Normal Stress (1.1-1.5)

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Normal Stress (1.1-1.5)
MAE 314 – Solid Mechanics
Yun Jing
Normal Stress
1
Pretest
This is a structure which was designed to support a 30kN load, it
consists of a boom AB and of a rod BC. The boom and rod are
connected by a pin at B and are supported by pins and brackets at A
and C. (1) Is there a reaction moment at A? why? (2) What is the
reaction force in the vertical direction at A? (3) What is the internal 2
force in AB? (4) What is the internal force in BC?
Statics Review
Pins = no rxn moment
Normal Stress
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Statics Review
• Solve for reactions at A & C:
 M C  0  Ax 0.6 m   30 kN 0.8 m 
Ax  40 kN
 Fx  0 Ax  C x
C x   Ax  40 kN
 Fy  0  Ay  C y  30 kN  0
Ay  C y  30 kN
• Ay and Cy can not be determined from
these equations.
Normal Stress
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Statics Review
• Consider a free-body diagram for the boom:
 M B  0   Ay 0.8 m 
Ay  0
substitute into the structure equilibrium
equation
Bx
C y  30 kN
• Results:
A  40 kN  Cx  40 kN  C y  30 kN 
See section 1.2 in text for complete static analysis and review of method of joints.
Normal Stress
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Normal Stress
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Introduction to Normal Stress

Methods of statics allow us to determine forces
and moments in a structure, but how do we
determine if a load can be safely supported?

Factors: material, size, etc.

Need a new concept….Stress
Normal Stress
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Introduction to Normal Stress
Stress = Force per unit area
F

A
Normal Stress
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Introduction to Normal Stress

If stress varies over a cross-section, the resultant of the
internal forces for an axially loaded member is normal to
a section cut perpendicular to the axis. Thus, we can write
the stress at a point as
  lim
A0

F
A
We assume the force F is evenly distributed
over the cross-section of the bar. In reality
F = resultant force over the end of the bar.
 dA  F
A
Normal Stress
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Introduction to Normal Stress
Sign convention
 0
 0
Tensile (member is in tension)
Compressive (member is in compression)
Units (force/area)
English: lb/in2 = psi
kip/in2 = ksi
SI:
Tensile
N/m2 = Pa (Pascal)
kN/m2 = kPa
MPa, GPa, etc.
Compressive
Normal Stress
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Definitions and Assumptions


Homogenous: material is the same throughout the bar
Cross-section: section perpendicular to longitudinal axis of bar
A
F’

F
Prismatic: cross-section does not change along axis of bar
Prismatic
Non-Prismatic
Normal Stress
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Definitions and Assumptions

Uniaxial bar: a bar with only one axis

Normal Stress (σ): stress acting perpendicular to the cross-section.

Deformation of the bar is uniform throughout. (Uniform Stress State)

Stress is measured far from the point of application.

Loads must act through the centroid of the cross-section.
Normal Stress
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Definitions and Assumptions

The uniform stress state does not apply near the ends of
the bar.

Assume the distribution of normal
stresses in an axially loaded
member is uniform, except in the
immediate vicinity of the points of
application of the loads
(Saint-Venant’s Principle).
Normal Stress
“Uniform” Stress
13
Definitions and Assumptions

How do we know all loads must act through the centroid
of the cross-section?

Let us represent P, the resultant force, by a uniform stress
over the cross-section (so that they are statically
equivalent).
Normal Stress
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Definitions and Assumptions

Moments due to σ:
Mx 
 ydA
A

M y  xdA
A

M
M
Set M
My y
Mx =
Mx x and MMy =
y
x
1
1
y   ydA   ydA
PA
AA
Equations for the centroid
1
1
x   xdA   xdA
PA
AA
Normal Stress
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Example Problem

Can the structure we used for our statics review safely support a
30 kN load? (Assume the entire structure is made of steel with a maximum
allowable stress σall=165 MPa.)
Cross-section 30 mm x 50 mm
Normal Stress
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Example Problem

Two cylindrical rods are welded together and loaded as shown. Find
the normal stress at the midsection of each rod.
d1  50 mm
d 2  30 mm
Normal Stress
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Shearing and Bearing
Stress (1.6-1.8, 1.12)
MAE 314 – Solid Mechanics
Yun Jing
Shearing and Bearing Stress
18
What is Shearing Stress?


