Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending...

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Analysis of Basic Load Cases • Axial Stress • Tension and Compression Shear Stress Examples • Bending •Tension/Compression & Shear • Torsion Shear stress
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Transcript of Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending...

Page 1: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Analysis of Basic Load Cases

• Axial Stress• Tension and Compression

• Shear Stress• Examples

• Bending•Tension/Compression & Shear

• Torsion• Shear stress

Page 2: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Axial Stress

axial F

A

FL

AE;

F= Axial Force (Newtons, N)A = Cross-Sectional Area Perpendicular to “F” (mm2)E = Young’s Modulus of Material, MPaL = Original Length of Component, mm

= Average Stress (N/mm2 or MPa)

= Total Deformation (mm)

AE = “Axial Stiffness of Component”

Page 3: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Direct Shear Stress

average

P

A

average = Avearge Shear Stress (MPa)P = Shear LoadA = Area of Material Resisting “P”

Page 4: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Examples of Direct Shear Stress

Bolted Joint withTwo Shear Planes.

P = 50 KND = 13 mm

avg = ?

Area of bolt (Ab) = D2 / 4 = 13)2 / 4 = 132.7 mm2

A resisting shear = 2 Ab

avg = P / 2Ab = 50000 N/ 2(132.7) mm2 = 188.4 MPa

D

Page 5: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Direct Shear II

175

150

1313

Fillet Weld

Find the load P, such that the stress in the weld does not exceed the allowable stress limit of 80 MPa.

9.2

Page 6: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Solution:

avg = P / Aw = 80 MPa

Aw = Throat x Total Length = (9.2)(175)(2) = 3217 mm2

P / 3217 = 80 MPaP = (3217)(80) N = 257386 N = 257 kN

Page 7: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

BENDING

C

TNeutral Plane

Before

After

Page 8: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Displacement in Beam

y

xv

Curvature of Beam = d v

dx

M

EI

2

2

M = Moment; EI = Bending stiffness of Beam

Page 9: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Bending Stress

M y

I

max (C)

cc

ct

max (T)

y

max M c

I

Page 10: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Source of Internal Moment

F1

F21 m M

V

x

Fc

Ft

d

Fc = Ft

M= Ft d

Page 11: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Equivalent FBD

F1

F21 m

V

x

d

F

F

M = Fd

Page 12: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Bending Stiffness

EI = Bending StiffnessE = Young’s Modulus (Material Dependant)I = Moment of Inertia (2nd Moment of Area)

XN eu tra l A x is

X Ixx = bh3 / 12 (mm4)h

Iyy= hb3 / 12 (mm4)

y

yb

Page 13: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Moment of Inertia

X

A

yx ' x '

X

Parallel Axis Theorem

IXX = Ix’x’ + A y2

AIf X-X is the neutral axis:

A y = 0

Page 14: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Locate the Neutral Axis and find the Moment of Inertia for the “T” section shown below. Consider the XX axis, all dimensions are mm.

X X

300

50

60

250

Ct

Ans: Ct = 200 mm Ixx = 2.50x108 mm4

Try it!

Page 15: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Neutral Axis

XX

300

50

60

250

Ct

Ok...

Ct/2

2/60 tt CC

2/25060250 tt CC

0

2502550300

tC

mmCt 200

Page 16: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Mech 422 – Stress and Strain Analysis

XX

300

5060

250

200

Ok...

75

48

7677

23

23

23

1050.2

1044.81013.31044.81081.7

)75)(50)(300(12/)50)(300(

)75)(250)(60(12/)250)(60(

12/

mmx

xxxx

AybhI xx

75

Page 17: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Determine the Bending Stiffness of beamswith this cross section made of:

1) Steel, E=203 000 MPa 2) Aluminum Alloy, E= 72 000 MPa

3) Glass Reinforced Polyester, E = 30 000 MPa

Page 18: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Determine the Bending Stiffness of beamswith this cross section made of:

1) Steel, E=203 000 MPa2) Aluminum Alloy, E= 72 000 MPa3) Glass Reinforced Polyester, E = 30 000 MPa

Ans:1) EI = 5.08 x 1013 Nmm2

2) EI = 1.80 x 1013 Nmm2

3) EI = 0.75 x 1013 Nmm2

In the ratio 1 : 0.36 : 0.15

Page 19: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Bending Stress:

3 m

1 m

3 0 0 0 K N

100 KN200 KN

M max = 200 KN.m max M c

I

tension = 200x106 N.mm (200) mm / 2.50 x108 mm4

= 160 MPa

compression = 200x106 N.mm (100) mm / 2.50 x108 mm4

= 80 MPa

MAX

300 kN

Page 20: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Bending Shear

VA

F1

F21 m

V

x

Page 21: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Bending Shear

VA

F1

F21 m

V

x Fy = 0

V

Page 22: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Bending Shear

VV

V

V

A

F1

F21 m

V

x

Fy = 0 Fx = 0 M = 0

Page 23: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Shear Stress - Bending

b

h

max

max = 1.5 V / A avg = V / A

A = b h

y

For a Rectangle:

Max Shear Stress is at N.A.

avg

Page 24: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Shear in Bending

b = width of the X-section at the plane of interest.

y = V Q / I bQ = 1st Moment of Area

In General:

X X

b @ N.A.

b @ top of web

Page 25: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

1st Moment of Area

y

A=zw

w

z

Qy = A d

• Consider all of the X -section above (or below)

the plane of interest.d

NA

Page 26: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Try It!

Find The 1st Moment of Area at the Neutral Axis.(dimensions are mm.)

X X

300

50

60

250

Ct

Page 27: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Solution:The 1st Moment of Area at the Neutral Axis:(dimensions are mm.)

X X

300

50

60

250

200

75

25

Q = Ay

= (300)(50)(75) + (60)(50)(25)= 1.2 x 106 mm3

Note: At the N.A. b=60 mm

Page 28: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Shear Stress:

3 m

1 m

3 0 0 0 K N

100 kN200kN

V max = 2000 KN y = V Q / I b

max = (200x103 N) (1.2x106 mm3) (2.50x108 mm4 (60) mm

avg = V/Aweb = (200x103 N) / (60)(300) mm2

= 11.1 MPa

= 16 MPa

300 kN

Page 29: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

SummaryThe “Maximum” Stress Distributions in the beam are:

X X

300

50

60

250

Bending Shear

80 MPa

160 MPaTension

Compression

max=16 MPa

N.A.

Page 30: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Torsion

Tr

J

TL

JG

r

T = Torque, G = Shear Modulus of ElasticityL = Length of Shaft, J = Polar Moment of Inertia is in Radians!

Page 31: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Shear Stress Distribution

max

r

D

Page 32: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Polar Moment of Inertia

Circle: J = 32

Tube: J = 32

D

D Do i

4

4 4( )

DoDi

Page 33: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Shear Modulus, G

GE

2 1( )

E = Young’s Modulusv = Poisson’s Ratio

Example: SteelE = 203 000 MPa, v = 0.3G = 78 000 MPa

Page 34: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Shear Stress-Strain Curve

She

ar S

tres

s,

Shear Strain,

Shear Modulus, G

Page 35: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Try it!

Determine the angle of twist for the steel shaft shownbelow. Calculate the safety factor against yield. The shearstrength of the steel is 200 MPa, and the Young’s Modulusis 203000 MPa.

500 mm

25 KN

150 mm

Torque

50 mm

Page 36: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Solution:T = 25x103 N (150 mm) = 3.75x106 N.mm

L = 500 mm

J = 32

D4

= (50)4 / 32 = 6.14x105 mm4

TL

JG3.75x106 N.mm (500) mm 6.14x105 mm4 (78000 N/mm2)

= = 3.9x10-2 = 2.2o

Tr

J=

3.75x106 N.mm (25 mm)

6.14x105 mm4= 152.7 MPa

FS = 200/152.7 = 1.31

Page 37: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Superposition

Assume the beam in our example is made of steel with a yield stress of 350 MPa. If it is subjected to an additional Axial Tension of 5000 kN along it N.A., will it yield ?

3 m

1 m

100 kN200 kN

F = 5000 KN

300 kN

Page 38: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Result:

Bending

80 MPa

160 MPaTension

Compression

N.A.

AxialF/A = 167 MPaTension

+ =

83 MPa (Tension)

327 MPa

Tension

Net

< 350 MPaNo Yield!

Page 39: Analysis of Basic Load Cases Axial Stress Tension and Compression Shear Stress Examples Bending Tension/Compression & Shear Torsion Shear stress.

Rule for Adding Stresses:

Like stresses at a point acting in the same direction and on the same plane can

be added algebraically.

• You can’t add a shear stress to a tensile/compressive stress.• You can’t add a stress in one location to one at another.• The effects of combined shear and tension/compression are covered later in this course.