Stress Strain Relationship for Mild Steel

7/28/2019 Stress Strain Relationship for Mild Steel

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Professional Publications, Inc. FERC

13-1Mechanics of MaterialsStress-Strain Curve for Mild Steel


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13-2aMechanics of MaterialsDefinitions

Hookes Law

Shear Modulus:

Stress:

Strain:

Poissons Ratio:

Normal stress or strain =

Shear stress = || to the surface

" to the surface


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13-2bMechanics of MaterialsDefinitions

Uniaxial Load and Deformation

Thermal Deformation


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13-3aMechanics of MaterialsStress and Strain

Thin-Walled Tanks

Hoop Stress:

Axial Stress:


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13-3bMechanics of MaterialsStress and Strain

Transformation of Axes


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13-3cMechanics of MaterialsStress and Strain

Five simplified steps to construct Mohrs circle

1. Determine the applied stresses (x, y, xy).2. Draw a set of - axes.

3. Locate the center:

4. Find the radius (ormax):

5. Draw Mohrs circle.

"c =1

2("

x+"

y).


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13-3d1Mechanics of MaterialsStress and Strain

For examples 1 and 2, use the following illustration.

Example 1 (FEIM)The principal stresses (2, 1) are most nearly

(A) 62400 kPa and 14400 kPa

(B) 84000 kPa and 28000 kPa

(C) 70000 kPa and 14000 kPa

(D) 112000 kPa and 28000 kPa


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The center of Mohrs circle is at

"max = (30000 kPa)2+ (24000 kPa)2 = 38419 kPa

"

1

= "c

# $max

= (#24000 kPa # 38419 kPa)= #62419 kPa

"2 = "c+ #max = ($24000 kPa+38419 kPa)=14418 kPa

Therefore, (D) is correct.

Using the Pythagorean theorem, the radius of Mohrs circle (max) is:"c =

1

2("

x+"

y) = 1

2(#48000 kPa+0)= #24000 kPa


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Example 2 (FEIM):

The maximum shear stress is most nearly

(A) 24000 kPa

(B) 33500 kPa

(C) 38400 kPa

(D) 218000 kPa

Therefore, (C) is correct.

In the previous example problem, the radius of Mohrs circle (max) was

"max = (30000 kPa)2+ (24000 kPa)2

= 38419 kPa (38400 kPa)


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13-3eMechanics of MaterialsStress and Strain

General Strain

Note that "x is no longer proportional to #x.


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13-3fMechanics of MaterialsStress and Strain

Static Loading Failure Theory

Maximum Normal Stress: A material fails if

Or

" # St

" # Sc

This is true of brittle materials.

For ductile materials:

Maximum Shear

Distortion Energy (von Mises Stress)

Tmax

=max"1 # "2

2,

"1 # "3

2,

"2 # "3

2

$

%

&&

'

(

))>Syt

2

"# = 12

#1

$ #2( )

2

+ #1

$ #3( )

2

+ #2

$ #3( )

2%&' (

)*>S

yt


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13-3gMechanics of MaterialsStress and Strain

Torsion

For a body with radius rbeingstrained to an angle , the shear

strain and stress are:

" = r

d#

dz " =G# =Gr

d$

dz

For a body with polar moment of

inertia (J), the torque (T) is:

T=G

d"

dzr

2dA

A# =GJ

d"

dz

The shear stress is:

"#z=Gr T

GJ= Tr

J

For a body, the general angulardisplacement () is:

" =

T

GJdz

0

L

#

For a shaft of length (L), the totalangular displacement () is:

Torsional stiffness:


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13-3hMechanics of MaterialsStress and Strain

Hollow, Thin-Walled Shafts


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13-4aMechanics of MaterialsBeams


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13-4bMechanics of MaterialsBeams

Load, Shear, and Moment Relations

Load:

Shear:

For a beam deflected to a radius of curvature (), the axial strain at a

distance (y) from the neutral axis is "x = #y/$.


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13-4c1Mechanics of MaterialsBeams

Shear and Bending Moment Diagrams

Example 1 (FEIM):

Draw the shear and bending moment diagrams for the following beam.


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Shear is undefined at concentrated force points, but just short ofx= 12 m

Rl+Rr = 100 Nm

"

#$

%

&'16 m( ) =1600 N

Rl= (8)(R

r(4)= 0

Therefore, Rl= 533.3 N and Rr= 1066.7 N

So the shear diagram is:

From 0 m to 12 m, V= Rl" 100

N

m

#

$

%&

'

(x= 533.3 N" 100N

m

#

$

%&

'

(x; 0 m < x < 12 m

V(12") = 533.3 N" 100N

m

#

$%

&

'((12 m) = "666.7 N

V=1600 N" 100N

m

#

$%

&

'(x; 12 < x ) 16 m

From 12 m to 16 m, V=V(12")+R

r" (100 N)(x"12)


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The bending moment is the integral of the shear.

M= Sdx0

12

" + Sdx12

x

" = #800+ 1600 N#100N

mx

$

%&

'

()

0

12

" dx

= "800 N #m+ 1600 N( )x" 50N

m

$

%

&'

(

)x2

$

%

&'

(

)

12

x

M= 533.3x 50x2; 0 m < x< 12 m

M= "800 N #m+ 1600 N( )x" 50N

m

$

%&

'

()x

2 " 1600 N( ) 12 m( )+ 50N

m

$

%&

'

()12 m( )

2

M= "12800 N #m+ 1600 N( )x" 50N

m

$

%&

'

()x

2

12 m


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The bending moment diagram is:


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13-4d1Mechanics of MaterialsBeams

Example 2 (FEIM):The vertical shear for the section at the midpoint of the beam shown is

(A) 0

(B)

(C) P

(D) none of these

Drawing the force diagram and the shear diagram,

Therefore, (A) is correct.

1

2P


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13-4d2Mechanics of MaterialsBeams

Example 3 (FEIM):For the shear diagram shown, what is the maximum bending moment? The

bending moment at the ends is zero, and there are no concentrated couples.

(A) 8 kNm

(B) 16 kNm

(C) 18 kNm

(D) 26 kN

m

Starting from the left end of the beam, areas begin to cancel after 2 m. Starting

from the right end of the beam, areas begin to cancel after 4 m. The rectangle on

the right has an area of 16 kNm. The trapezoid on the left has an area of

(1/2)(12 kN + 14 kN) (2 m) = 26 kNm. The trapezoid has the largest bending

moment.

Therefore, (A) is correct.


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13-4eMechanics of MaterialsBeams

Bending Stress

Deflection

Shear Stress

Note: Beam deflection formulas are given in the NCEES Handbook for

any situation that might be on the exam.


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13-4fMechanics of MaterialsBeams

Example (FEIM):

Find the tip deflection of the beam shown. EI is 3.47 106 Nm2, the

load is 11379 N/m, and the beam is 3.7 m long.

From the NCEES Handbook:

" =w

oL

4

8EI=

11379N

m

#

$%

&

'(3.7 m( )

4

8( ) 3.47)106 N *m2( )= 0.077 m


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13-5aMechanics of MaterialsColumns

Beam-Columns (Axially Loaded Beams)

Maximum and minimum stresses in an eccentrically loaded column:


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13-5cMechanics of MaterialsColumns

Elastic Strain Energy:

Strain energy per unit volume for tension:

Documents

Stress Strain Relationship for Mild Steel