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1. Stress

CHAPTER OBJECTIVES

• Review important principles of

statics

• Use the principles to determine

internal resultant loadings in a

body

• Introduce concepts of normal

and shear stress

• Discuss applications of analysis and design

members subjected to an axial load or direct shear

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1. Stress

CHAPTER OUTLINE

1.Introduction

2.Equilibrium of a deformable body

3.Stress

4.Average normal stress

5.Average shear stress

6.Allowable stress

7.Design of Simple Connections

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1. Stress

1.1 Mechanics

BRANCHES OF MECHANICS

Statics

Dynamics

Rigid Bodies

(Things that do not change shape)

Deformable Bodies

(Things that do change shape)

Incompressible

Compressible

Fluids

Mechanics

Type title here

• Statics – Equilibrium of bodies

At rest

Move with constant velocity

• Dynamics – Accelerated motion of bodies

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1. Stress

Mechanics of materials

• A branch of mechanics

• It studies the relationship of

– External loads applied to a deformable body,

and

– The intensity of internal forces acting within the

body

• are used to compute deformations of a body

• Study body’s stability when external forces are

applied to it

1.1 INTRODUCTION

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1. Stress

1.2 EQUILIBRIUM OF A DEFORMABLE BODY

External loads

• Surface forces

– Area of contact

– Concentrated force

– Linear distributed force

– Centroid C

• Body force (e.g., weight)

Center of gravity

A review of some main principles of static

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1. Stress

Support reactions

• for 2D problems

1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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1. Stress

Equations of equilibrium

• For equilibrium

– balance of forces

– balance of moments

• Draw a free-body diagram to account for all

forces acting on the body

• Apply the two equations to achieve equilibrium

state

∑

F = 0

∑

M

O

= 0

1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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1. Stress

Internal resultant loadings

• Define resultant force (F

R

) and moment (M

Ro

) in 3D:

– Normal force, N

– Shear force, V

– Torsional moment

or torque, T

– Bending moment, M

1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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1. Stress

Internal resultant loadings

• For coplanar loadings:

– Normal force, N

– Shear force, V

– Bending moment, M

1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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1. Stress

Internal resultant loadings

• For coplanar loadings:

– Apply ∑ F

x

= 0 to solve for N

– Apply ∑ F

y

= 0 to solve for V

– Apply ∑ M

O

= 0 to solve for M

1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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1. Stress

Procedure for Analysis

• Support Reactions: Method of sections

1.Choose segment to analyze

2.Determine Support Reactions

3.Draw free-body diagram for whole body

4.Apply equations of equilibrium

1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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1. Stress

Procedure for analysis

• Free-body diagram

1.Keep all external loadings in exact locations

before “sectioning”

2.Indicate unknown resultants, N, V, M, and T

at the section, normally at centroid C of

sectioned area

3.Coplanar system of forces only include N, V,

and M

4.Establish x, y, z coordinate axes with origin at

centroid

1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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1. Stress

Procedure for analysis

• Equations of equilibrium

1.Sum moments at section, about each

coordinate axes where resultants act

2.This will eliminate unknown forces N and V,

with direct solution for M (and T)

3.Resultant force with negative value implies

that assumed direction is opposite to that

shown on free-body diagram

1.2 EQUILIBRIUM OF A DEFORMABLE BODY

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1. Stress

EXAMPLE 1.1

Determine resultant loadings acting on cross

section at C of beam.

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1. Stress

EXAMPLE 1.1 (SOLN)

Support Reactions

• Consider segment CB

Free-Body Diagram:

• Keep distributed loading exactly where it is on

segment CB after “cutting” the section.

• Replace it with a single resultant force, F.

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1. Stress

EXAMPLE 1.1 (SOLN)

Intensity (w) of loading at C (by proportion)

w/6 m = (270 N/m)/9 m

w = 180 N/m

F = ½ (180 N/m)(6 m) = 540 N

F acts 1/3(6 m) = 2 m from C.

Free-Body Diagram:

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1. Stress

EXAMPLE 1.1 (SOLN)

Equilibrium equations:

∑ F

x

= 0;

∑ F

y

= 0;

∑ M

c

= 0;

− N

c

= 0

N

c

= 0

V

c

− 540 N = 0

V

c

= 540 N

−M

c

− 504 N (2 m) = 0

M

c

= −1080 N∙m

+

+

+

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1. Stress

EXAMPLE 1.1 (SOLN)

Equilibrium equations:

Negative sign of M

c

means it acts in the opposite

direction to that shown below

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1. Stress

EXAMPLE 1.5

Mass of pipe = 2 kg/m,

subjected to vertical

force of 50 N and couple

moment of 70 N∙m at

end A. It is fixed to the

wall at C.

Determine resultant internal loadings acting on cross

section at B of pipe.

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1. Stress

EXAMPLE 1.5 (SOLN)

Support Reactions:

• Consider segment AB,

which does not involve

support reactions at C.

Free-Body Diagram:

• Need to find weight of

each segment.

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1. Stress

EXAMPLE 1.5 (SOLN)

W

BD

= (2 kg/m)(0.5 m)(9.81 N/kg)

= 9.81 N

W

AD

= (2 kg/m)(1.25 m)(9.81 N/kg)

= 24.525 N

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1. Stress

EXAMPLE 1.5 (SOLN)

Equilibrium equations:

∑ F

x

= 0;

∑ F

y

= 0;

(F

B

)

x

= 0

(F

B

)

y

= 0

∑ F

z

= 0;

(F

B

)

z

− 9.81 N − 24.525 N − 50 N = 0

(F

B

)

z

= 84.3 N

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1. Stress

EXAMPLE 1.5 (SOLN)

Equilibrium Equations:

∑ (M

B

)

x

= 0;

(M

c

)

x

+ 70 N∙m− 50 N (0.5 m) − 24.525 N (0.5 m)

− 9.81 N (0.25m) = 0

(M

B

)

x

= − 30.3 N∙m

∑ (M

B

)

y

= 0;

(M

c

)

y

+ 24.525 N (0.625∙m) + 50 N (1.25 m) = 0

(M

B

)

y

= − 77.8 N∙m

∑(M

B

)

z

= 0;

(M

c

)

z

= 0

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1. Stress

EXAMPLE 1.5 (SOLN)

N

B

= (F

B

)

y

= 0

V

B

= √ (0)

2

+ (84.3)

2

= 84.3 N

T

B

= (M

B

)

y

= 77.8 N∙m

M

B

= √ (30.3)

2

+ (0)

2

= 30.3 N∙m

The direction of each moment is determined

using the right-hand rule: positive moments

(thumb) directed along positive coordinate axis

Equilibrium Equations:

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1. Stress

1.3 STRESS

Concept of stress

• To obtain distribution of force acting over a

sectioned area

• Assumptions of material:

1.It is continuous (uniform distribution of matter)

2.It is cohesive (all portions are connected

together)

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1. Stress

1.3 STRESS

Concept of stress

• Consider ∆A in figure below

• Small finite force, ∆F acts on ∆A

• As ∆A →0, ∆ F →0

• But stress (∆F / ∆A) →finite limit (∞)

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1. Stress

Normal stress

• Intensity of force, or force per unit area, acting

normal to ∆A

• Symbol used for normal stress,is σ (sigma)

• Tensile stress: normal force “pulls” or “stretches”

the area element ∆A

• Compressive stress: normal force “pushes” or

“compresses” area element ∆A

1.3 STRESS

σ

z

=

lim

∆A →0

∆F

z

∆A

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1. Stress

Shear stress

• Intensity of force, or force per unit area, acting

tangent to ∆A

• Symbol used for normal stress is τ (tau)

1.3 STRESS

τ

zx

=

lim

∆A →0

∆F

x

∆A

τ

zy

=

lim

∆A →0

∆F

y

∆A

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1. Stress

General state of stress

• Figure shows the state of stress

acting around a chosen point in a

body

Units (SI system)

• Newtons per square meter (N/m

2

)

or a pascal (1 Pa = 1 N/m

2

)

• kPa = 10

3

N/m

2

(kilo-pascal)

• MPa = 10

6

N/m

2

(mega-pascal)

• GPa = 10

9

N/m

2

(giga-pascal)

1.3 STRESS

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1. Stress

1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

Examples of axially loaded bar

• Usually long and slender structural members

• Truss members, hangers, bolts

• Prismatic means all the cross sections are the same

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1. Stress

Assumptions

1.Uniform deformation: Bar remains straight before

and after load is applied, and cross section

remains flat or plane during deformation

2.In order for uniform deformation, force P be

applied along centroidal axis of cross section

1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

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1. Stress

Average normal stress distribution

σ = average normal stress at any

point on cross sectional area

P = internal resultant normal force

A = x-sectional area of the bar

1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

F

Rz

= ∑ F

xz

∫ dF = ∫

A

σ dA

P = σA

+

P

A

σ =

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1. Stress

Equilibrium

• Consider vertical equilibrium of the element

∑ F

z

= 0

σ (∆A) − σ’ (∆A) = 0

σ = σ’

Above analysis

applies to members

subjected to tension

or compression.

1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

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1. Stress

Maximum average normal stress

• For problems where internal force P and x-

sectional A were constant along the longitudinal

axis of the bar, normal stress

σ

= P/A is also

constant

• If the bar is subjected to several external loads

along its axis, change in x-sectional area may

occur

• Thus, it is important to find the maximum

average normal stress

• To determine that, we need to find the location

where ratio P/A is a maximum

1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

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1. Stress

Maximum average normal stress

• Draw an axial or normal force diagram (plot of

P vs. its position x along bar’s length)

• Sign convention:

– P is positive (+) if it causes tension in the

member

– P is negative (−) if it causes compression

• Identify the maximum average normal stress

from the plot

1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

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1. Stress

Procedure for Analysis

Average normal stress

• Use equation of σ = P/A for x-sectional area of a

member when section subjected to internal

resultant force P

Axially loaded members

• Internal Loading:

• Section member perpendicular to its longitudinal

axis at pt where normal stress is to be determined

• Draw free-body diagram

• Use equation of force equilibrium to obtain

internal axial force P at the section

1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

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1. Stress

Procedure for Analysis

Axially loaded members

• Average Normal Stress:

• Determine member’s x-sectional area at the

section

• Compute average normal stress σ = P/A

1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR

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1. Stress

EXAMPLE 1.6

Bar width = 35 mm, thickness = 10 mm

Determine max. average normal stress in bar when

subjected to loading shown.

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1. Stress

EXAMPLE 1.6 (SOLN)

Internal loading

Normal force diagram

By inspection, largest

loading area is BC,

where P

BC

= 30 kN

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1. Stress

EXAMPLE 1.6 (SOLN)

Average normal stress

σ

BC

=

P

BC

A

30(10

3

) N

(0.035 m)(0.010 m)

=

= 85.7 MPa

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1. Stress

EXAMPLE 1.8

Specific weight γ

st

= 80 kN/m

3

Determine average compressive stress acting at

points A and B.

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1. Stress

EXAMPLE 1.8 (SOLN)

Internal loading

Based on free-body diagram,

weight of segment AB determined from

W

st

= γ

st

V

st

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1. Stress

EXAMPLE 1.8 (SOLN)

Average normal stress

+

∑ F

z

= 0;

P − W

st

= 0

P − (80 kN/m

3

)(0.8 m)π(0.2 m)

2

= 0

P = 8.042 kN

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1. Stress

EXAMPLE 1.8 (SOLN)

Average compressive stress

Cross-sectional area at section is:

A = π(0.2)m

2

8.042 kN

π(0.2 m)

2

P

A

=

σ =

σ = 64.0 kN/m

2

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1. Stress

1.5 AVERAGE SHEAR STRESS

• Shear stress is the stress component that act

in the plane of the sectioned area.

• Consider a force F acting to the bar

• For rigid supports, and F is large enough, bar

will deform and fail along the planes identified

by AB and CD

• Free-body diagram indicates that shear force,

V = F/2 be applied at both sections to ensure

equilibrium

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1. Stress

1.5 AVERAGE SHEAR STRESS

Average shear stress over each

section is:

τ

avg

= average shear stress at

section, assumed to be same

at each pt on the section

V =internal resultant shear force at

section determined from

equations of equilibrium

A = area of section

P

A

τ

avg

=

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1. Stress

1.5 AVERAGE SHEAR STRESS

• Case discussed above is example of simple or

direct shear

• Caused by the direct action of applied load F

• Occurs in various types of simple connections,

e.g., bolts, pins, welded material

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1. Stress

Single shear

• Steel and wood joints shown below are

examples of single-shear connections, also

known as lap joints.

