Mechanics of Materials 13-1

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Professional Publications, Inc.
FERC
13-1
Mechanics of Materials
Stress-Strain Curve for Mild Steel
Professional Publications, Inc.
FERC
13-2a
Mechanics of Materials
Definitions

Hooke

s
Law

Shear Modulus:

Stress:

Strain:

Poisson

s Ratio:

Normal stress or strain =

Shear stress = || to the surface




to the surface
Professional Publications, Inc.
FERC
13-2b
Mechanics of Materials
Definitions
Uniaxial
Load and Deformation
Thermal Deformation
Professional Publications, Inc.
FERC
13-3a
Mechanics of Materials
Stress and Strain
Thin-Walled Tanks
Hoop Stress:
Axial Stress:
Professional Publications, Inc.
FERC
13-3b
Mechanics of Materials
Stress and Strain
Transformation of Axes
Professional Publications, Inc.
FERC
13-3c
Mechanics of Materials
Stress and Strain
Five simplified steps to construct Mohr

s circle
1.
Determine the applied stresses (
σ
x
,
σ
y
,
τ
xy
).
2.
Draw a set of
σ
-
τ
axes.
3.
Locate the center:
4.
Find the radius (or
τ
max
):
5.
Draw Mohr

s circle.





c

1
2
(

x


y
).
Professional Publications, Inc.
FERC
13-3d1
Mechanics of Materials
Stress and Strain
For examples 1 and 2, use the following illustration.
Example 1 (FEIM)
The principal stresses (
σ
2
,
σ
1
) are most nearly
(A)

62

400 kPa and 14

400 kPa
(B)
84

000 kPa and 28

000 kPa
(C)
70

000 kPa and 14

000 kPa
(D)
112

000 kPa and

28

000 kPa
Professional Publications, Inc.
FERC
13-3d2
Mechanics of Materials
Stress and Strain
The center of Mohr

s circle is at




max

(
30
000
kPa)
2

(
24
000
kPa
)
2

38
419
kPa





1


c


max

(

24
000
kPa

38
419
kPa)


62419
kPa





2


c


max

(

24
000
kPa

38
419
kPa)

14
418
kPa
Therefore, (D) is correct.
Using the Pythagorean theorem, the radius of Mohr

s circle (
τ
max
) is:





c

1
2
(

x


y
)

1
2
(

48
000
kPa

0
)


24
000
kPa
Professional Publications, Inc.
FERC
13-3d3
Mechanics of Materials
Stress and Strain
Example 2 (FEIM):
The maximum shear stress is most nearly
(A)
24

000 kPa
(B)
33

500 kPa
(C)
38

400 kPa
(D)
218

000 kPa
Therefore, (C) is correct.
In the previous example problem, the radius of Mohr

s circle (
τ
max
) was




max

(
30
000
kPa)
2

(
24
000
kPa
)
2




38
419
kPa
(
38
400
kPa)
Professional Publications, Inc.
FERC
13-3e
Mechanics of Materials
Stress and Strain
General Strain




Note that

x
is no longer proportional to

x
.
Professional Publications, Inc.
FERC
13-3f
Mechanics of Materials
Stress and Strain
Static Loading Failure Theory
Maximum Normal Stress: A material fails if

Or






S
t





S
c
This is true of brittle materials.
For ductile materials:
Maximum Shear
Distortion Energy (von
Mises
Stress)




T
max

max

1


2
2
,

1


3
2
,

2


3
2

















S
yt
2









1
2

1


2


2


1


3


2


2


3


2













S
yt
Professional Publications, Inc.
FERC
13-3g
Mechanics of Materials
Stress and Strain
Torsion

For a body with radius
r
being
strained to an angle
φ
, 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

dz
r
2
dA
A


GJ
d

dz
The shear stress is:





z

Gr
T
GJ

Tr
J

For a body, the general angular
displacement (
φ
) is:






T
GJ
dz
0
L


For a shaft of length (
L
), the total
angular displacement (
φ
) is:
Torsional stiffness:
Professional Publications, Inc.
FERC
13-3h
Mechanics of Materials
Stress and Strain
Hollow, Thin-Walled Shafts
Professional Publications, Inc.
FERC
13-4a
Mechanics of Materials
Beams
Professional Publications, Inc.
FERC
13-4b
Mechanics of Materials
Beams
Load, Shear, and Moment Relations
Load:
Shear:
For a beam deflected to a radius of curvature (
ρ
), the axial strain at a
distance (
y
) from the neutral axis is





x


y
/

.
Professional Publications, Inc.
FERC
13-4c1
Mechanics of Materials
Beams
Shear and Bending Moment Diagrams
Example 1 (FEIM):
Draw the shear and bending moment diagrams for the following beam.
Professional Publications, Inc.
FERC
13-4c2
Mechanics of Materials
Beams
Shear is undefined at concentrated force points, but just short of
x
= 12 m





R
l

R
r

100
N
m












16
m



1600
N
R
l

(
8
)

R
r
(4
)

0
Therefore,
R
l
= 533.3 N and
R
r
= 1066.7 N
So the shear diagram is:





From 0 m to 12 m,

V

R
l

100
N
m












x

533.3
N

100
N
m












x
;
0 m
<

x

<
12 m




V
(
12

)

533.3
N

100
N
m












(
12
m
)


666.7
N




V

1600
N

100
N
m












x
;

12

<

x


16 m




From 12 m to 16 m,

V

V
(
12

)

R
r

(
100
N)(
x

12
)
Professional Publications, Inc.
FERC
13-4c3
Mechanics of Materials
Beams
The bending moment is the integral of the shear.




M

Sdx
0
12


Sdx
12
x



800

1600
N

100
N
m
x












0
12

dx






800
N

m

1600
N


x

50
N
m












x
2












12
x
M
= 533.3
x


50
x
2
; 0 m <
x
< 12 m




M


800
N

m

1600
N


x

50
N
m












x
2

1600
N


12 m



50
N
m












12 m


2




M


12
800
N

m

1600
N


x

50
N
m












x
2




12
m

x

16
m
Or, let the right end of the beam be
x
= 0 m




Then,

S


100
N
m












x
;


4
m

x

0
m




M

Sdx
x
0



100
N
m












x
0

x


50
N
m












x
2
Professional Publications, Inc.
FERC
13-4c4
Mechanics of Materials
Beams
The bending moment diagram is:
Professional Publications, Inc.
FERC
13-4d1
Mechanics of Materials
Beams
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
2
P
Professional Publications, Inc.
FERC
13-4d2
Mechanics of Materials
Beams
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 kN



m
(B) 16 kN



m
(C) 18 kN



m
(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 kN



m. The trapezoid on the left has an area of
(1/2)(12 kN + 14 kN) (2 m) = 26 kN



m. The trapezoid has the largest bending
moment.
Therefore, (A) is correct.
Professional Publications, Inc.
FERC
13-4e
Mechanics of Materials
Beams
Bending Stress
Deflection
Shear Stress
Note: Beam deflection formulas are given in the NCEES Handbook for
any situation that might be on the exam.
Professional Publications, Inc.
FERC
13-4f
Mechanics of Materials
Beams
Example (FEIM):
Find the tip deflection of the beam shown.
E
I
is 3.47 × 106 N



m
2
, the
load is 11

379 N/m, and the beam is 3.7 m long.
From the NCEES Handbook:







w
o
L
4
8
E
I

11379
N
m












3.7
m


4
8


3.47

10
6
N

m
2



0.077
m
Professional Publications, Inc.
FERC
13-5a
Mechanics of Materials
Columns
Beam-Columns (Axially Loaded Beams)
Maximum and minimum stresses in an eccentrically loaded column:
Professional Publications, Inc.
FERC
13-5b
Mechanics of Materials
Columns
Euler

s Formula
Critical load that causes a long column to buckle:
r
= the radius of gyration
k
= the end-resistant coefficient
k
l
= the effective length

= slenderness ratio




l
r
Professional Publications, Inc.
FERC
13-5c
Mechanics of Materials
Columns
Elastic Strain Energy:
Strain energy per unit volume for tension: