Download - 020614 - Session
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Strength of Materials
Stress vs. Strain
Stress = Force / Area
Strain = Change in Length / Original Length
σ
εAP
=σ
LL∆
=ε
2
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Stress – Strain Diagram Typical for Ductile Materials
Stre
ss
Strain
∆σ
∆ε
3
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Stress – Strain Diagram
Stre
ss
Strain
Proportional limit
∆σ
∆ε
4
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Stress – Strain Diagram
Stre
ss
Strain
Proportional limit
Elastic limit
∆σ
∆ε
5
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Stress – Strain Diagram
Stre
ss
Strain
Proportional limit
Yield point
Elastic limit
∆σ
∆ε
6
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Stress – Strain Diagram
Stre
ss
Strain
∆σ
∆ε
7
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Stress – Strain Diagram
Stre
ss
Strain
Proportional limit
Yield point
Elastic limit
Ultimate strength
∆σ
∆ε
8
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Stress Strain Diagram
Stre
ss
Strain
Proportional limit
Yield point
Elastic limit
Ultimate strength
Rupture
∆σ
∆ε
9
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Stress- Strain Relationship
Stre
ss
Strain
Proportional limit
Yield point
Elastic limit
Ultimate strength
Rupture
True rupture
∆σ
∆ε
10
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Stress – Strain Relationship Typical for Ductile Materials
Stre
ss
Strain
∆σ
∆ε Hooke’s Law εσ E=
ExyportionlineartheofSlope =
∆∆
=∆∆
=εσ
11
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Areas under stress-strain curves: (a) modulus of resilience, (b) toughness, and (c) high-strength and high-toughness materials.
Stre
ss
(a)
Strain
Modulus of resilience
Stre
ss
(b)
Strain
Stre
ss
(c)
Strain
High strength
Toughness
Fracture
High toughness
12
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Methods for estimating yield stress: (a) offset method and (b) extension method
Stre
ss
(a)
Strain, %
Proportional limit
0.2% offset yield strength
Elastic limit
0.2%
Stre
ss
(b)
Strain, %
0.5% extension yield strength
0.5%
13
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Problem 1)
For the beam shown, compute the maximum bending stress in compression and maximum bending stress in tension. Also compute the shear stress at the neutral axis as well as at a section 6 in. from the bottom of the cross-section. Calculate the shear stresses at a point along the beam where the maximum shear force occurs.
1500 lb
1 in
8 in
4 in
3.5 in 1 in
INA = 97.0 in4
NA
1500 lb 4500 lb
4 ft 4 ft 6 ft 6 ft
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Problem 1 (continued)
For the beam shown, compute the shear stress at the NA and at 6 in. from the bottom of the cross section at the point where the maximum shear force occurs.
1500 lb
1 in
8 in
4 in
3.5 in 1 in
INA = 97.0 in4
NA
1500 lb 4500 lb
4 ft 4 ft 6 ft 6 ft
-1500 lb
1500 lb
3750 lb 3750 lb
V (Ib)
2250
- 2250 - 2250
1500
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Problem 1 (continued) - For the beam shown, compute the max. bending stress in tension and in compression.
1500 lb
1 in
8 in
4 in
3.5 in 1 in
INA = 97.0 in4
NA
1500 lb 4500 lb
4 ft 4 ft 6 ft 6 ft
-6000 -6000
+7500
M (ft-lb)
------ ---i---
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Bending Stress in beams :
In the negative moment region In the positive moment region
inCc 5.3= inCc 5.5=
inCt 5.5= inCt 5.3=
psit 324797
)5.3)(12)(7500(==σpsit 4082
97)5.5)(12)(6000(==σ
psic 510397
)5.5)(12)(7500(==σpsic 2598
97)5.3)(12)(6000(==σ
ICM c
c =σ
ICM
=σ
ICM t
t =σ
17
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CISModulusSection =
ICM
=σ =
SMσ
36.175.5
97 inCIS ===
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Problem 1 (continued)
For the beam shown, compute the shear stress at the NA and at 6 in. from the bottom of the cross section at the point where the maximum shear force occurs.
1500 lb
1 in
8 in
4 in
3.5 in 1 in
INA = 97.0 in4
NA
1500 lb 4500 lb
4 ft 4 ft 6 ft 6 ft
-1500 lb
1500 lb
3750 lb 3750 lb
V (Ib)
2250
- 2250 - 2250
1500
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Shear stress in bending bIQV
=τ3
sec6 12]4[)1()3( inQ tioncrossofbottomfromin ==−
312.15]75.2[)1()5.5( inQ axisneturalat ==
psitioncrossofbottomfromin 3.278)1)(97(
)12()2250(sec6 ==−τ
psiaxisneturalat 7.350)1)(97(
)12.15()2250(==τ
20
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What is the value of Q if shear stress is being calculated at:
1 in
8 in
4 in
3.5 in 1 in
INA = 97.0 in4
NA NA
3 in 4.0 in
NA 2.75 in 2.5 in
5.5 in
312]4[)1()3( inQ ==
Neutral axis a section 6 in. above the base
312.15]75.2[)1()5.5( inQ ==
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Problem 2.
The vertical shear force acting on the I-section shown is 100 kN. Compute the maximum shear stress in bending acting on the this beam section.
20 mm
20 mm
20 mm
160 mm
120 mm
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20 mm
20 mm
20 mm
160 mm
120 mm
NA
TheoremAxisParallelAdbhI 23
12+=
2
3
1
423
1 000,520,19]90)[20)(120(12
)20)(120( mmI NAabout =+=
43
2 667,826,612
)160)(20( mmI NAabout ==
423
3 000,520,19]90)[20)(120(12
)20)(120( mmI NAabout =+=
4666,866,45 mmI NAabouttotal =20 mm
23
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Problem 2 (continued) Maximum shearing stress occurs at the neutral axis.
3000,280]90)[20)(120(]40)[80)(20( mmNAatQ =+=
NAatmmb 20=
KNV 100=
4666,866,45 mmI NAabouttotal =
2/03.0)20()45866666(
)280000()100( mmKN==τ
bIQV
=τ
24
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Problem 3.
The simply supported beam has the T-shaped cross section shown. Determine the values and locations of the maximum tensile and compressive bending stresses.
400 lb/ft 1000 lb
A
B
D
10 ft 4 ft
y
x
RA = 1600 lb RB = 3400 lb
6 in
8 in
0.8 in
0.8 in
25
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Problem 3 (continued) Shear Diagram
400 lb/ft 1000 lb
A
B
D 10 ft 4 ft
y
x
RA = 1600 lb RB = 3400 lb
1600 1000
-2400
4 ft V (lb)
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Moment Diagram 400 lb/ft
1000 lb
A
B
D 10 ft 4 ft x
RA = 1600 lb RB = 3400 lb
1600 1000
-2400
4 ft V (lb)
M (lb-ft) 4 ft
-4000
3200
2°
27
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Find y-bar.
21
2211
AAyAyAbary
++
=−
( ) ( )8.44.6
4.88.444.6++
=
in886.5=
6 in
0.8 in
8 in
A2
A1 y1
y2
y-bar
N.A.
C
0.8 in
28
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Compute Moment of Inertia, I.
6 in
0.8 in
8 in
A2
A1 y1
y2
y-bar
N.A.
C
0.8 in
( ) ( ) ( ) ( )
−++
−+= 2
32
3
886.54.88.412
8.06886.544.612
88.0I
449.87 inI =29
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Stresses at x = 4 ft
8.8 in
y1 y-bar
N.A. C
cbot = 5.886 in.
ctop = 2.914 in.
4 ft
( )( ) psiI
Mcbotbot 2580
49.87886.5123200
=×
==σ
3200
-4000
( )( ) psiI
Mctoptop 1279
49.87914.2123200
=×
==σ
---i---
Compression on top
Tension in bottom 30
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Stresses at x = 10 ft
8.8 in
y1 y-bar
N.A. C
cbot = 5.886 in.
ctop = 2.914 in.