• Since we assume members are thin, there are

no moments caused by F

1.5 AVERAGE SHEAR STRESS

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1. Stress

Single shear

• For equilibrium, x-sectional area of bolt and

bonding surface between the two members are

subjected to single shear force, V = F

• The average shear stress equation can be

applied to determine average shear stress

acting on colored section in (d).

1.5 AVERAGE SHEAR STRESS

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1. Stress

1.5 AVERAGE SHEAR STRESS

Double shear

• The joints shown below are examples of double-

shear connections, often called double lap joints.

• For equilibrium, x-sectional area of bolt and

bonding surface between two members

subjected to double shear force, V = F/2

• Apply average shear stress equation to

determine average shear stress acting on

colored section in (d).

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1. Stress

1.5 AVERAGE SHEAR STRESS

Procedure for analysis

Internal shear

1.Section member at the pt where the τ

avg

is to be

determined

2.Draw free-body diagram

3.Calculate the internal shear force V

Average shear stress

1.Determine sectioned area A

2.Compute average shear stress τ

avg

= V/A

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1. Stress

EXAMPLE 1.10

Depth and thickness = 40 mm

Determine average normal stress and average

shear stress acting along (a) section planes a-a,

and (b) section plane b-b.

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1. Stress

EXAMPLE 1.10 (SOLN)

Part (a)

Internal loading

Based on free-body diagram, Resultant loading

of axial force, P = 800 N

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1. Stress

EXAMPLE 1.10 (SOLN)

Part (a)

Average stress

Average normal stress,

=

P

A

800 N

(0.04 m)(0.04 m)

= 500 kPa

=

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1. Stress

EXAMPLE 1.10 (SOLN)

Part (a)

Internal loading

No shear stress on section, since shear force at

section is zero.

τ

avg

= 0

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1. Stress

EXAMPLE 1.10 (SOLN)

Part (b)

Internal loading

+

∑ F

x

= 0;

− 800 N + N sin 60° + V cos 60° = 0

+

∑ F

y

= 0;

V sin 60° − N cos 60° = 0

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1. Stress

EXAMPLE 1.10 (SOLN)

Part (b)

Internal loading

Or directly using x’, y’ axes,

∑ F

x’

= 0;

∑ F

y’

= 0;

+

+

N − 800 N cos 30° = 0

V − 800 N sin 30° = 0

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1. Stress

EXAMPLE 1.10 (SOLN)

Part (b)

Average normal stress

=

N

A

692.8 N

(0.04 m)(0.04 m/sin 60)

= 375 kPa

=

30

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1. Stress

EXAMPLE 1.10 (SOLN)

Part (b)

Average shear stress

τ

avg

=

V

A

400 N

(0.04 m)(0.04 m/sin 60)

= 217 kPa

=

Stress distribution shown below

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1. Stress

1.6 ALLOWABLE STRESS

• When designing a structural member or

mechanical element, the stress in it must be

restricted to safe level

• Choose an allowable load that is less than the

load the member can fully support

• One method used is the factor of safety (F.S.)

F.S.=

F

fail

F

allow

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1. Stress

1.6 ALLOWABLE STRESS

• If load applied is linearly related to stress

developed within member, then F.S. can also

be expressed as:

F.S.=

fail

allow

F.S.=

τ

fail

τ

allow

• In all the equations, F.S. is chosen to be greater than 1,

to avoid potential for failure

• Specific values will depend on types of material used

and its intended purpose

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1. Stress

1.7 DESIGN OF SIMPLE CONNECTIONS

• To determine area of section subjected to a

normal force, use

A =

P

allow

A =

V

τ

allow

• To determine area of section subjected to a shear

force, use

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1. Stress

1.7 DESIGN OF SIMPLE CONNECTIONS

Cross-sectional area of a tension member

Condition:

The force has a line of action that passes

through the centroid of the cross section.

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1. Stress

1.7 DESIGN OF SIMPLE CONNECTIONS

Cross-sectional area of a connecter subjected to

shear

Assumption:

If bolt is loose or clamping force of bolt is unknown,

assume frictional force between plates to be

negligible.

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1. Stress

Assumptions:

1.(σ

b

)

allow

of concrete <

(σ

b

)

allow

of base plate

2.Bearing stress is

uniformly distributed

between plate and

concrete

1.7 DESIGN OF SIMPLE CONNECTIONS

Required area to resist bearing

• Bearing stress is normal stress produced by the

compression of one surface against another.

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1. Stress

1.7 DESIGN OF SIMPLE CONNECTIONS

• Although actual shear-stress distribution along rod

difficult to determine, we assume it is uniform.

• Thus use A = V/ τ

allow

to calculate l, provided d and

τ

allow

is known.

Required area to resist shear caused by axial load

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1. Stress

1.7 DESIGN OF SIMPLE CONNECTIONS

Procedure for analysis

When using average normal stress and shear stress

equations, consider first the section over which the

critical stress is acting

Internal Loading

1.Section member through x-sectional area

2.Draw a free-body diagram of segment of

member

3.Use equations of equilibrium to determine

internal resultant force

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1. Stress

1.7 DESIGN OF SIMPLE CONNECTIONS

Procedure for Analysis

Required Area

• Based on known allowable stress, calculate

required area needed to sustain load from

A = P/τ

allow

or A = V/τ

allow

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1. Stress

EXAMPLE 1.13

The two members pinned together at B. If the pins

have an allowable shear stress of

τ

allow

= 90 MPa,

and allowable tensile stress of rod CB is

(σ

t

)

allow

= 115 MPa

Determine to nearest

mm the smallest

diameter of pins A

and B and the

diameter of rod CB

necessary to support

the load.

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1. Stress

EXAMPLE 1.13 (SOLN)

Draw free-body diagram:

=

P

A

800 N

(0.04 m)(0.04 m)

= 500 kPa

=

No shear stress on section, since shear force at

section is zero

τ

avg

= 0

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1. Stress

EXAMPLE 1.13 (SOLN)

Diameter of pins:

d

A

= 6.3 mm

A

A

=

V

A

T

allow

2.84 kN

90 × 10

3

kPa

=

= 31.56 × 10

6

m

2

= π(d

A

2

/4)

d

B

= 9.7 mm

A

B

=

V

B

T

allow

6.67 kN

90 × 10

3

kPa

=

= 74.11 × 10

6

m

2

= π(d

B

2

/4)

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1. Stress

EXAMPLE 1.13 (SOLN)

Diameter of pins:

d

A

= 7 mm d

B

= 10 mm

Choose a size larger to nearest millimeter.

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1. Stress

EXAMPLE 1.13 (SOLN)

Diameter of rod:

d

BC

= 8.59 mm

A

BC

=

P

(

t

)

allow

6.67 kN

115 × 10

3

kPa

=

= 58 × 10

6

m

2

= π(d

BC

2

/4)

d

BC

= 9 mm

Choose a size larger to nearest millimeter.

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1. Stress

CHAPTER REVIEW

• Internal loadings consist of

1.Normal force, N

2.Shear force, V

3.Bending moments, M

4.Torsional moments, T

• Get the resultants using

1.method of sections

2.Equations of equilibrium

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1. Stress

CHAPTER REVIEW

• Assumptions for a uniform normal stress

distribution over x-section of member (σ = P/A)

1.Member made from homogeneous isotropic

material

2.Subjected to a series of external axial loads

that,

3.The loads must pass through centroid of

cross-section

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1. Stress

CHAPTER REVIEW

• Determine average shear stress by using

τ = V/A equation

– V is the resultant shear force on cross-

sectional area A

– Formula is used mostly to find average

shear stress in fasteners or in parts for

connections

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1. Stress

CHAPTER REVIEW

• Design of any simple connection requires that

– Average stress along any cross-section not

exceed a factor of safety (F.S.) or

– Allowable value of σallow or τallow

– These values are reported in codes or

standards and are deemed safe on basis of

experiments or through experience

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