Machine Design Lecture 4

mec 391-mechanical design
BY
ENGR. FAZLAR RAHMAN
Assistant Professor
ME Faculty, IUBAT
LECTURE NOTE-4

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LOAD AND STRESS ANALYSIS

Plane Stress:
 Plane stress is defined to be a state of stress in which the
normal and shear stresses are zero in the direction of
perpendicular to a plane.
 Stresses in the Z-direction will be zero for a XY-plane. But
strain in Z-direction is not zero.
Plane Strain:
 It is defined to be a state of strain in which the strain in
the direction normal to a plane is assumed to be zero.
 It is normally happened if dimension of the structure in
one direction say Z-direction is large in comparison with

dimensions in the other two directions or XY-directions.
 Strain in the Z-direction will be zero for a XY plane.
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mec -391 (mechanical design)
STRESS AT A POINT IN A BODY

 The normal stress (perpendicular to the surface) is
denoted by ‘σ ’ and shear stress (parallel to the surface) is
denoted by ‘τ ’ .
 ‘σ ‘ outward to the surface is considered tensile stress
and positive. ‘σ‘ into the surface is compressive &
negative.
 (τx)net is resolved in two components τxy and τxz .
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mec -391 (mechanical design)
STRESS AT A POINT IN A BODY (Continue)

 To determine the state of stress at a point in the body, it

would be necessary to examine all surfaces by making
different planar slices through the point.
 The state of stress at point can be examined by only three
mutually perpendicular surfaces.
 Double subscripts are used to denote the shear stresses.
 First subscript indicates direction of surface normal and
second subscript indicates direction of shear stress.

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STRESS AT A POINT IN A BODY (Continue)

 There total nine stress components: σx , σy , σz, τxy , τxz ,
τyx , τyz , τzx and τzy .
 For equilibrium cross-shear are equal to each other.
τxy = τyx ; τxz = τzx and τzy = τyz .
 Reduces stress component from nine to six components:
σx , σy , σz, τxy , τyz and τzx .
 Very common state of stress occurs when stresses on one

surface are zero. When this occurs the state of stress is
called plane stress.
 Assuming the stress-free surface is the Z-direction such
that σz = τzx = τzy = 0 .
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mec -391 (mechanical design)
STRESS AT A POINT IN A BODY (Continue)

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mec -391 (mechanical design)
PLANE STRESS TRANSFORMATION EQUATIONS

 Consider an element ‘dx dy dz’ is cut by an oblique plane.
 Cutting plane stress element at an arbitrary angle ‘φ’ and
balancing stresses gives plane-stress transformation
equations.
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mec -391 (mechanical design)
PLANE STRESS TRANSFORMATION EQUATIONS (Continue)

 The stress σ and τ are found to be

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mec -391 (mechanical design)
PLANE STRESS TRANSFORMATION EQUATIONS (Continue)

 Normal stress σ is maximum at,

 The two values of 2fp are the
principal directions.
 Substitute value of Cos2φp and Sin2φp in both equations.
 The stresses in the principal directions are the principal
stress (max/min) and shear stress are

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mec -391 (mechanical design)
PLANE STRESS TRANSFORMATION EQUATIONS (Continue)

 Shear stress τ is maximum at,
 The two values of 2fs are the directions max/min shear
stress.
 Substitute value of Cos2φs and Sin2φs in both equations.
 The max shear stress and principal stress in max shear
stress plane are

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mec -391 (mechanical design)
PLANE STRESS TRANSFORMATION EQUATIONS (Continue)
Problem-1:

A 12.0 in long beam support load of 15,000 lbf at 3.0 in
from the left end. A designer has selected a 3.0 in Cchannel, which is made of 7075-T6 aluminium extrusion.
Find margin of safety of the beam. Assume beam is simply
supported at both ends. Assume Fty = 66 ksi and Fcy = 72 ksi

, Fsu = 42 ksi and E = 10.4 x106 psi for Al.

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mec -391 (mechanical design)
Material
Property
Card

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PLANE STRESS TRANSFORMATION EQUATIONS (Continue)
Problem-2:

A 15.0 in long beam support a load of 19,500 lbf at 5.0 in
from the left end of the beam. A designer has selected an IBeam, which is made of 7075-T6 aluminium extrusion.
Cross-section of the beam is shown below. The points of
interest are labelled (a, b, c and d) at distances y from the
neutral axis of 0 in, 1.240- in, 1.240+ in and 1.50 in.

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mec -391 (mechanical design)
PLANE STRESS TRANSFORMATION EQUATIONS (Continue)
Problem-2 (continue):

At the critical axial location along the beam find the
following information,
(a). Determine transverse shear stress at each interested
point and find profile of shear distribution.
(b). Determine bending stress at each interested point.
(c). Determine the maximum shear stresses at the point of
interest and compare them.
(d). Find margin of safety.
Assume beam is simply supported at both ends. Assume Fty
= 72 ksi and Fcy = 66 ksi , Fsu = 42 ksi and E = 10.4 x106 psi
for Al.
(SEE HAND ANALYSIS)
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mec -391 (mechanical design)
MOHR’S CIRCLE DIAGRAM

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mec -391 (mechanical design)
MOHR’S CIRCLE DIAGRAM (Continue)

Example-1:

 Two value of principal stress and two value of φp but how
do we know which value of φp corresponds to which
value of principal stress. To find it we need to substitute
value of φp to the corresponding equation.
 A graphical method called Mohr’s circle diagram is very
effective to find directions of the various stress
components associated with plane stress.
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mec -391 (mechanical design)
MOHR’S CIRCLE DIAGRAM (Continue)

Convention of Mohr’s Circle:
 A graphical method for visualizing the stress state at a
point by plotting normal stress along X-axis and shear
stress along Y-axis.
 It represents relation between σx, σy stresses and
principal stresses σ1 , σ2 .
 Parametric relationship between  and  (with 2f as
parameter).
 Relationship is a circle with center at ‘C’ and radius ‘R’
2

 x   y 
2


C = (, ) = [( x +  y)/2, 0 ] and R  
xy


2


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mec -391 (mechanical design)
MOHR’S CIRCLE DIAGRAM (Continue)

Sign Convention of Mohr’s Circle:
 Normal tensile stress (σx, σy positive) is plotted along
positive X-axis and normal compressive stress (σx, σy
negative) is plotted along negative X-axis.
 Shear stress that tending to rotate the element in
clockwise (cw) is plotted along positive Y-axis;
 Shear stress that tending to rotate the element in
counterclockwise (ccw) is plotted along negative Y-axis.
 φ is counted along counterclockwise direction.

Fazlar Rahman; ME Faculty; MEC 391; LEC_4;

mec -391 (mechanical design)
MOHR’S CIRCLE DIAGRAM (Continue)

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mec -391 (mechanical design)
MOHR’S CIRCLE DIAGRAM (Continue)

Problem-3:
A stress element has σx= 80 Mpa and τxy = -50 Mpa as
shown in the figure below.
(a). Find principal & shear stresses and their direction by
using Mohr’s circle diagram; also show alignment of each
element respect to xy coordinates.
(b). Find part (a) by algebraic approach.
(SEE HAND ANALYSIS)

Fazlar Rahman; ME Faculty; MEC 391; LEC_4;

mec -391 (mechanical design)
MOHR’S CIRCLE DIAGRAM (Continue)

Problem-4:
A plane stress element is subject to stress as shown in the
figure below.

(SEE HAND ANALYSIS)

(a). Find the principal & maximum shear stresses and their
direction by using Mohr’s circle diagram; also show
alignment of each element respect to xy coordinates.
(b). Find part (a) by algebraic approach.
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mec -391 (mechanical design)
MOHR’S CIRCLE DIAGRAM (Continue)

Assignmet-1:
A plane stress element is subject to stress as shown in the
figure below.

(a). Find the principal & maximum shear stresses and their
direction by using Mohr’s circle diagram; also show
alignment of each element respect to xy coordinates.
(b). Find part (a) by algebraic approach.
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mec -391 (mechanical design)
THREE DIMENSIONAL STRESSES

 If there is six stress components (σx , σy , σz , τxy , τyz and
τzx , in an element, then there will be three principal
stresses (σ1 , σ2 , σ3) and three shear stresses ( τ1/2 , τ2/3 ,
τ1/3 ).
 Finding roots of the cubic equation,

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mec -391 (mechanical design)
THREE DIMENSIONAL STRESSES

 In plotting Mohr’s circles for three-dimensional stress,
the principal normal stresses are ordered so that
 There are always three extreme-value shear stresses,
 The maximum shear stress is always the greatest of these
three.
 If If principal stresses are ordered so that 1 > 2 > 3,
then max = 1/3 .
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mec -391 (mechanical design)
BENDING IN CURVED BEAM

 Crane hooks, U-shaped Frame and Frame of clamp etc
are example of curved beam (beam initially curved) .
 Neutral axis and centroidal axis are not coincident.
 Bending stress does not vary linearly with distance from
the neutral axis.
 Assume, ro = radius of outer fiber; ri = radius of inner
fiber ; rn = radius of neutral axis ; rc = radius of centroidal
axis ; h = depth of section; always rn < rc and e = rc – rn.
 And co= distance from neutral axis to outer fiber; ci =
distance from neutral axis to inner fiber ; e = distance
from centroidal axis to neutral axis ; M = bending
moment (positive M decreases curvature).
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BENDING IN CURVED BEAM (Continue)

