Strength of Materials

Imam Khomeini International University Department of Mechanical Engineering

MECHANICS OF MATERIALS Lecture Notes:

M. Ghadiri September 2010

Stress Analysis

Stress Analysis

Axial Loading: Normal Stress

Centric & Eccentric Loading

Shearing Stress

Shearing Stress Single Shear

 ave 

P F  A A

Double Shear

 ave 

P F  A 2A

Bearing Stress in Connections

• Corresponding average force intensity is called the bearing stress, b 

P P  A td

Kinds of Stress

Stress Analysis & Design Example • Would like to determine the stresses in the members and connections of the structure shown. • From a statics analysis: FAB = 40 kN (compression) FBC = 50 kN (tension) • Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection

Rod & Boom Normal Stresses • The rod is in tension with an axial force of 50 kN. • At the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2) is sBC = +159 MPa. • At the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline, A  20 mm 40 mm  25 mm   300  10 6 m 2 P 50  103 N  BC ,end    167 MPa A 300  10 6 m 2

• The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa. • The minimum area sections at the boom ends are unstressed since the boom is in compression.

Pin Shearing Stresses • The cross-sectional area for pins at A, B, and C, 2

 25 mm  6 2 A r    491 10 m  2  2

• The pin at A is in double shear with a total force equal to the force exerted by the boom AB,

 A, ave 

P 20 kN   40.7 MPa A 49110 6 m 2

Pin Shearing Stresses • Divide the pin at B into sections to determine the section with the largest shear force, PE  15 kN PG  25 kN (largest)

• Evaluate the corresponding average shearing stress,  B, ave 

PG 25 kN   50.9 MPa A 491 10 6 m 2

Pin bearing Stresses • To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm, b 

P 40kN   53.3MPa td 30mm25mm

• To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm, b 

P 40kN   32.0 MPa td 50mm25mm

Stress in Two Force Members • Axial forces on a two force member result in only normal stresses on a plane cut perpendicular to the member axis.

• Transverse forces on bolts and pins result in only shear stresses on the plane perpendicular to bolt or pin axis.

Stress on an Oblique Plane • Resolve P into components normal and tangential to the oblique section, F  P cos

V  P sin 

• The average normal and shear stresses on the oblique plane are  

F P cos P   cos 2  A A0 A0 cos



V P sin  P   sin  cos A A A0 0 cos

Maximum Stresses • Normal and shearing stresses on an oblique plane 

P cos 2  A0



P sin  cos A0

• The maximum normal stress occurs when the reference plane is perpendicular to the member axis, m 

P A0

  0

• The maximum shear stress occurs for a plane at + 45o with respect to the axis, m 

P P sin 45 cos 45    A0 2 A0

Stress Under General Loadings

• The distribution of internal stress components may be defined as, F x  x  lim  A  0 A

 xy  lim A 0

V yx A

V zx  xz  lim  A  0 A

State of Stress • The combination of forces generated by the stresses must satisfy the conditions for equilibrium:  Fx   Fy   Fz  0

Mx  My  Mz  0

• Consider the moments about the z axis:  M z  0   xy Aa   yx Aa  xy   yx similarly,  yz   zy

and  yz   zy

• It follows that only 6 components of stress are required to define the complete state of stress

Stress & Strain: Axial Loading Normal Strain

P  stress A     normal strain L



2P P  2A A   L



P A 2    2L L



Stress-Strain Test

Stress-Strain Diagram: Ductile Materials

Stress-Strain Diagram: Brittle Materials

Stress-Strain Diagram

Hooke’s Law: Modulus of Elasticity • Below the yield stress

  E E  Youngs Modulus or Modulus of Elasticity • Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not.

Impact Testing Toughness is usually measured by an impact test. Charpy test is most commonly used in the US. Charpy Test

Izod Test

Pendulum

Impact

Release Height Height after Impact

Notched Specimen

Deformation and Fracture Section

Deformation and Fracture Section

Deformation and Fracture Section

Elastic vs. Plastic Behavior • If the strain disappears when the stress is removed, the material is said to behave elastically. • The largest stress for which this occurs is called the elastic limit. • When the strain does not return to zero after the stress is removed, the material is said to behave plastically.

Loop Hysteresis

Bauschinger Effect

Fatigue • Fatigue properties are shown on S-N diagrams. • A member may fail due to fatigue at stress levels significantly below the ultimate strength if subjected to many loading cycles. • When the stress is reduced below the endurance limit, fatigue failures do not occur for any number of cycles.

Fatigue

Deformations Under Axial Loading • From Hooke’s Law:

  E



 P  E AE

• From the definition of strain:   L • Equating and solving for the deformation, PL   AE • With variations in loading, cross-section or material properties, PL   i i i Ai Ei

Deformations Under Axial Loading Example 1 Determine the deformation of the steel rod shown under the given loads.

E  29  10 6 psi D  1.07 in. d  0.618 in.

Deformations Under Axial Loading SOLUTION: • Divide the rod into three components:

Static Indeterminacy Example 2

Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied.

Static Indeterminacy SOLUTION: • Solve for the displacement at B due to the applied loads with the redundant constraint released,

P1  0 P2  P3  600  103 N A1  A2  400  10 6 m 2

A3  A4  250  10 6 m 2

L1  L2  L3  L4  0.150 m Pi Li 1.125  109 L    A E E i i i

P4  900  103 N

Static Indeterminacy • Solve for the displacement at B due to the redundant constraint,

P1  P2   RB A1  400  10 6 m 2

A2  250  10 6 m 2

L1  L2  0.300 m 1.95  103 RB Pi Li äR    A E E i i i





Static Indeterminacy   L R  0 1.125  109 1.95  103 RB    0 E E





RB  577  103 N  577 kN

 Fy  0  R A  300 kN  600 kN  577 kN R A  323 kN

R A  323 kN RB  577 kN

Thermal Stresses • Treat the additional support as redundant and apply the principle of superposition. PL  T   T L P  AE   thermal expansion coef.

• The thermal deformation and the deformation from the redundant support must be compatible.

  T   P  0  T L 

PL 0 AE

  T   P  0 P   AE T  P     E T  A

Poisson’s Ratio • For a slender bar subjected to axial loading:  x  x  y z  0 E • The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence),

 y  z  0 • Poisson’s ratio is defined as y  lateral strain    z axial strain x x

Generalized Hooke’s Law • For an element subjected to multi-axial loading, the normal strain components resulting from the stress components may be determined from the principle of superposition. This requires: 1) strain is linearly related to stress 2) deformations are small

 x  y  z x     E E E  x  y  z y     E E E  x  y  z z     E E E

Dilatation: Bulk Modulus • Relative to the unstressed state, the change in volume is











e  1  1   x  1   y 1   z   1  1   x   y   z



 x  y  z 

1  2  x  y  z E





 dilatation (change in volume per unit volume)

• For element subjected to uniform hydrostatic pressure, e  p k

31  2  p  E k

E  bulk modulus 31  2 

• Subjected to uniform pressure, dilatation must be negative, therefore 0    12

Shearing Strain • A cubic element subjected to a shear stress will deform into a rhomboid. The corresponding shear strain is quantified in terms of the change in angle between the sides,

 xy  f  xy 

• A plot of shear stress vs. shear strain is similar the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains,

 xy  G  xy  yz  G  yz  zx  G  zx where G is the modulus of rigidity or shear modulus.

Shearing Strain Example 3 A rectangular block of material with modulus of rigidity G = 90 ksi is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P. Knowing that the upper plate moves through 0.04 in. under the action of the force, determine a) the average shearing strain in the material, and b) the force P exerted on the plate.

Shearing Strain SOLUTION:

• Determine the average angular deformation or shearing strain of the block.  xy  tan  xy 

0.04 in. 2 in.

 xy  0.020 rad

• Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.  xy  G xy  90  103 psi 0.020 rad   1800 psi





• Use the definition of shearing stress to find the force P. P   xy A  1800 psi 8 in.2.5 in.  36  103 lb P  36.0 kips

Relation Among E, n, and G • An axially loaded slender bar will elongate in the axial direction and contract in the transverse directions. • An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped. The axial load produces a normal strain. • If the cubic element is oriented as in the bottom figure, it will deform into a rhombus. Axial load also results in a shear strain. • Components of normal and shear strain are related, E  1    2G

Relation Among E, n, and G Example 4 A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses sx = 12 ksi and sz = 20 ksi. For E = 10x106 psi and n = 1/3, determine the change in: a) the length of diameter AB, b) the length of diameter CD, c) the thickness of the plate, and d) the volume of the plate.

Relation Among E, n, and G SOLUTION: • Apply the generalized Hooke’s Law to find the three components of normal strain. x  

 x  y  z   E E E

1       12 ksi 0 20 ksi     3 10  106 psi 

• Evaluate the deformation components. B

A

  x d   0.533  10 3 in./in. 9 in.



B C

D

 x  y  z y     E E E  1.067  103 in./in.

 x  y  z z     E E E  1.600  103 in./in.

A

 4.8  10 3 in.

  z d   1.600  10 3 in./in. 9 in.





C

1

 0.533  10 3 in./in.



D

 14.4  10 3 in.

 t   y t   1.067  103 in./in. 0.75 in.





 t  0.800  10  3 in.

• Find the change in volume e   x   y   z  1.067  103 in 3/in 3 V  eV  1.067  10 3 15  15  0.75in 3 V  0.187 in 3

Stress & Strain Distribution under Axial Loding

Saint-Venant’s Principle

Stress Concentration

Stress-Concentration Factor

Stress Concentration

Stress-Concentration Factor

Chapter 3

Jet engine of plane

Automotive power train

Torsion

Torsional Loads on Circular Shafts • Interested in stresses and strains of circular shafts subjected to twisting couples or torques torques. • Turbine exerts torque T on the shaft. • Shaft transmits the torque to the generator. • Generator creates an equal and opposite torque T’.

Stresses in a Shaft • Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque, torque

T    dF     dA

• Although the net torque due to the shearing stresses is known, the distribution of the stresses is not. • Distribution of shearing stresses is statically indeterminate – must consider shaft deformations.

• Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed uniform uniform.

Axial Shear Components • Torque applied to shaft produces shearing stresses on the faces perpendicular to the axis. • Conditions of equilibrium require the existence of equal stresses on the faces of the two planes containing the axis of the shaft.

• The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats.

• The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.

Shaft Deformations • From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length.  T L

• When subjected to torsion, every cross-section of a circular shaft remains plane and undistorted undistorted.

• Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric.

• Cross-sections of noncircular (non-axisymmetric) shafts are distorted when subjected to torsion.

Shearing Strain • Since the ends of the element remain planar, the shear strain is equal to angle of twist.

• It follows that L   or  

 L

• Shear strain is proportional to twist and radius c   max  and    max L

c

Stresses in Elastic Range • Multiplying the previous equation by the shear modulus,  G   G  max c

From Hooke’s Law,     max

  G  , so

c

The shearing stress varies linearly with the radial position in the section. section

J  12  c 4

• Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section,   T    dA  max   2 dA  max J c c

J  12  c24  c14





• The results are known as the elastic torsion formulas, Tc T  max c 2  max  ;  and  J J  min c1

Shearing Strain Example 1: What is the largest torque that can be applied to the shaft if the shearing stress is not exceed 120 Mpa? What is the corresponding minimum value of the shearing stress in the shaft?

Hollow Cylindrical Steel Shaft

Normal Stresses • Elements with faces parallel and perpendicular to the shaft axis are subjected to shear stresses only. Normal stresses, shearing stresses or a combination of both may be found for other orientations.

• Consider an element at 45o to the shaft axis, F  2 max A0  cos 45   max A0 2

 45o 

F  max A0 2    max A A0 2

• Element a is in pure shear. shear • Element c is subjected to a tensile stress on two faces and compressive stress on the other two. • Note that all stresses for elements a and c have the same magnitude

Torsional Failure Modes • Ductile materials generally fail in shear. Brittle materials are weaker in tension than shear.

• When subjected to torsion, a ductile specimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis.

• When subjected to torsion, a brittle specimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft axis.

Shearing Strain Example 2: Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, respectively. Shafts AB and CD are solid of diameter d. For the loading shown, determine (a) the minimum and maximum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD if the allowable shearing stress in these shafts is 65 MPa.

Shearing Strain SOLUTION: • Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings.

 M x  0  6 kN  m   TAB TAB  6 kN  m  TCD

 M x  0  6 kN  m   14 kN  m   TBC TBC  20 kN  m

Shearing Strain • Apply elastic torsion formulas to find minimum and maximum stress on shaft BC.

J

 4 4  c2  c1  0.0604  0.0454 2 2

 



 13.92 10

 max   2 

6

m



• Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter.

 max 

2

4

TBC c2 20 kN  m 0.060 m   J 13.92  10  6 m 4

 min 45 mm  86.2 MPa 60 mm

 min  64.7 MPa

65MPa 

6 kN  m  c3 2

c  38.9  103 m

 86.2 MPa

 min c1   max c2

Tc Tc  J  c4

 max  86.2 MPa  min  64.7 MPa

d  2c  77.8 mm

Angle of Twist in Elastic Range • Recall that the angle of twist and maximum shearing strain are related,  max 

c L

• In the elastic range, the shearing strain and shearing stress are related by Hooke’s Law,  max 

 max Tc  G JG

• Equating the expressions for shearing strain and solving for the angle of twist, 

TL JG

Angle of Twist in Elastic Range • If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations Ti Li i J i Gi



• If the shaft cross-section changes along the length, the angle of rotation is found as following L

 

0

T dx JG

Angle of Twist in Elastic Range

T A D  2T rA A  rB B

Determining the Modulus of Rigidity

Torsion Testing Machine

Statically Indeterminate Shafts • Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B. • From a free-body analysis of the shaft, TA  TB  90 lb  ft

which is not sufficient to find the end torques. The problem is statically indeterminate. • Divide the shaft into two components which must have compatible deformations,   1   2 

T A L1 TB L2  0 J1G J 2G

LJ TB  1 2 TA L2 J1

• Substitute into the original equilibrium equation, LJ TA  1 2 TA  90 lb  ft L2 J1

Stress Concentrations in Circular Shafts

Stress Concentrations in Circular Shafts Example 3:

Stepped Shaft

Stress Concentrations in Circular Shafts Solution:

Torsion of Noncircular Members

Torsion of Noncircular Members

Torsion of Noncircular Members

Several Thin-Walled Members

Thin-Walled Hollow Shafts

Thin-Walled Hollow Shafts