2. Analyzing stress: Definitions and Concepts Definitions and Concepts

2. Analyzing stress:

Definitions and Concepts

Definitions and Concepts

2.1 Introduction

• Stress and strain

- The fundamental and paramount subject in mechanics of materials - Chapter 2: Stress

- Chapter 3: Strain - Chapter 3: Strain

2.2 Normal stress under axial loading

Force Stress =

Area

- (Magnitude of) stress = force intensity:

- Axial loading: tension or compression load acting along the long axis of

- Normal stress: stress normal to a plane (σ) acting along the long axis of a straight structural member

0

, lim

avg A

F F

A A

s s

D ®

= = D

D

- Sign convention of normal stress of the textbook:

tensile - positive, compressive - negative

2.3 Shearing stress in connections

- Shear stress: stress tangential to a plane (σ)

0

, lim

avg A

V V

A A

t t

D ®

= = D

D

- Punching shear: ex) an action of a punch in forming rivet holes in a metal plate

2.4 Bearing stress

- Bearing stress: compressive normal stress occurring on the surface of contact between two interacting members

ex) stress on the surfaces of contact between a bolt head and a top plate

2.5 Units of stress

- In SI units (newtons and meters): N/m² = Pa (pascal), kPa, MPa - In US units: lb/in.² = psi, ksi (= 1,000 psi = kip per square inch)

ex) 1 lb ≈ 4.45 N, 1 inch = 2.54 cm = 1/12 feet, 1 psi ≈ 6.89 kPa

• Example Problem 2-1

- Normal stress (ksi) on mid-points between A-B, B-C, and C-D A cross-sectional area is 3.0 in.²

2.5 Units of stress

• Example Problem 2-2

- Diameter required for each of the sections when the allowable axial stresses are 125 MPa in the steel, 70MPa in the brass, and 85MPa in the aluminum.

2.5 Units of stress

• Example Problem 2-5

- Rigid: The minimum collar diameter when the bearing stress of collar is 10 ksi and force on the bearing plate is 50 kip

- Flexible: Calculate and plot σ_max vs. collar diameter (2.5 in. ≤ d_c ≤ 5.0 in.)

2.6 Stresses on an inclined plane in an axially loaded member

- Equilibrium of force:

( )

2

,

cos , sin

/ cos

/ cos 1 cos 2

2

n

n n

P F F N V

N P V P

A A

P P

N A A A

q q

q

s q q

= = -

= = -

=

= = = +

r r r

( )

/ cos 1 cos 2

2

/ sin cos sin 2

2

n n

n n

N A A A

P P

V A A A

s q q

t q q q

= = = +

= = - = -

® Stress depends on force and area:

stress is not a vector quantity and therefore, the law of vector addition cannot be applied to stresses on

different planes.

2.6 Stresses on an inclined plane in an axially loaded member

max min

0 when ,

q q

t

s

s s

- Maximum shear stress is a half of the maximum normal stress:

- Shear stress is zero on the planes of max. or min. normal stress:

max max 2 t =s

- Maximum shear stress is 45° apart from max. or min. normal stress.

- Magnitude of shear stress at θ is the same as that at 90+ θ.

2.6 Stresses on an inclined plane in an axially loaded member

( ) ( )

Moment equilibrium ( Mz 0):

yx xy

yx xy

dx dz dy dy dz dx

t t

t t

= =

® =

å

- Equality of shearing stress on orthogonal planes:

2.6 Stresses on an inclined plane in an axially loaded member

• Example Problem 2-6

- Normal and shear stresses on section a-a.

2.6 Stresses on an inclined plane in an axially loaded member

• Example Problem 2-7

When A = 200´ 100-mm, normal stress on section a-a = 12 Mpa, - Load P

- Shear stress on section a-a - Shear stress on section a-a

- Maximum normal and shear stresses in the block

2.7 Stress at a general point in an arbitrarily loaded member

- Stress (force) at a point in an arbitrary plane can be resolved into a normal and shear components.

- When an x – axis is set to be normal to the plane the shear component can be resolved into two shear stresses coinciding with y- or z-axis.

- It is customary to show the stresses on positive and negative surfaces through a point using a small element.

(tensile/compressive is positive/negative)

2.8 Two-dimensional or plane stress

- A state of plane stress occurs where forces can be resolved into only two components: within thin plates where z-dimension of the body is small and

the z-components of force are zero.

ex) sz =t_zx(t_xz) =t_zy(t_yz) = 0.

2.9 The stress transformation equations for plane stress

- Normal and shear stresses on a plane can be calculated by using the force equilibrium of a free-body diagram.

0 : cos 2 sin 2

2 2

0 : sin 2 cos 2

2

x y x y

n n xy

x y

t nt xy

F F

s s s s

s q t q

s s

t q t q

+ -

= = + +

= = - - +

å

å

2.9 The stress transformation equations for plane stress

• Example Problem 2-8

- Normal and shear stresses on section a-b.

- Normal and shear stresses on section c-d.

- Stresses on section a-b and c-d using a small element.

- Stresses on section a-b and c-d using a small element.

2.9 The stress transformation equations for plane stress

• Example Problem 2-9

- Normal and shear stresses on section b-b.

