in the figure

below.

4. The built-up shaft consists of a pipe AB and solid rod BC. The pipe has

an inner diameter of 20 mm and an outer diameter of 28 mm. The rod

has a diameter of 12 mm. Determine the average normal stress at points

D and E.

78

5. Two solid cylindrical rods AB and BC are welded together at B and

loaded as shown. Knowing that d1 = 30 mm and d2 = 50 mm, find the

average normal stress at rods AB and BC.

6. A strain gage located at C on the surface of bone AB indicates that the

average normal stress in the bone is 3.80 MPa when the bone is

subjected to two 1200-N forces as shown. Assuming the cross section

of the bone at C to be annular and knowing that its outer diameter is 25

mm, determine the inner diameter of the bone’s cross section at C.

7. The cross-sectional area of bar ABCD is 600 mm2. Determine the

maximum normal stress in the bar

8. For the Pratt bridge truss and

loading shown, the average

normal stress in member CE not

exceed 21 Ksi, Determine the

cross-sectional area of the

member.

79

9. Two steel plates are to be held together by means of 16 mm diameter

high-strength steel bolts fitting snugly inside cylindrical brass spacers.

Knowing that the average normal stress must not exceed 200 MPa in

the bolts and 130 MPa in the spacers, determine the outer diameter of

the spacers that yields the most economical and safe design.

10. The 1000-kg uniform bar AB is suspended from two cables AC and

BD; each with cross-sectional area 400 mm2. Find the magnitude P and

location x of the largest additional vertical force that can be applied to

the bar. The stresses in AC and BD are limited to 100 MPa and 50 MPa,

respectively.

80

Sheet (2)

1. What force is required to punch a 20

mm diameter hole in a plate that is 25

mm thick? The shear strength is 350

MN/m².

2. Compute the shearing stress in the pin

at B for the member supported as shown. The pin diameter is 20 mm.

3. A circular hole is to be punched o\in a plate having

a shearing strength of 275 MPa. The compressive

stress in the punch is limited to 345 MPa. (a)

Compute the maximum thickness of plate in

which a hole 64 mm diameter can be punched.

(b) If the plate is 6.5 mm thick, determine the

diameter of the smallest hole that can be

punched.

4. A load P is applied to a steel rod supported as shown by an aluminum

plate into which a 12-mm-diameter hole has been drilled. Knowing that

81

the shearing stress must not exceed 180 MPa

in the steel rod and 70 MPa in the aluminum

plate, determine the largest

load P that can be applied to the rod.

5. The yoke and rod connection are subjected to

a tensile force of 5 kN. Determine the average

normal stress in each rod and the average shear

stress in the pin at A.

6. An axial load P is supported by a short W250 x 67 column of cross-

sectional area A = 8580 mm2 and is distributed to a concrete foundation

by a square plate as shown. Knowing that the average normal stress in

the column must not exceed 150 MPa and that the bearing stress on the

concrete foundation must not exceed 12.5 MPa, determine the side a of

the plate that will provide the most economical and safe design.

7. Assume that the axial load P applied to the lap joint is distributed

equally among the three 20-mm-diameter

rivets. What is the maximum load P that can be

applied if the allowable stresses are 40 MPa for

shear in rivets, 90 MPa for bearing between a

plate and a rivet, and 120 MPa for tension in

the plates?

82

8.

9. Each of the four vertical links has an 8x36 mm uniform rectangular

cross section and each of the four pins has a 16-mm diameter.

Determine (a) the average shear stress at pin C (b) the average bearing

stress at pin C in link CE (c) the average bearing stress in member ABC

with uniform rectangular area 10x50 mm.

83

Sheet (3)

1. Link BC is 6 mm thick and is made of a steel with

a 450-MPa ultimate strength in tension. What

should be its width w if the structure shown is being

designed to support a 20-kN load P with a factor of

safety of 3?

2. Each of the two vertical links CF connecting the

two horizontal members AD and EG has a

10*40-mm uniform rectangular cross section and

is made of a steel with an ultimate strength in

tension of 400 MPa, while each of the pins at C

and F has a 20-mm diameter and are made of a

steel with an ultimate strength in shear of 150

MPa. Determine the overall factor of safety for

the links CF and the pins connecting them to the

horizontal members.

3. Link AB is to be made of a steel for which

the ultimate normal stress is 450 MPa.

Determine the cross-sectional area for AB

for which the factor of safety will be 3.5.

Assume that the link will be adequately

reinforced around the pins at A and B.

84

4. The steel bolts are to be used to attach the steel

plate shown to a wooden beam. Knowing that

the plate will support a load P = 110 KN, that

the ultimate shearing stress for the steel used

is 360 MPa, and that a factor of safety of 3.35

is desired, determine the required diameter of

the bolts.

5. Member ABC, which is supported by a pin

and bracket at C and a cable BD, was

designed to support the 16-kN load P as

shown. Knowing that the ultimate load for

cable BD is 100 kN, determine the factor of

safety with respect to cable failure.

6. Members AB and AC of the truss shown consist of

bars of square cross section made of the same

alloy. It is known that a 20-mm square bar of the

same alloy was tested for failure and that an

ultimate load of 120 kN was recorded. If bar AB

has a 15-mm square cross section, determine (a)

the factor of safety of bar AB, (b) the dimensions

of the cross section of bar AC if it is to have the

same factor of safety as bar AB.

85

7. Two wooden members shown, which support

20 KN load, are joined by plywood splices

fully glued on the surfaces in contact. The

ultimate shearing stress in the glue is 2.8 MPa

and the clearance between the members is 8

mm. Determine the required length L of each

splice if a factor of safety of 3.5 is to be

achieved.

8. Each of the steel links AB and CD is connected

to a support and to member BCE by 0.5-in.

diameter steel pins acting in single shear.

Knowing that the ultimate shearing stress is 30 ksi

for the steel used in the pins and that the ultimate

normal stress is 70 ksi for the steel used in the

links, determine the allowable load P if an overall

factor of safety of 3 is desired. (Note that the links

are not reinforced around the pin holes.

9. The hydraulic cylinder CF, which partially

controls the position of rod DE, has been locked

in the position shown. Member BD is 5/8 in. thick

and is connected to the vertical rod by a 3/8 -in.-

diameter bolt. Determine (a) the average shearing

stress in the bolt, (b) the bearing stress at C in

member BD.

86

Sheet (4)

1. An 80-m-long wire of 5-mm diameter is made of a steel with E = 200 GPa and an

ultimate tensile strength of 400 MPa. If a factor of safety of 3.2 is desired,

determine (a) the largest allowable tension in the wire, (b) the corresponding

elongation of the wire.

2. The aluminum rod ABC E = (10.1×106 psi), which consists of

two cylindrical portions AB and BC, is to be replaced with a

cylindrical steel rod DE, E = (29×106 psi) of the same overall

length. Determine the minimum required diameter d of the steel

rod if its vertical deformation is not to exceed the deformation

of the aluminum rod under the same load and if the allowable

stress in the steel rod is not to exceed 24 ksi.

3. The specimen shown is made from a 1 in. diameter

cylindrical steel rod with two 1.5 in. outer-diameter

sleeves bonded to the rod as shown. Knowing that

E=29×106 psi, determine (a) the load P so that the total

deformation is 0.002 in., (b) the corresponding

deformation of the central portion BC.

4. Both portions of the rod ABC are made of an aluminum for which

E = 70 GPa. Knowing that the magnitude of P is 4 kN, determine

(a) the value of Q so that the deflection at A is zero, (b) the

corresponding deflection of B.

87

5. For the steel truss (E = 200 GPa) and loading

shown, determine the deformations of the

members AB and AD, knowing that their cross-

sectional areas are 2400 mm2 and 1800 mm2,

respectively.

88

Sheet (5)

1. For the 60-mm-diameter solid cylinder and

loading shown,

(a) Determine the maximum shearing stress.

(b) Determine the inner diameter of the

hollow cylinder, of 80-mm outer diameter, for

which the maximum stress is the same as in

part (a).

2. A circular steel shaft (G = 800

GPa) and aluminum shaft (G = 28

GPa) are attached and loaded as

shown, determine (a) the angle of

rotation of the section D with

respect to section A. (b) the

torsional shear stress at point E.

3. The torques shown are exerted on

pulleys A, B, and C. Knowing that

both shafts are solid, determine the

maximum shearing stress in (a) shaft

AB, (b) shaft BC.

89

4. Knowing that each of the shafts AB,

BC and CD consists of a solid circular

rod, determine (a) the shaft in which

the maximum shearing stress occurs,

(b) the magnitude of that stress.

5. A torque T = 3 kN.m is applied to the

solid bronze cylinder shown. Determine (a)

the maximum shearing stress, (b) the

shearing stress at point D, which lies on a

15-mm-radius circle drawn on the end of the

cylinder, (c) the percent of the torque

carried by the portion of the cylinder

within the 15-mm radius.

6. The steel shaft of a socket wrench has

a diameter of 8 mm. and a length of

200 mm. If the allowable stress in

shear is 60 MPa, what is the

maximum permissible torque Tmax

that may be exerted with the wrench?

Through what angle of twist in

degree will the shaft twist under the

action of the maximum torque? (Assume G = 78 GPa).

90

Sheet (6)

1. Knowing that the couple shown acts in a vertical

plane, determine the stress at (a) point A, (b) point

B.

2. A beam of the cross section shown is extruded from

an aluminum alloy for which σu = 450 MPa. Using a

factor of safety of 3.00, determine the largest couple

that can be applied to the beam when it is bent about

the z-axis.

3. Knowing that the couple shown acts in a vertical

plane, determine the stress at (a) point A, (b)

point B.

4. A nylon spacing bar has the cross section shown.

Knowing that the allowable stress for the grade of nylon

used is 24 MPa, determine the

largest couple Mz that can be applied to the bar.

91

5. Knowing that the couple shown acts in a vertical plane, determine the stress at

(a) point A, (b) point B.

6. Knowing that for the extruded beam shown the

allowable stress is 120 MPa in tension and 150 MPa

in compression, determine the largest couple M that

can be applied.