10 ft
3200
-4000
( )( ) psiI
Mcbotbot 3230
49.87886.5124000
=×
==σ
( )( ) psiI
Mctoptop 1599
49.87914.2124000
=×
==σ
------
Tension on top
Compression in bottom 31
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Problem 3 results.
Identify maximum tensile and compressive stresses in the beam.
( ) psiT 2580max =σ
( ) psiC 3230max =σ
(bottom of the section at x = 4 ft)
(bottom of the section at x = 10 ft)
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Original State of Stress
y
x
a
a
θxσ
yσ
xyτ
xyτ
xσ
yσ
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x
y
R
σ
τ
xσ
yσ
xyτ
xyτ
σ
y
x xσ
yσ
xyτConstruction of Mohr’s circle from given stress components.
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Convention for plotting shear stress on Mohr’s circle.
Shear plotted up Shear plotted down
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y’
σ
τ
xσ
yσ
yx ′′τ
σ
xyτ
yσ
xσ x
y
y′σ
Θ
yx ′′τy’
x’
x′σ
2σ
Θ1
2 1
1σ
y′σ
1σ
y
a
2
1 2σ
x′σ
xyτ
b
x’
x
36
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Problem 4 )
For the state of stress shown, determine (a) the principal stresses; and (b) the maximum in-plane shear stress. Show the results on properly oriented elements.
4 ksi
8 ksi
6 ksi
y
x
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τ (y-axis)
σ (x-axis)
(4,-6)
(-8, 6)
4 ksi
8 ksi
6 ksi
y
x
y
x
Problem 4 (continued)
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R
4 ksi
8 ksi
6 ksi
y
x
(-8, 6)
(4,-6)
τ (y-axis)
σ (x-axis)
y
x
Problem 4 (continued)
39
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R
4 ksi
8 ksi
6 ksi
y
x
(-8, 6)
(4,-6)
τ (y-axis)
σ (x-axis) -2
y
x
Problem 4 (continued)
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R
2Θ1
-2
4 ksi
8 ksi
6 ksi
y
x
(-8, 6)
(4,-6)
τ (y-axis)
σ (x-axis)
y
x
Problem 4 (continued)
41
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τmax
y
1 2
R
2Θ1
-2
4 ksi
8 ksi
6 ksi
y
x
(-8, 6)
x (4,-6)
τ (y-axis)
σ (x-axis)
Problem 4 (continued)
42
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(a)
τ
σ
τmax
x
y
1 2
R
4 2Θ1
6 -8 -2
6
(b)
22.5°
10.49
6.49
22.5°
8.49 2
2
ksiR 7266 22 =+=( ) °=Θ⇒°==Θ − 5.22456/6tan2 1
11
ksi
ksi
49.10722
49.6722
2
1
−=−−=
=+−=
σ
σ
ksiR 49.872max ===τ
Problem 4 (continued)
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Problem 5)
The radius of the 15-in. long bar in Fig. (a) is 3/8 in. Determine the maximum normal stress in the bar at (1) point A; and (2) point B.
z x
y
30 lb
540 lb-in
A B
z x
y
B A
M
T
V
(a) (b) 44
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Problem 5 (continued)
The internal force system acting on the cross section at the base of the rod is shown in Fig. (b). It consists of the torque T = 540 lb-in, the bending moment M = 15P = 15(30) = 450 lb-in (acting about the x-axis), and the transverse shear force V = P = 30 lb.
The cross-sectional properties of the bar are:
( ) 4344
10532.154
8/34
inrI −×===ππ
( ) 433 1006.3110532.1522 inIJ −− ×=×==
2
4rJInertiaofmomentPolar π=
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z
x
y
ksi520.6=τksi865.10=σ
τ (ksi)
σ (ksi)
x
y
6.520
10.865
5.433
Problem 5 (continued) PART 1
State of stress at point A with the corresponding Mohr’s circle.
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Problem 5 (continued)
The bending stress is calculated:
z
x
y
ksi520.6=τksi865.10=σ
τ (ksi)
x
y
6.520
10.865
5.433
( )( ) ksipsiI
Mr 865.108651010532.15
8/34503 ==
×== −σ
47
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z
x
y
ksi520.6=τksi865.10=σ σ (ksi)
τ (ksi)
x
y
6.520
10.865
5.433
Problem 5 (continued)
And the shear stress is calculated:
( ) ksipsiJ
TrT 520.65206
1006.318/3540
3 ==×
== −τ
The shear stress due to the transverse shear force V is zero at A.
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z
x
y
ksi520.6=τksi865.10=σ σ (ksi)
τ (ksi)
x
y
6.520
10.865
5.433
Problem 5 (continued)
The maximum normal stress at point A is:
ksi92.13487.8433.5max =+=σ
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Problem 5 (continued) PART 2
State of stress at point B with the corresponding Mohr’s circle.
z
x
y
ksi611.6=τ
B
τ (ksi)
σ (ksi)
z
y
6.611
σmax
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Problem 5 (continued)
The shear stress due to torque is τT = 6.520 ksi, as before. But because the point lies on the neutral axis, the bending stress is zero. There is, however, an additional shear stress caused by the transverse shear force V. The magnitude of this shear stress is τV = VQ/(Ib), where b = 2r = ¾ in. and Q is the first moment of half the cross-sectional area about the neutral axis. Referring to the figure below, Q is calculated:
==
ππ
34
2''
2 rrZAQ
x
z
C
π34' rZ =( )
38/32
32 33
=
=
r
331016.35 in−×=
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Problem 5 (continued)
Therefore:
z
x
y ksi611.6=τ
B
τ (ksi)
σ (ksi)
z
y
6.611
σmax
( )( )( ) ksipsi
IbVQ
V 091.05.904/310532.15
1016.35303
3
==×
×== −
−
τ
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Problem 5 (continued)
Since τT and τV act on the same planes they can be added. So the total shear
stress is:
z
x
y ksi611.6=τ
B
τ (ksi)
σ (ksi)
z
y
6.611
σmax
ksiVT 611.6091.0520.6 =+=+= τττ
Vτ
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Problem 5 (continued)
The Mohr’s circle for this state of pure shear yields for the maximum normal stress at B:
z
x
y ksi611.6=τ
B
τ (ksi)
σ (ksi)
z
y
6.611
σmax
ksi61.6max =σ
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Poisson’s ratio
If a structural member is subjected to axial tension, the material elongates and there is a reduction in its cross-section. When the member is under compression, the opposite happens. The ratio of the lateral strain to axial strain is Poisson’s ratio.
axial
lateral
εεν −
=
Most solids have a Poisson’s ratio between 0.10 and 0.45 55
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General State of Stress
z
x
y
1σ
2σ
3σ
τ
σ
1σ
3σ
2σ
−−−=
2,
2,
2max 133221 σσσσσσ
τ abs
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Problem 6)
For the state of stress shown, determine the maximum in-plane shear stress and the absolute maximum shear stress.
50 ksi
20 ksi
y
x
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Problem 6 (continued)
The given stresses are:
50 ksi
20 ksi
y
x
501 == xσσ
202 == yσσ
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Problem 6 (continued)
The maximum in-plane shear stress τmax is equal to the radius of the circle that represents transformation in the xy-plane.
τ (ksi)
σ (ksi)
501 =σ202 =σ
10
25
15
Transformation in xy-plane
Transformation in zx-plane
Transformation in yz-plane
ksi15max =τ
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Problem 6 (continued)
The absolute maximum shear stress equals the radius of the largest circle, which represents the transformation in the zx-plane.
τ (ksi)
σ (ksi)
501 =σ202 =σ
10
25
15
Transformation in xy-plane
Transformation in zx-plane
Transformation in yz-plane
ksiabs 25=τ
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References:
Pytel and Kiusalaas, “Mechanics of Materials,” Thomson, 2003. McCormac and Nelson, “Design of Reinforced Concrete,” Wiley, 2003.
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Thank You
and
GOOD LUCK!!!
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