ppt lecture ch 9

STRENGTH OF MATERIALS

CHAPTER 9

STRESSES AND STRAINS

STATICS

STRENGTH OF MATERIALS

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

PP

sP

x-sectional area

P

AP

s

Internal stresses Revealed

DETERMINING LOAD CAPACITY

AsPall

DETERMINING REQUIRED AREA

ALLsP

A

Direct Stress is always:

1 2

0%0%

1. Parallel to the cross section

2. Perpendicular to the cross section

Normal or Direct stress is a function of:

1 2 3

0% 0%0%

1. Force/Length

2. Area/Force

3. Force/Area

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

AP

sP

Area = 2in x 2in = 4 in2

Part a

Part b

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

Areacolumn = 5.5 in x 5.5 in = 30.25 in2

Bearing stress between post and concrete

Bearing stress at base of footing

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

2

2

4914

)25(mmA

Determine x-sec Area

Determine max load

Determine required Area

Bearing Area

AT A-A

Area = 0.5in x 4in = 2in2

AT B-B

sP P

SHEAR

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

AREA (PER JOINT) = 75MM X 100 MM = 7,500MM2

ASSUMPTIONS1. EACH JOINT RESISTS HALF THE LOAD

ASSUMPTIONS1. 2 PLANES OF SHEAR2. EACH PLAN RESISTS HALF OF LOAD

Direct or Normal Strain

Direct Strain ( ) = Change in Length

Original Length

i.e. = dl/L

dl

FF

L

Strain Deformation

Stretching a copper wire

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

Shear Animation

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

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

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

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

. .

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.

RELATIONSHIP OF STRESS AND STRAIN

s

E E – Modulus of Elasticity (Young’s Modulus)

s

ss

strainShearstressShear

G

__

MODULUS OF ELASTICITY IN SHEAR

(OR MODULUS OF RIGIDITY)

calculating stess

Strain - Steel

Strain - Aluminum

Strain - Titanium

Total Elongation

Cross sectional area

Axial Deformation

Tensile Stress

Area 1

Area 2

larger of the 2 areas must be used to find diameter

verify stress

BAR WITH INTERMEDIATE AXIAL LOADS

δ =δAB + δBC +δCD

Total Elongation

AB

BC

CD

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?