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”

Direct Shear Stress

average

P

A

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

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

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

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

BENDING

C

TNeutral Plane

Before

After

Displacement in Beam

y

xv

Curvature of Beam = d v

dx

M

EI

2

2

M = Moment; EI = Bending stiffness of Beam

Bending Stress

M y

I

max (C)

cc

ct

max (T)

y

max M c

I

Source of Internal Moment

F1

F21 m M

V

x

Fc

Ft

d

Fc = Ft

M= Ft d

Equivalent FBD

F1

F21 m

V

x

d

F

F

M = Fd

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

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

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!

Neutral Axis

XX

300

50

60

250

Ct

Ok...

Ct/2

2/60 tt CC

2/25060250 tt CC

0

2502550300

tC

mmCt 200

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

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

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

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

Bending Shear

VA

F1

F21 m

V

x

Bending Shear

VA

F1

F21 m

V

x Fy = 0

V

Bending Shear

VV

V

V

A

F1

F21 m

V

x

Fy = 0 Fx = 0 M = 0

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

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

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

Try It!

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

X X

300

50

60

250

Ct

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

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

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.

Torsion

Tr

J

TL

JG

r

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

Shear Stress Distribution

max

r

D

Polar Moment of Inertia

Circle: J = 32

Tube: J = 32

D

D Do i

4

4 4( )

DoDi

Shear Modulus, G

GE

2 1( )

E = Young’s Modulusv = Poisson’s Ratio

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

Shear Stress-Strain Curve

She

ar S

tres

s,

Shear Strain,

Shear Modulus, G

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

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

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

Result:

Bending

80 MPa

160 MPaTension

Compression

N.A.

AxialF/A = 167 MPaTension

+ =

83 MPa (Tension)

327 MPa

Tension

Net

< 350 MPaNo Yield!

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.