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BENDING IN CURVED BEAM (Continue)

 Location of the neural axis with respect to the centre of
curvature is given by,
 Stress distribution due to resisting internal moment,
 Stress distribution is hyperbolic. Critical stresses at the
inner and outer surfaces, where y = ci and y = co
respectively,
 Above equations are valid for pure bending.
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mec -391 (mechanical design)
BENDING IN CURVED BEAM (Continue)

 In the usual or more general case, such as crane hook, U
frame of press, frame of clamp, bending moment is
calculated about centroidal axis but not the neutral axis.
 Additional axial tensile or compressive stress must be
added to the bending stresses,
 Where M is positive if decreases the curvature;
e = distance in between neutral and centroidal axis;
rn = radius of neutral axis and y = distance from neutral
axis; F = axial force (tension or compressive); A = cross
sectional area.
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mec -391 (mechanical design)
BENDING IN CURVED BEAM (Continue)

Problem-5: (SEE HAND ANLAYSIS)
Plot stress distribution across section A-A of the crane hook
shown in the figure below. The cross section is rectangular
with b = 0.75 in and h = 4.0 in. Load on the hook is 5000 lbf.

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BENDING IN CURVED BEAM (Continue)

Problem-6: (SEE HAND ANLAYSIS)
A machine component has T-shaped cross section and
loaded as shown in the figure below. Find largest value of P
which may apply to the component (allowable compressive
stress is 50 Mpa).

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



PRESSURE VESSELS
There are two kind of pressure vessels:
o Thin-walled pressure vessel :
o Thick-walled pressure vessel.
Example of pressure vessel: Cylindrical Pressure vessels,
Spherical pressure vessel, hydraulic cylinders, gun barrels
and pipe carrying fluids at high pressure etc.
There are three types of stresses developed in the
pressure vessels depending on wall thickness, shape
(cylindrical or spherical).
o Tangential or hoop stress, σt
o Radial stress, σr
o Longitudinal stress, σl
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mec -391 (mechanical design)
PRESSURE VESSELS

 If

inner radius ri ,

 Other


types are Cylindrical Pressure vessels, Spherical
pressure vessel, Closed and Opened Pressure vessels.
Hoop and radius stresses,

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mec -391 (mechanical design)
THICK-WALLED PRESSURE VESSELS
 Cylinder with inside radius ri , outside radius ro , internal
pressure pi , and external pressure po

 Tangential and radial stresses,

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mec -391 (mechanical design)
THICK-WALLED PRESSURE VESSELS
 If outside pressure or external pressure po = 0 then

 If ends are closed, then
longitudinal stress σl
also exist ,

 Shear stress,

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mec -391 (mechanical design)
THIN-WALLED PRESSURE VESSELS
 Cylindrical pressure vessel with wall thickness 1/10 or less
of the radius.
 Radial stress is quite small compared to tangential stress.
 Average and maximum tangential stresses are,

 Longitudinal stress (if ends are closed),

 For Spherical pressure vessel,
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mec -391 (mechanical design)
PRESSURE VESSELS PROBLEM
Problem-7:
A pressure vessel has an outside diameter of 24 cm and wall
thickness of 10 mm. If the internal pressure is 2400 Kpa,
what is the maximum shear stress in the vessel wall?
Problem-8:
A cylinder with 0.30 m internal and 0.40 m external
diameters is fabricated of a material whose elastic limit is
250 Mpa.
Determine the following:
(a). Maximum internal pressure if po = 0 and
(b). Maximum external pressure if pi = 0.
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PRESS AND SHRINK FITS
 A contact pressure ‘p’ exists between the members at the
nominal transition radius ‘R’, causing radial stresses σr = p in each member at the contact surface.
 Pressure for cylinder,
 For the inner member, po = p and pi = 0
 For the outer member, po = 0 and pi = p

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PRESS AND SHRINK FITS (Continue)
 The difference in dimensions in radial direction called
radial interference.
 δi and δo is the change in radii of outer and inner member.

 Tangential strain,
 As per biaxial stress,
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PRESS AND SHRINK FITS (Continue)
 Radial stress of outer cylinder, (σr)o = - p then

 Similarly,
 Total deformation,
 If both cylinders are made of same materials

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PRESS AND SHRINK FITS (Continue)
Problem-9:
A compound cylinder with a = 150 mm, b = 200 mm and
c = 250 mm as shown in the figure below. E = 200 Gpa, and
δ = 0.1 mm is subjected to an internal pressure of 140 Mpa.
Determine the distribution of tangential stress throughout the
cylinder.

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