We learned about normal
stress (σ), which acts
perpendicular to the
cross-section.
Normal stress results
in a volume change.
Shear stress (τ) acts
tangential to the surface
of a material element.
Shear stress results
in a shape change.
Shearing and Bearing Stress
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Where Do Shearing Stresses Occur?

Shearing stresses are commonly found in bolts, pins, and
rivets.
Bolt is in “single” shear
Free Body Diagram of Bolt
Force P results in shearing stress
Force F results in bearing stress
(will discuss later)
Shearing and Bearing Stress
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Shear Stress Defined

We do not assume τ is uniform over the cross-section,
because this is not the case.

τ is the average shear stress.
 ave 

P F

A A
The maximum value of τ may be considerably greater
than τave, which is important for design purposes.
Shearing and Bearing Stress
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Double Shear
Bolt is in “double” shear
Free Body Diagram of Bolt
Free Body Diagram of Center
of Bolt
 ave
Shearing and Bearing Stress
F
P
F
  2
A
A 2A
22
Bearing Stress

Bearing stress is a normal stress, not a shearing stress.

Thus,
P P
b 

Ab td
Single shear case
where
Ab = projected area where bearing pressure is applied
P = bearing force
Read section 1.8 in text for a detailed stress analysis of a structure.
Shearing and Bearing Stress
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Would like to determine the
stresses in the members
and connections of the
structure shown.
From a statics analysis:
FAB = 40 kN (compression)
FBC = 50 kN (tension)
Must consider maximum
normal stresses in AB and
BC, and the shearing stress
and bearing stress at each
pinned connection
Rod & Boom Normal Stresses
The rod is in tension with an axial force of 50 kN.
At the rod center, the average normal stress in the circular
cross-section (A = 314x10-6m2) is BC = +159 MPa.
At the flattened rod ends, the smallest cross-sectional
area occurs at the pin centerline,
A  20 mm 40 mm  25 mm   300 10 6 m 2
P
50 103 N
 BC ,end  
 167 MPa
A 300 10 6 m 2
The boom is in compression with an axial force of 40 kN
and average normal stress of –26.7 MPa.
The minimum area sections at the boom ends are
unstressed since the boom is in compression.
Pin Shearing Stresses
The cross-sectional area for pins at A, B,
and C,
2
 25 mm 
6 2
A  r  
  491 10 m
 2 
2
The force on the pin at C is equal to the
force exerted by the rod BC,
P
50 103 N
 C , ave  
 102 MPa
A 491 10 6 m 2
The pin at A is in double shear with a total
force equal to the force exerted by the
boom AB,
 A,ave 
P
20 kN

 40.7 MPa
A 491 10 6 m 2
Pin Shearing Stresses
Divide the pin at B into sections to determine the
section with the largest shear force,
PE  15 kN
PG  25 kN (largest)
Evaluate the corresponding average shearing
stress,
 B,ave 
PG
25 kN

 50.9 MPa
A 491 10 6 m2
Pin Bearing Stresses
To determine the bearing stress at A in the boom AB, we
have t = 30 mm and d = 25 mm,
b 
P
40 kN

 53.3 MPa
td 30 mm 25 mm 
To determine the bearing stress at A in the bracket, we
have t = 2(25 mm) = 50 mm and d = 25 mm,
b 
P
40 kN

 32.0 MPa
td 50 mm 25 mm 
Equilibrium of Shear Stresses


Consider an infinitesimal element of material. Apply a single shear
stress, τ1.
Total shear force on surface is (τ1)bc.
2
1

For equilibrium in the y-direction, apply
τ1 on (-) surface.

For moment equilibrium about the z-axis,
apply τ2 on top and bottom surfaces.

Moment equilibrium equation about z-axis:

Thus, a shear stress must be balanced by three other stresses for the
element to be in equilibrium.
1
2
1 (bc)a   2 (ac)b
 1   2
Shearing and Bearing Stress
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Equilibrium of Shear Stresses

What does this tell us?



Shear stresses on opposite (parallel) faces of an element are equal in
magnitude and opposite in direction.
Shear stress on adjacent (perpendicular) faces of an element are equal in
magnitude and both point towards or away from each other.
Sign convention for shear stresses


Positive face – normal is in (+) x, y, or z direction
Negative face - normal is in (-) x, y, or z direction
Face
Direction
Shear
Stress
+
+
+
+
-
-
-
-
+
-
+
Shearing and Bearing Stress
2
1
-
-
+
1
+
2
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Define General State of Stress
y
z







x
σx = stress in x-direction applied in the plane normal to x-axis
σ y = stress in y-direction applied in the plane normal to y-axis
σ z = stress in z-direction applied in the plane normal to z-axis
τxy = stress in y-direction applied in the plane normal to x-axis
τ xz = stress in z-direction applied in the plane normal to x-axis
τ zy = stress in y-direction applied in the plane normal to z-axis
And so on…
Shearing and Bearing Stress
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Define General State of Stress
y
z
x

There are 9 components of stress:
σx, σ y, σ z, τxy, τ xz, τ yx, τ yz, τ zx, τ zy

As shown previously, in order to maintain equilibrium:
τ xy= τ yx, τ xz = τ zx, τ yz = τ zy

There are only 6 independent stress components.
Shearing and Bearing Stress
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Example Problem

A load P = 10 kips is applied to a rod supported as shown
by a plate with a 0.6 in. diameter hole. Determine the shear
stress in the rod and the plate.
Shearing and Bearing Stress
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Example Problem

Link AB is used to support the end of a horizontal beam. If link AB is
subject to a 10 kips compressive force determine the normal and
bearing stress in the link and the shear stress in each of the pins.
b  2 in
t  1 4 in
d  1 in
Shearing and Bearing Stress
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Oblique Planes and Design
Considerations (1.11, 1.13)
MAE 314 – Solid Mechanics
Yun Jing
Oblique Planes and Design Considerations
35
Stress on an Oblique Plane

What have we learned so far?

Axial forces in a two-force member cause normal stresses.

Transverse forces exerted on bolts and pins cause shearing
stresses.
Oblique Planes and Design Considerations
36
Stress on an Oblique Plane

Axial forces cause both normal and shearing stresses on planes which
are not perpendicular to the axis.

Consider an inclined section of a uniaxial bar.

The resultant force in the axial direction
must equal P to satisfy equilibrium.

The force can be resolved into components perpendicular to the
section, F, and parallel to the section, V.
F  P cos 
V  P sin 

The area of the section is A0  A cos  A  A0 / cos
Oblique Planes and Design Considerations
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Stress on an Oblique Plane

We can formulate the average normal stress on the section as
F
P cos
P



cos2 
A A0 / cos A0

The average shear stress on the section is


V
P sin 
P

 sin  cos
A A0 / cos A0
Thus, a normal force applied to a bar on an inclined section
produces a combination of shear and normal stresses.
Oblique Planes and Design Considerations
38
Stress on an Oblique Plane

Since σ and τ are functions of sine and cosine, we know the
maximum and minimum values will occur at θ = 00, 450, and
900.


P
sin  cos
A0
P
cos2 
A0
At θ=±900
σ=0
At θ=±900
τ=0
At θ=±450
σ=P/2A0
At θ=±450
τ=P/2A0 (max)
At θ=00
σ=P/A0 (max)
At θ=00
τ=0
Oblique Planes and Design Considerations
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Stress on an Oblique Plane

What does this mean in reality?
Oblique Planes and Design Considerations
Design Considerations

From a design perspective, it is important to know the
largest load which a material can hold before failing.

This load is called the ultimate load, Pu.

Ultimate normal stress is denoted as σu and ultimate shear
stress is denoted as τu.
Pu
u 
A
Pu
 u  A
Oblique Planes and Design Considerations
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Design Considerations

Often the allowable load is considerably smaller than the
ultimate load.

It is a common design practice to use factor of safety.
F.S. 
ultimateload
allowableload

F .S. 
 u
allowable stress  all
ultimatestress

F .S. 
Pu
Pall
ultimate stress
allowable stress
Oblique Planes and Design Considerations

u
 all
42
Example Problem

Two wooden members are spliced as shown. If the maximum
allowable tensile stress in the splice is 75 psi, determine the
largest load that can be safely supported and the shearing stress
in the splice.
Oblique Planes and Design Considerations
43
Example Problem

A load is supported by a steel pin inserted into a hanging
wooden piece. Given the information below, determine the load
P if an overall factor of safety of 3.2 is desired.
 u _ wood  60 MPa (in tension)
 u _ wood  7.5 MPa
 u _ steel  145 MPa
b  40 m m
c  55 m m
d  12 m m
Oblique Planes and Design Considerations
44

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