ppt lecture ch 9
TRANSCRIPT
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STRENGTH OF MATERIALS
CHAPTER 9
STRESSES AND STRAINS
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STATICS
STRENGTH OF MATERIALS
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Direct Stress Direct stress results from forces acting
perpendicular to the plane of the cross-section
may be tensile, st or compressive, sc
Tension
Compression
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PP
sP
x-sectional area
P
AP
s
Internal stresses Revealed
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DETERMINING LOAD CAPACITY
AsPall
DETERMINING REQUIRED AREA
ALLsP
A
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Direct Stress is always:
1 2
0%0%
1. Parallel to the cross section
2. Perpendicular to the cross section
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Normal or Direct stress is a function of:
1 2 3
0% 0%0%
1. Force/Length
2. Area/Force
3. Force/Area
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The units of Normal Stress are:
1 2 3
0% 0%0%
1. lbs/ft or N/m
2. lbs2/ft or N2/m
3. lbs/ft2 or N/m2
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AP
sP
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Area = 2in x 2in = 4 in2
Part a
Part b
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Rework the previous example assuming the steel bar to be 50mm by 50 mm in cross section a loaded with an axial load of 400 KiloNewtons (KN)
1 2 3
0% 0%0%
1. 160 KPa
2. 1600 Pa
3. 160 MPa
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Areacolumn = 5.5 in x 5.5 in = 30.25 in2
Bearing stress between post and concrete
Bearing stress at base of footing
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TSTC got money to build a new sports arena for Basketball. The new scoreboard isto be hung with 4 steel cables, one at each corner. The scoreboard itself is going toweigh 350,000 lbs. If maximum allowable stress in steel cable is 20 ksi, what diameter must the cables be to support the scoreboard?
1 2 3 4
0% 0%0%0%
1. 1.18 inches2. 2.75 inches3. 2.36 inches4. 3.27 inches
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2
2
4914
)25(mmA
Determine x-sec Area
Determine max load
Determine required Area
Bearing Area
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AT A-A
Area = 0.5in x 4in = 2in2
AT B-B
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sP P
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SHEAR
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1. LOAD RESISTED EQUALLY BY EACH BOLT2. STRESS IS DISTRIBUTED EQUALLY ACROSS EACH CROSS SECTION3. ONE PLANE OF SHEAR IN EACH BOLT4. EACH BOLT RESISTS 9,000 LB
ASSUMPTIONS
Area(each bolt) = π x (.75in)2 = .442in2
psiinA
PsS
362,20)442(.2
000,182
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AREA (PER JOINT) = 75MM X 100 MM = 7,500MM2
ASSUMPTIONS1. EACH JOINT RESISTS HALF THE LOAD
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ASSUMPTIONS1. 2 PLANES OF SHEAR2. EACH PLAN RESISTS HALF OF LOAD
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Direct or Normal Strain
Direct Strain ( ) = Change in Length
Original Length
i.e. = dl/L
dl
FF
L
Strain Deformation
Stretching a copper wire
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Shear Stress and Shear Strain Contd.
P Q
S R
FD D’
A B
C C’
L
x
Shear strain is the distortion produced by shear stress on an element or rectangular block as above. The shear strain, (gamma) is given as:
= x/L = tan
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Shear Animation
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Shear Stress and Shear Strain Concluded
For small , Shear strain then becomes the change in the
right angle. It is dimensionless and is measured in
radians.
Φarc length = radians radius
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1.3 Complementary Shear Stress
a
1
1
2
2
P Q
S R
Consider a small element, PQRS of the material in the last diagram. Let the shear stress created on faces PQ and RS be 1
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Complimentary Shear Stress Contd.
The element is therefore subjected to a couple and for equilibrium, a balancing couple must be brought into action.
This will only arise from the shear stress on faces QR and PS.
Let the shear stresses on these faces be
. 2
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Complimentary Shear Stress Contd.
Let t be the thickness of the material at right angles to the paper and lengths of sides of element be a and b as shown.
For equilibrium, clockwise couple = anticlockwise couple
i.e. Force on PQ (or RS) x a = Force on QR (or PS) x b
1 2
1 2
x b t x a x a t x b
i e
. .
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Complimentary Shear Stress Concluded
Thus: Whenever a shear stress occurs on a plane within a material, it is automatically accompanied by an equal shear stress on the perpendicular plane.
The direction of the complementary shear stress is such that their couple opposes that of the original shear stresses.
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RELATIONSHIP OF STRESS AND STRAIN
s
E E – Modulus of Elasticity (Young’s Modulus)
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s
ss
strainShearstressShear
G
__
MODULUS OF ELASTICITY IN SHEAR
(OR MODULUS OF RIGIDITY)
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calculating stess
Strain - Steel
Strain - Aluminum
Strain - Titanium
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Total Elongation
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Cross sectional area
Axial Deformation
Tensile Stress
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Area 1
Area 2
larger of the 2 areas must be used to find diameter
verify stress
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BAR WITH INTERMEDIATE AXIAL LOADS
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δ =δAB + δBC +δCD
Total Elongation
AB
BC
CD
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P1 = 10kNP2 = 26kNa=bL1=.5mA1=160mm2L2=.8mA2=100mm2E=200GPa for steel
vertical steel barP3 x a = P2 x b
P3 = P2 x b a
P3 = 26kN
LOAD ON AB
P3 – P1 = 16kNDISPLACEMENT AT POINT C
What is displacement ofpoint C?