(1.1-1.5)
MAE 314 – Solid Mechanics
Yun Jing
Normal Stress 1

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 force in AB? (4) What is the internal force in BC?
2

Statics Review
Pins = no rxn moment
Normal Stress 3

Statics Review
• Solve for reactions at A & C:

M
C

0

A x

0 .
6 m
 
30 kN

0 .
8 m

A x

F x

40 kN

0

A x

C x
C x

F y
 
A x
 
40 kN

0

A y

C y

30 kN

0
A y

C y

30 kN
• A y and C y can not be determined from these equations.
Normal Stress 4

Statics Review
• Consider a free-body diagram for the boom:

M
B

0
 
A y

0 .
8 m

A y

0 substitute into the structure equilibrium
B x equation
C y

30 kN
• Results:
A

40 kN

C x

40 kN

C y

30 kN

See section 1.2 in text for complete static analysis and review of method of joints.
Normal Stress 5

Normal Stress 6

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….
Normal Stress 7

Introduction to Normal Stress
Stress = Force per unit area
 
F
A
Normal Stress 8

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.
A
  dA

F
Normal Stress 9

Introduction to Normal Stress
Sign convention




0
0
Tensile (member is in tension)
Compressive (member is in compression)
Units (force/area)
English: lb/in 2 = psi kip/in 2 = ksi
Tensile
SI: N/m 2 = Pa (Pascal) kN/m 2 = kPa
MPa, GPa, etc.
Compressive
Normal Stress 10

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
Normal Stress
Non-Prismatic
11

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 12

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).
“ Uniform” Stress
Normal 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 14

Definitions and Assumptions


Moments due to σ :
M x
M y


A
 y
 dA
A
 x
 dA
Set
M x x
=
M x x and M y
= M y y y

1
P
A
 y
 dA

1
A
A
 ydA x

1
P
A
 x
 dA

1
A
A
 xdA
Equations for the centroid
Normal Stress 15

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 σ all
=165 MPa.)
Cross-section 30 mm x 50 mm
Normal Stress 16

Example Problem
 Two cylindrical rods are welded together and loaded as shown. Find the normal stress at the midsection of each rod.
d
1

50 d
2

30 mm mm
Normal Stress 17

(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.
 Shear stress (τ) acts tangential to the surface of a material element.
Shear stress results in a shape change.
Shearing and Bearing Stress
Normal stress results in a volume change.
19

 Shearing stresses are commonly found in bolts, pins, and rivets.
Bolt is in “single” shear
Free Body Diagram of Bolt
Shearing and Bearing Stress
Force P results in shearing stress
Force F results in bearing stress
(will discuss later)
20


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
A

F
A
 The maximum value of τ may be considerably greater than τ ave
, which is important for design purposes.
Shearing and Bearing Stress 21

Bolt is in “double” shear
Free Body Diagram of Bolt Free Body Diagram of Center of Bolt
 ave

P
A

F
A
2 
F
2 A
Shearing and Bearing Stress 22

 Bearing stress is a normal stress , not a shearing stress.
 Thus,
 b

P
A b

P td
Single shear case where
A b
= 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 23

Would like to determine the stresses in the members and connections of the structure shown.
From a statics analysis:
F
AB
= 40 kN (compression)
F
BC
= 50 kN (tension)
Must consider maximum normal stresses in AB and
BC , and the shearing stress and bearing stress at each pinned connection

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 -6 m 2 ) 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

BC , end

P
A

50
300

10
3

10

6
N m
2

167 MPa
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.

The cross-sectional area for pins at A , B , and C ,
A
  r
2  
25 mm
2
2

491

10

6 m
2
The force on the pin at C is equal to the force exerted by the rod BC ,

C , ave

P
A

50

10
3
491

10

6
N m
2

102 MPa
The pin at A is in double shear with a total force equal to the force exerted by the boom AB ,

A , ave

P
A

20 kN
491

10

6 m 2

40 .
7 MPa

Divide the pin at B into sections to determine the section with the largest shear force,
P
E

15 kN
P
G

25 kN (largest)
Evaluate the corresponding average shearing stress,

B , ave

P
G
A

25 kN
491

10

6 m
2

50 .
9 MPa

To determine the bearing stress at A in the boom AB , we have t = 30 mm and d = 25 mm,
 b

P td


30
40 mm kN

25 mm


53 .
3 MPa
To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm,
 b

P td


50
40 mm kN

25 mm


32 .
0 MPa

 Consider an infinitesimal element of material. Apply a single shear stress, τ
1
.





Total shear force on surface is (τ
1
) bc .
For equilibrium in the y-direction, apply
τ
1 on (-) surface.

1

2

2

1
For moment equilibrium about the z-axis, apply τ
2 on top and bottom surfaces.
Moment equilibrium equation about z-axis:

1
( bc ) a
 
2
( ac ) b
Thus, a shear stress must be balanced by three other stresses for the element to be in equilibrium.
  
1

2
Shearing and Bearing Stress 29



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

2
+ 
1
+ + +
-

1
+
+ 
2
+
-
+ -
Shearing and Bearing Stress 30

y






 z x
σ x
σ y
= stress in x-direction applied in the plane normal to x-axis
= stress in y-direction applied in the plane normal to y-axis
σ
τ xy
τ z xz
= stress in z-direction applied in the plane normal to z-axis
= stress in
= stress in y-direction z-direction applied in the plane normal to applied in the plane normal to x-axis x-axis
τ zy
= stress in
And so on… y-direction applied in the plane normal to z-axis
Shearing and Bearing Stress 31

y


 z x
There are 9 components of stress:
σ x
, σ y
, σ z
, τ xy
, 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 32

 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 33

 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 34

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

The area of the section is A
0

A
 cos

P sin

A



A
0
/ cos

Oblique Planes and Design Considerations 37

Stress on an Oblique Plane


We can formulate the average normal stress on the section as
 
F
A


P cos

A
0
/ cos


P
A
0 cos
2

The average shear stress on the section is
 
V
A


P
A
0
/ sin
 cos


P
A
0 sin
 cos

 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 = 0 0
90 0 .
, 45 0 , and
 
P
A
0 sin
 cos

 
P
A
0 cos
2

At θ = ± 90 0
At θ = ± 45 0
At θ =0 0
σ =0
σ =P/2A
0
σ =P/A
0
(max)
At θ = ±
At θ =0 0
90
At θ = ± 45 0
Oblique Planes and Design Considerations
0 τ =0
τ =P/2A
0
(max)
τ =0
39


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 , P u
.
 Ultimate normal stress is denoted as σ u stress is denoted as τ u
.
and ultimate shear
 u

P u
A

 u

P u
A

Oblique Planes and Design Considerations 41

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 .
 ultimate allowable load load

P u
P all
F .
S .
 ultimate allowable stress stress


 u all
F .
S .
 ultimate allowable stress stress

 u
 all
Oblique Planes and Design Considerations 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
 u _ wood
 u _ steel

7 .
5

145 b

40 mm
MPa
MPa
MPa
( in tension ) c

55 mm d

12 mm
Oblique Planes and Design Considerations 44