Chapter 3 Load and Stress Analysis -...

Chapter 3

Load and Stress

Analysis

Tuesday, February 11, 2014

Dr. Mohammad Suliman Abuhaiba, PE

1

Chapter Outline



2

Equilibrium & Free-Body

Diagrams

Shear Force and Bending

Moments in Beams

Singularity Functions

Stress

Cartesian Stress

Components

Mohr’s Circle for Plane

Stress

General 3-D Stress

Elastic Strain

Uniformly Distributed

Stresses

Normal Stresses for Beams

in Bending

Shear Stresses for Beams in

Bending

Torsion

Stress Concentration

Stresses in Pressurized

Cylinders

Stresses in Rotating Rings

Press and Shrink Fits

Temperature Effects

Curved Beams in Bending

Contact Stresses

Figure 3–1a shows a simplified rendition of a gear

reducer where the input & output shafts AB and

CD are rotating at constant speeds ωi & ωo,

respectively. The input & output torques are Ti =

240 lbf.in & To, respectively. The shafts are

supported in the housing by bearings at A, B, C,

and D. The pitch radii of gears G1 and G2 are r1 =

0.75 in & r2 = 1.5 in, respectively. Draw FBD of

each member and determine the net reaction

forces and moments at all points.

Free-Body Diagram: Example 3–1



3

FBD: Example 3-1



4



5

FBD: Example 3-1



6

FBD: Example 3-1



7

FBD: Example 3-1

Shear Force and Bending

Moments in Beams

Internal shear force V & bending moment M

must ensure equilibrium



8

Fig. 3−2


Sign Conventions for Bending and

Shear

Fig. 3−3

Distributed Load on Beam

Distributed load q(x) = load intensity

Units of force per unit length



10

Fig. 3−4

Relationships between Load, Shear,

and Bending



11


A notation useful for integrating across

discontinuities

Angle brackets indicate special function

to determine whether forces and

moments are active

Table 3-1



12




13

Table 3−1




14

Table 3−1

Example 3-2


Fig. 3-5

Derive the loading, shear-force, and bending-

moment relations for the beam of Fig. 3–5a.

Example 3-2


Example 3-3


Figure 3–6a shows the loading diagram for a

beam cantilevered at A with a uniform load of 20

lbf/in acting on the portion 3 in ≤ x ≤ 7 in, and a

concentrated ccw moment of 240 lbf.in at x = 10

in. Derive the shear-force and bending moment

relations, and the support reactions M1 and R1.

Example 3-3


Stress

Normal stress, s

Tangential shear stress, t

Tensile Normal stress

Compressive Normal stress

Units of stress: psi, N/m2 = Pa



20

HomeWork Assignment #3-1

3, 6 Due Monday 17/2/2014



21

Stress element Choosing coordinates which result in zero

shear stress will produce principal stresses



22

Cartesian Stress Components

Shear stress is resolved into perpendicular

components

1st subscript = direction of surface normal

2nd subscript = direction of shear stress



23

Plane stress occurs = stresses on one surface

are zero



24

Fig. 3−8

Cartesian Stress Components

Plane-Stress Transformation

Equations



25

Fig. 3−9

Principal Stresses for Plane Stress

principal directions

principal stresses

Zero shear stresses at principal surfaces

Third principal stress = zero for plane stress



26

Extreme-value Shear Stresses for

Plane Stress

Max shear stresses are found to be on

surfaces that are ±45º from principal

directions.

The two extreme-value shear stresses are



27

Maximum Shear Stress

Three principal stresses. One is zero for

plane stress.

Three extreme-value shear stresses.

Max shear stress is greatest of these three

If principal stresses are ordered so that

s1>s2>s3,

then tmax = t1/3



28

Mohr’s Circle Diagram

A graphical method for visualizing stress

state at a point

Relation between x-y stresses and principal

stresses

Relationship is a circle with center at

C = (s, t) = [(sx + sy)/2, 0]

and radius of



29

2

2

2

x y

xyRs s

t

Mohr’s

Circle

Diagram



30

Fig. 3−10

Example 3-4


A stress element has σx =

80 MPa & τxy = 50 MPa cw. a. Using Mohr’s circle, find the

principal stresses & directions,

and show these on a stress

element correctly aligned wrt

xy coordinates. Draw another

stress element to show τ1 & τ2,

find corresponding normal

stresses, and label the

drawing completely.

b. Repeat part a using

transformation equations only.



32

Principal

stress

orientation

Max shear

orientation

Example 3-4

General 3-D Stress

Principal stresses are found

from the roots of the cubic

equation



33

Fig. 3−12

Principal stresses are ordered such that

s1 > s2 > s3, in which case tmax = t1/3



34

Fig. 3−12

General 3-D Stress


15 (a, d), 20 Due Saturday 22/2/2014



35

Elastic Strain Hooke’s law

E = modulus of elasticity

For axial stress in x direction,

n = Poisson’s ratio

Table A-5: values for common materials



36

For a stress element undergoing sx, sy, and

sz, simultaneously,



37

Elastic Strain

Hooke’s law for shear:

Shear strain g = change in a right angle of a stress element when subjected to pure

shear stress.

G = shear modulus of elasticity or modulus

of rigidity

For a linear, isotropic, homogeneous

material,



38

Elastic Strain

Uniformly Distributed Stresses

Uniformly distributed stress distribution is

often assumed for pure tension, pure

compression, or pure shear.

For tension and compression,

For direct shear (no bending present),



39

Normal Stresses for Beams in

Bending Straight beam in positive bending

x axis = neutral axis

xz plane = neutral plane

Neutral axis is coincident with centroidal

axis of the cross section



40

Fig. 3−13

Bending stress varies linearly with distance

from neutral axis, y



41

Fig. 3−14


Bending

Maximum bending stress is where y is

greatest.

c = magnitude of greatest y

Z = I/c = section modulus



42


Bending

Assumptions for Normal Bending

Stress

Pure bending

Material is isotropic and homogeneous

Material obeys Hooke’s law

Beam is initially straight with constant cross section

Beam has axis of symmetry in the plane of bending

Proportions are such that failure is by bending

rather than crushing, wrinkling, or sidewise buckling

Plane cross sections remain plane during bending



43


Example 3-5

Fig. 3−15

A beam having a T section

with the dimensions shown

in Fig. 3–15 is subjected to

a bending moment of 1600

N·m, about the negative z

axis, that causes tension at

the top surface. Locate the

neutral axis and find the

maximum tensile and

compressive bending

stresses.


25, 29, 34.C, 44 Due Monday 24/2/2014



45

Two-Plane Bending

Consider bending in both xy & xz planes

Cross sections with one or two planes of

symmetry only

For solid circular cross section, the

maximum bending stress is



46

Example 3-6


As shown in Fig. 3–16a, beam OC is loaded in

the xy plane by a uniform load of 50 lbf/in, and in

the xz plane by a concentrated force of 100 lbf at

end C. The beam is 8 in long.

a. For the cross section shown determine the

max tensile and compressive bending

stresses and where they act.

b. If the cross section was a solid circular rod of

diameter, d = 1.25 in, determine the

magnitude of the maximum bending stress.

Example 3-6


Dr. Mohammad Suliman Abuhaiba, PE Example 3-6

Example 3-6


Example 3-6

(b) For a solid circular cross section of

diameter, d = 1.25 in, the maximum

bending stress at end O is given by Eq. (3–

28) as


Shear Stresses for Beams in

Bending



53

Fig. 3−17

Transverse Shear Stress (TSS)

TSS is always accompanied

with bending stress



54

Fig. 3−18

Transverse Shear Stress in a

Rectangular Beam



55

Maximum Values of TSS



56

Table 3−2

Maximum Values of TSS



57

Table 3−2

Significance of TSS Compared

to Bending

Figure 3–19: Plot of max shear stress for a

cantilever beam, combining the effects of

bending and TSS.

Max shear stress, including bending stress

(My/I) and transverse shear stress (VQ/Ib),



58


to Bending Figure 3–19



59

Critical stress element (largest tmax) will always be either due to:

bending, on outer surface (y/c=1), TSS = 0

TSS at neutral axis (y/c=0), bending is zero

Transition at some critical value of L/h

Valid for any cross section that does not

increase in width farther away from the

neutral axis.

Includes round & rectangular solids, but

not I beams and channels



60


to Bending

Example 3-7 A beam 12” long is to support a load of 488 lbf

acting 3” from the left support, as shown in Fig. 3–20.

The beam is an I beam with cross-sectional

dimensions shown. Points of interest are labeled a, b,

c, and d (Fig. 3–20c). At the critical axial location

along the beam, find the following information.

a. Determine the profile of the distribution of the

transverse shear stress, obtaining values at each

of the points of interest.

b. Determine bending stresses at points of interest.

c. Determine max shear stresses at the points of

interest, and compare them.


Example 3-7


Torsion

Angle of twist, in radians, for a solid round

bar



64

Fig. 3−21

Assumptions for Torsion

Equations

Pure torque

Remote from any discontinuities or point of

application of torque, Material obeys Hooke’s

law

Adjacent cross sections originally plane &

parallel remain plane & parallel

Radial lines remain straight: Depends on axi-

symmetry, so does not hold true for noncircular

cross sections

Consequently, only applicable for round cross

sections



65

Torsional Shear in Rectangular

Section

Shear stress does not vary

linearly with radial distance

Shear stress is zero at the corners

Max shear stress is at the middle

of the longest side



66

Torsional Shear in Rectangular

Section

For rectangular bxc bar, where b

is longest side



67

Power, Speed, and Torque

Power equals torque times speed

A convenient conversion with speed in rpm

H = power, W

n = angular velocity, rpm



68

Power, Speed, and Torque

In U.S. Customary units, with unit conversion

built in



69

Example 3-8 Figure 3–22 shows a crank loaded by a force F = 300 lbf that

causes twisting and bending of a 34”diameter shaft fixed to a support at the origin of the reference system.

In actuality, the support may be an inertia that we wish to

rotate, but for the purposes of a stress analysis we can consider

this a statics problem.

a. Draw separate FBDs of the shaft AB and the arm BC, and

compute the values of all forces, moments, and torques

that act. Label the directions of the coordinate axes on

these diagrams.

b. Compute maxima of the torsional stress and bending stress

in the arm BC and indicate where these act.

c. Locate a stress element on top surface of shaft at A, and

calculate all stress components that act upon this element.

d. Determine max normal and shear stresses at A. Dr. Mohammad Suliman Abuhaiba, PE

Example 3-8


Fig. 3−22

Example 3-8


Example 3-9

The 1.5”-diameter solid steel shaft shown in

Fig. 3–24a is simply supported at the ends.

Two pulleys are keyed to the shaft where

pulley B is of diameter 4.0 in and pulley C is

of diameter 8.0 in. Considering bending and

torsional stresses only, determine the

locations and magnitudes of the greatest

tensile, compressive, and shear stresses in

the shaft.


Example 3-9


Fig. 3−24


Example 3-9

Closed Thin-Walled Tubes

t << r

t × t = constant

t is inversely proportional to t



79

Fig. 3−25

Total torque T is

Am = area enclosed by section median line

Solving for shear stress



80


Angular twist (radians) per unit length

Lm = length of the section median line



81


Example 3-10

A welded steel tube is 40 in long, has a 1/8

-in wall thickness, and a 2.5-in by 3.6-in

rectangular cross section as shown in Fig. 3–

26. Assume an allowable shear stress of 11500

psi and a shear modulus of 11.5 Mpsi.

a. Estimate the allowable torque T.

b. Estimate the angle of twist due to the

torque.


Example 3-10


Fig. 3−26

Example 3-11

Compare the shear stress on a circular

cylindrical tube with an outside diameter of

1 in and an inside diameter of 0.9 in,

predicted by Eq. (3–37), to that estimated

by Eq. (3–45).


Open Thin-Walled Sections When the median wall line is not closed,

the section is said to be an open section

Torsional shear stress

T = Torque, L = length of median line, c =

wall thickness, G = shear modulus, and q1 = angle of twist per unit length



85

Open Thin-Walled Sections

For small wall thickness, stress and

twist can become quite large

Example:

Compare thin round tube with and

without slit

Ratio of wall thickness to outside

diameter of 0.1

Stress with slit is 12.3 times greater

Twist with slit is 61.5 times greater



86

Example 3-12

A 12” long strip of steel is 1/8”

thick and 1” wide, as shown in

Fig. 3–28. If the allowable shear

stress is 11500 psi and the shear

modulus is 11.5 Mpsi, find the

torque corresponding to the

allowable shear stress and the

angle of twist, in degrees,

a. using Eq. (3–47)

b. using Eqs. (3–40) and (3–41) Dr. Mohammad Suliman Abuhaiba, PE


49, 51, 52, 57, 64 Due Saturday 1/3/2014



88

Stress Concentration

Localized increase

of stress near

discontinuities

Kt = Theoretical

(Geometric) Stress

Concentration

Factor



89

Theoretical Stress

Concentration Factor

Graphs available for standard

configurations

A-15 and A-16 for common examples

Many more in Peterson’s Stress-

Concentration Factors

Higher Kt at sharper discontinuity radius,

and at greater disruption



90

Theoretical Stress




91

Theoretical Stress




92

Stress Concentration for Static

and Ductile Conditions

With static loads and ductile materials

Highest stressed fibers yield (cold work)

Load is shared with next fibers

Cold working is localized

Overall part does not see damage unless

ultimate strength is exceeded

Stress concentration effect is commonly

ignored for static loads on ductile

materials



93

Techniques to Reduce Stress

Concentration

Increase radius

Reduce disruption

Allow “dead zones” to shape flow lines

more gradually



94


Concentration



95


Concentration



96

Example 3-13 The 2-mm-thick bar shown in Fig. 3–30 is

loaded axially with a constant force of 10 kN.

The bar material has been heat treated and

quenched to raise its strength, but as a

consequence it has lost most of its ductility. It

is desired to drill a hole through the center of

the 40-mm face of the plate to allow a cable

to pass through it. A 4-mm hole is sufficient for

the cable to fit, but an 8-mm drill is readily

available. Will a crack be more likely to

initiate at the larger hole, the smaller hole, or

at the fillet? Dr. Mohammad Suliman Abuhaiba, PE

Example 3-13


Fig. 3−30


Example 3-13

Fig. A−15 −1


Fig. A−15−5

Example 3-13


68, 72, 84 Due Monday 3/3/2014



101

Stresses in Pressurized Cylinders

Tangential and radial stresses,



102

Fig. 3−31

Special case of zero outside pressure, po = 0



103


If ends are closed, then longitudinal stresses

also exist



104


Thin-Walled Vessels

Cylindrical pressure vessel with wall

thickness 1/10 or less of the radius

Radial stress is quite small compared to

tangential stress

Average tangential stress



105

Thin-Walled Vessels

Maximum tangential stress

Longitudinal stress (if ends are closed)



106

Example 3-14 An aluminum-alloy pressure vessel is made

of tubing having an outside diameter of 8 in

and a wall thickness of ¼ in.

a. What pressure can the cylinder carry if

the permissible tangential stress is 12 kpsi

and the theory for thin-walled vessels is

assumed to apply?

b. On the basis of the pressure found in part

(a), compute the stress components

using the theory for thick-walled

cylinders. Dr. Mohammad Suliman Abuhaiba, PE


Rotating rings: flywheels, blowers, disks, etc.

Tangential and radial stresses are similar to

thick-walled pressure cylinders, except

caused by inertial forces

Conditions:

Outside radius is large compared with

thickness (>10:1)

Thickness is constant

Stresses are constant over the thickness



108


Stresses are



109


Two cylindrical parts are assembled with

radial interference d

Pressure at interface



110

Fig. 3−33


If both cylinders are of the same material



111

Eq. (3-49) for pressure cylinders applies



112


For the inner member, po = p and pi = 0

For the outer member, po = 0 and pi = p



113



94, 104, 110 Due Wensday 5/3/2014



114

Temperature Effects

Normal strain due to expansion from

temperature change

a = coefficient of thermal expansion



115

Temperature Effects

Thermal stresses occur when members are

constrained to prevent strain during

temperature change

For a straight bar constrained at ends,

temperature increase will create a

compressive stress

Flat plate constrained at edges



116

Coefficients of Thermal

Expansion

Table 3–3: Coefficients of Thermal Expansion (Linear

Mean Coefficients for the Temperature Range 0 –

100°C)



In thick curved beams

Neutral axis and centroidal axis are not

coincident

Bending stress does not vary linearly with

distance from the neutral axis



118




119

Fig. 3−34

Location of neutral axis

Stress distribution



120


Stress at inner and outer surfaces



121


Example 3-15

Plot the distribution of stresses across

section A–A of the crane hook shown in Fig.

3–35a. The cross section is rectangular, with

b = 0.75 in and h = 4 in, and the load is F =

5000 lbf.


Example 3-15


Fig. 3−35

Example 3-15


Formulas for Sections of

Curved Beams (Table 3-4)



Formulas for Sections of Curved Beams (Table 3-4)

Alternative Calculations for e

Approximation for e, valid for large curvature where e is small with rn rc

Substituting Eq. (3-66) into Eq. (3-64), with rn

– y = r, gives



129

Example 3-16

Consider the circular section in Table 3–4

with rc = 3 in and R = 1 in. Determine e by

using the formula from the table and

approximately by using Eq. (3–66).

Compare the results of the two solutions.


Contact Stresses

Two bodies with curved surfaces pressed

together

Point or line of contact changes to area

contact

Stresses developed are 3-Dl

Called contact stresses or Hertzian stresses

Common examples

Wheel rolling on rail

Mating gear teeth

Rolling bearings



131

Spherical Contact Stress

Two solid spheres of diameters d1 and d2

are pressed together with force F

Circular area of contact of radius a



132

Spherical Contact

Stress

Pressure distribution is

hemispherical

Max pressure at the

center of contact area



133

Fig. 3−36


Maximum stresses on the z axis

Principal stresses



134


From Mohr’s circle, maximum shear stress is



135


Plot of three principal stress & max shear

stress as a function of distance below the

contact surface

tmax peaks below the contact surface

Fatigue failure below the surface leads to

pitting & spalling

For poisson ratio of 0.30,

tmax = 0.3 pmax at depth of z = 0.48a



136




137

Fig. 3−37

Cylindrical Contact Stress

Two right circular

cylinders with length l

and diameters d1 & d2

Area of contact is a

narrow rectangle of

width 2b and length l

Pressure distribution is

elliptical

Maximum pressure



138

Fig. 3−38


Half-width b

Maximum pressure



139

Fig. 3−38

Maximum stresses on z axis



140


Plot of stress components and max shear

stress as a function of distance below the

contact surface

For poisson ratio of 0.30,

tmax = 0.3 pmax at depth of z = 0.786b



141




142

Fig. 3−39



129, 133, 138 Due Saturday 8/3/2014



143

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