2.10 Principal stresses and maximum shearing stress – plane stress

- Transformation equations for plane stress:

cos 2 sin 2 (2-12b)

2 2

sin 2 cos 2 2

x y x y

n xy

x y

nt xy

s s s s

s q t q

s s

t q t q

+ -

= + +

= - - +

- Max. and min. values of σ_n:

( )

= when sin 2 2 cos 2 0

tan 2 2 , 0 (2-14)

n

n p x y xy

xy

p nt

x y

s s s s s q t q

q

q t t

s s

¶ = - - + =

¶

® = =

-

2 2

1, 2 from (2-12b) and (2-14)

2 2

x y x y

p p xy

s s s s

s = ⁺ ± ^æç ^{- ö}÷ +t

è ø

2.10 Principal stresses and maximum shearing stress – plane stress

- Max. value of τ_nt:

( )

2 2

= when cos 2 2 sin 2 0

tan 2 , (2-17)

nt

nt p x y xy

x y x y

t t t s s q t q

q

s s s s

q t t

¶ = - - - =

¶

- æ - ö

® tan 2 = - , = ± ç ÷ + ² (2-17)

2 2

x y x y

p xy

xy

qt t t

t

æ ö

® = - = ± ç ÷ +

è ø

1 2

max min

max

from (2-15) and (2-17) 2

in 3D cases 2

p p

p

s s

t

s s

t

= -

® = -

- The direction of the maximum shear stress must oppose the larger of the two principal stresses.

2.10 Principal stresses and maximum shearing stress – plane stress

- Invariant of stress

2 2

1, 2

2 2

x y x y

p p xy

s s s s

s = ⁺ ± ^æç ^{- ö}÷ +t

è ø

1 2

p p x y

s s s s

® + = +

- In engineering problems, “maximum” will always refer to the largest absolute value (largest magnitude).

2.10 Principal stresses and maximum shearing stress – plane stress

• Example Problem 2-11

- Principal stresses and the maximum shear stress

- Locate the planes on which these stresses act and show the stresses

2.11 Mohr’s circle for plane stress

- Equation of Mohr’s circle

cos 2 sin 2

2 2

sin 2 cos 2 2

x y x y

n xy

x y

nt xy

s s s s

s q t q

s s

t q t q

+ -

= + +

= - - +

2 2

2 2

2 2

2

2 2

2

x y x y

n nt xy

x y

R xy

s s s s

s t t

s s

t

+ -

æ ö æ ö

®ç - ÷ + = ç ÷ +

è ø è ø

æ - ö

= ç ÷ +

è ø

- Horizontal coordinates: normal stress (V’,F’..) - Vertical coordinates: shear stress (F’F, V’V..)

(CW - above; CCW – below)

2.11 Mohr’s circle for plane stress

- Line CV, CH… : plane

- Angle on the circle: twice the angle for an actual body (CW - above; CCW – below)

- Procedure for drawing Mohr’s circle:

2.11 Mohr’s circle for plane stress

• Example Problem 2-13

- Principal stresses and the maximum shear stress - Normal and shear stress on plane a-a

2.12 General state of stress at a point

- Traction components on the oblique face

x x x yx zx

y y xy y zy

z z xz yz z

F S dA dA l dA m dA n F S dA dA l dA m dA n F S dA dA l dA m dA n

s t t

t s t

t t s

= = × + × + ×

= = × + × + ×

= = × + × + ×

- Force equilibrium on an oblique face

x y z

x x xy xz

y yx y yz

z zx zy z

S S S S

S l m n

S l m n

S l m n

s t t

t s t

t t s

= + +

= + +

= + +

= + +

- Normal/shear stress on the oblique face

( ) ^{( )}

2 2 2

2 2

, ˆ , , , ,

2 2 2

, n

n nt n x y z

n x y z xy yz zx

nt n nt

S S n S S S l m n

l m n lm mn nl

S S

s t s

s s s s t t t

t s t s

= + = =

\ = + + + + +

= - = -

rg g

r r r

2.12 General state of stress at a point

- Principal stress when the oblique face is a principal plane

( )

( )

( )

, , ,

0 0 0

p x p y p z p

x p x p xy xz

y p yx y p yz

z p zx zy z p

S S l S m S n

S l l m n

S m l m n

S n l m n

s s s s

s s s t t

s t s s t

s t t s s

= = = =

- = - + + =

- = + - + =

- = + +

(

)

( )

( )

( )

0

z p zx zy z p

x p xy xz

yx y p yz

zx zy z p

s s t t

t s s t

t t s s

-

- =

-

( ) ( )

( )

3 2 2 2 2

2 2 2

2 0

p x y z p x y y z z x xy yz zx p

x y z x yz y zx z xy xy yz zx

s s s s s s s s s s s t t t s

s s s s t s t s t t t t

® - + + + + + - - -

- - - - + =

2.12 General state of stress at a point

- Maximum shear stress when σ_x, σ_y, and σ_z are principal stresses

( )

( )

2 2 2 2 2 2 2

2 2 2 2 2

2 2 2 2 2 2 2 2 2 2

, ,

x x y y z z

x y z

n x y z

S l S m S n

S l m n

l m n

l m n l m n

s s s

s s s

s s s s

t s s s s s s

= = =

= + +

= + +

= ^{2 2} + ^{2 2} + ^{2 2} -

(

)

nt xl ym zn xl ym zn

t = s +s +s - s +s +s

2

1 2

2 2 max min

max 1 2 1 2 max

1 1 1 1

2 2 2 2 2 2

p p

p p p p

s s s s

t = s + s -^æç s + s ^ö÷ = ^- ® t = ^-

è ø

2.12 General state of stress at a point

14 ksi 12 ksi 10 ksi

8 ksi 10 ksi 6

x y x

xy yz zx

s s s

t t t

= = =

= = - =

• Example Problem 2-15

- Normal/shear stresses on a plane whose outward normal is oriented at angles - Normal/shear stresses on a plane whose outward normal is oriented at angles

of 61.3˚, 53.1˚, and 50.2 ˚ with x-, y-, and z-axes, respectively.

- Principal stresses and the maximum shearing stress at the point

• Home work problems: