mom summary points

8/17/2019 MoM Summary Points

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Ammar

Qaseem

BSME 13-17

Pakistan Institute of Engineering and Applied Scie

Mechanics of Materials -I


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Book Used: Mechanics of

Materials by R.C Hibbeler

Chapter 1: Stress

External Forces:

Surface forces:

Caused by direct contact of bodies.

Body forces:

Caused without contact e.g gravitational, magnetic

Support reactions:

Their cause(supports) aren’t included in free body diagram.

Always balance forces; not stresses since stress is per unit area (units are N/m2=Pa) and

will sum up to different force magnitudes depending on the area involved.

Stress has same units as that of pressure but pressure is on the outside while stress is in

the inside (in sections)(though it is still caused by external forces)

Types of Stresses/Stress Forces:

Shear Stress (V):

Directed tangent to the plane.

Normal Stress (N):

Directed perpendicular to the plane.

Principle Plane:

The plane that doesn’t have any shear stresses

Symbols:

Normal Stress: (Sigma) σ

Shear Stress: (Thao) τSubscripts: for normal, the normal direction is used. Forshear stresses, first subscript is the plane’s normal stress

direction while second shows their direction.


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Similarly, like shear and normal stress forces, there are

torsional (T) and bending (M) moments that are

components of the net stress moment.

Special Case of Coplaner Forces:

If a body is subjected to coplanar forces only,

then it will have only Normal, shear forces and

bending moment at the section point. (NO

TORSIONAL MOMENT)

If normal force or stress pulls on delta A, it is called tensile stress, if it

pushes, it is called compressive stress.

By convention, tensile forces are taken positive while compressive are taken

negative.

Average Normal Stress:

It is stress force divided by area of the plane on which it is applied on. σavg

Average Shear Stress:It is shear force divided by area of the plane on which it is applied on. τavg

Complementary property of shear:

It says all the shear forces acting on cube are

equal and have to be either pointing towards or

away from each other at the edges.

Here tzy and tyz are equal and vice versa.

(proved by balancing all moments about x axis)

But since tyz and t’zy are also equal, so all four shears become equal.

τzy = τyz = τ’zy = τ’yz

Fail Stress:

Ex 1.9

Ex 1.10


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The stress above which the material would break. Allowable stress is always kept

smaller than fail stress.

Allowable Stress:The stress which the device is supposed to bear. It is smaller.

Factor of Safety FS:The ratio of fail stress to allowable stress.

Area required for load bearing:Force divided by allowable stress gives the area required.

FS = σfail / σall A = F / σall

Chapter 2: Strain

Normal Strain:

Change in length of a line per unit length.

ϵavg =

The exact strain per unit length can be defined by applying limit that

s approaches zero. Although strain is a dimensionless quantity, it is sometimes expressed in units of length.

Shear Strain: If we consider two lines perpendicular to each other, than the change of angle between them

that occurs due to stress is called strain.

Denoted by

=

– lim(B->a along n)(C->a along t)

Small Strain Analysis:

Assumption is taken in most cases that the

deformations in the body are almost infinitesimal i.e


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Chapter 3: Mechanical Properties of Materials

Tension/Compression Test:Before testing, material is made into standard shape and size (constant circular cross section

with enlarged ends). Two small punch marks are placed along the length. Initial readings are

taken and the material is placed into the testing machine.

For measurement purposes, either extensometer (manual) or electrical resistance

strain gauge can be used.

ERSG It works on the resistance technology. It has an inner wire

placed in circular orientation which is attached to the

material being tested. When material length is changed,

the coil also changes cross sectional area and length

R=

This length is calculated by the change in the resistance of

the coil.

Caution: The elastic limit of coil should be taken in consideration.

Nominal/Engineering Stress:It assumes that Stress is constant over the cross sectional area and over the length

It always takes the original Area under consideration(even if it changes)

= Nominal/Engineering Strain:It assumes that Strain was constant throughout the region between gauge points. It can be

gotten by gauge or by equation

ϵ=

True Stress-Strain:During a Stress-Strain Diagram, we are calculating stress at every point by taking the A and L

as the original length and area. If we take the area and length of specimen at the instant we

measure load P and change , we would get true stress and true strain. Since in True Stress, the A has decreased (due to necking), so true stress is greater than

engineering stress. True Stress is equal to the nominal Stress as long as the strain is

small.

Note: All three examples are very

important for covering strain topic.

Ex 2.3 (normal

and shear

strain)

Ex 2.4 (shear

strain

proper)


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Stress-Strain Diagrams:There are two types of Stress-Strain Diagrams:

1) Conventional Stress-Strain Diagram

2) True

Stress-Strain

Diagram

Elastic Region:

In this region, the material has the capability to revert to original shape. The linear deformation

ends at proportional limit while the elastic region ends at elastic limit.

Yield Stress:

It has two points. Upper and lower. The material has a sudden change in length after which it

approaches lower yield stress. (elastic region).

Strain Hardening: (why true stresses rise)

After the yield stress elongation, plastic behavior is observed. The material has now increased

capacity to bear loads. Thus greater force causes small strain. So

Nominal stress increases.

Cross-sectional area decreases so true stress increases a lot.

This strain hardening ends at ultimate stress pointNecking:

After ultimate stress, a neck starts to form and stress concentrates at that point. Here the strain

hardening isn’t present anymore. So

Nominal stress decreases (normal force, normal strain)

cross-sectional area decreases a lot, so true stress increases significantly


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Ductile-Brittle Materials:Any material that can be subjected to large strains before it breaks is called ductile materials.

Otherwise it is brittle.

It’s measure is given by percent elongation. (formula is obvious)

It can also be given by percent decrease in area.

Offset yield point method:

When the yield point of material is not well defined e.g Almunium, we use offset

method (graphical). A 0.2 % (0.002 in/in usually) strain is chosen(strain is a percentage

itself) and a line is that drawn from point parallel to initial proportional line of material.

The point at which this line intersects the curve is

declared as the yield point.

At low temperature, materials become harder

and more brittle, while at high temperature theyare softer and more ductile

Hooke’s law: Stress is proportional to Strain within proportional limit.

=E This E is the Young’s modulus (modulus of elasticity). It

is the slope of stress-strain graph (within proportional

limit).

Strain Energy:As deformation occurs in a body, it tends to store energy throughout its volume. This is called

strain energy.

When strain is applied on a body, it undergoes deformation due to force. This constitutes work

done which is equal to the strain energy.

- Force is increased uniformly from zero to its final magnitude, so we take average

- Displacement is , where x can be any axis

Work=Strain Energy= =

= =

For convenience, we also use strain energy density which is denoted by u


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u=

Note: Strain energy density is the area under the curve at the desired point

Modulus of Elasticity: It is the young’s modulus. Read the Hook’s law above.

Modulus of Rigidity/Shear Elasticity:It is the ratio of shear stress to shear strain. See shear stress-strain diagram below.

Modulus of Resilience:When the material reaches its proportional limit(not yield point), the strain energy

density at that point is referred to as modulus of resilience

Def: Ability of material to absorb energy without any permanent deformation.

Modulus of Toughness:The strain energy density at the material’s fracture point is referred to as modulus of

toughness (entire area under stress-strain diagram)

Def: Ability of material to absorb energy without fracturing.

Poisson’s Ratio: When a deformable body is under axial stress, it deforms longitudinally as well as laterally.

Then the respective strains are

ϵlong=

, ϵlat=

Now the Poisson’s ratio is

(negative sign added)

Note: The formula is only valid for homogenous and isotropic materials


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The negative sign is added because if longitudinal is positive, latitudinal will be negative.

Thus the negatives cancel out and ratio always remain positive

1) If latitudinal strain is zero, Poisson’s ratio will be 0. (very least possible value)

2) Maximum possible value for this ratio is 0.5

Shear Stress-Strain Diagram:

Shear Modulus of Elasticity

G=shear stress/shear strain

G =

(both under proportional limit)

Related to young’s modulus as…

G=

Creep:When a material is subjected to load for a long period of time, it keeps deforming slowly until a

sudden fracture occurs or its usefulness is impaired. This time dependent effect is called creep.

- Generally creep strength (resistance to creep) decreases with high temperature and

high stress

In Diagram, as less strain is applied, the creep time of thematerial increases. Thus the stress at which time approaches

infinity is called endurance limit.

Fatigue:When a material is subjected to repeated cycles of stress or strain, it causes its structure to

break down, ultimately leading to fracture. This is called fatigue.

e.g crankshafts, train shafts, cyclic swing in amusement parks, breakage of wire by to and fro

bending etc. In all cases, the material breaks before its yield stress.

Reason: There are minute imperfections in the structure of material. When subjected to cyclic

loadings, cracks form in the material. These cracks lead to formation of other cracks and even a

ductile material fractures like a brittle one.


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In diagram, as lower stress is applied, the material can

endure greater number of cycles. The stress limit at

which the number of cycles approach infinity is called

endurance limit.

Numerical points:

While calculating strain from young’s modulus, we must

remember that young’s modulus will only be valid if there is no plastic deformation.(only

elastic)

Symbols So far…

σ

τ ϵ

Normal Stress

Shear Stress

Normal Strain

Shear Strain

E

Gu

v

Modulus of Elasticity

Modulus of Rigidity(shear elasticity)

Modulus of resilience

Poisson’s ratio

Chapter 4: Axial Load

Actual Deformation: Deformation more in the region near the point of force

application.

Saint Venants’s principle: stress and strain produced at points in a body sufficiently

removed from the region of load application will be the

same as the stress and strain produced by any applied

loadings that have the same statically equivalent

resultant, and are applied to the body within the sameregion.

Ex 3.3

Tension force is taken positive by

convention while compression is

taken negative


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Elastic deformation of axially loaded member:Since stress is force per unit area, if member hasn’t uniform cross section and the force is also

variable, then the stress will be a ration of their respective functions.

Special case: (Constant cross section and force)

Superposition principle:By subdividing the loading into components, the principle of superposition states that the

resultant stress or displacement at the point can be determined by algebraically summing the

stress or displacement caused by each load component applied separately to the member.

Conditions:

1) The loading must be linearly related to the stress or displacement that is to be

determined.

2) The loading must not significantly change the original geometry or configuration of the

member

Statically indeterminate axially loaded member:

Such members in which the reaction forces are not simply enough to find thestresses are called statically indeterminate.

Solution:

The bar is split in two at the point C. Now the change δ caused by both bars

must cancel each other out.

Flexibility or force method of analysis:


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In this method, one support reaction is assumed to be none (forced statically determinate) and

change in length is found, then the removed reaction is supposed to be another force causing

change in length. In the end, the net change is set equal to zero.

Thermal Stress:

If heat is applied to the body and the body isn’t allowed to expand, then thermal stress isproduced.

Where α is the linear coefficient of expansion.

Note: The formula is only valid for homogenous and isotropic materials

Stress Concentrations:

- At sharp bends or sudden change in cross sections, the stress gets concentrated- We are concerned with the maximum stress value that can accumulate in a material

- For this we have stress concentration factor K

Determining the K (concentration factor):

- The value of K can be seen from graphs- Graphs for K are different for different bar geometries.

- Different graphs need different variables for finding K due to different geometry of

materials

- Graphs for 2 kinds of geometries are given below.

Thermal Stress and Thermal strain

can never exist at the same time.


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Inelastic axial deformation:

- When load is increased, after a while the yield point is reached.

- Now the plastic deformation starts.

- In reality the strain at this point will be yield strain for a little while and then strain

hardening will start.

- But we assume the material to be elastic perfectly plastic or elastoplastic so that strain

hardening never occurs.

- This way, all our analysis will be safe since in reality the structure will have additional

capability to resist loads due to strain hardenings.


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When plastic deformation starts, the irregular stress distribution slowly converts to uniform

stress distribution.

Residual Stress:

- When stress is applied to a material and it deforms plastically, it reverts back to zero

stress when relaxed.

- Some strain still exists at this point.

- But If the member is statically indeterminate, then upon

removing the force, the reaction forces would be trying to

bring it to the zero strain.

- But the original stress had already equaled to zero, when

the strain wasn’t zero. Thus the reaction forces would take the material to a point of

zero strain and negative stress.

- This is called Residual Stress

Numerical Point:

First the parts of material deform; both

plastically or one plastically. The sum of change

in length of both parts equals to zero. PL/AEdoesn’t apply here

Then the force P is released. Now we

know that they both will revert to original shape

elastically. So we find the forces by using PL/AE.

The sum of change in length of both parts equals

to zero because extension of one is only possible

by compression of other.


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Numerically, residual stress is found because during plastic deformation, the maximum stress

can only be the yield stress but during the elastic recovery, stress is greater than yield stress.

This causes an overall residual stress since full recovery isn’t possible on removing the force.

Chapter 5: Torsion

Shear Strain:

Shear Strain increases with radius

linearly.

Thus according to τ=Gγ the shearstress also increases linearly


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The Torsion Formula:

and

Where T is the torque, J is the moment of inertia. Derivation can be seen from book. Max shear

stress is at the outer surface of the shaft. C is the total radius of the shaft. Rho is the radius at

the point at which we are finding stress.

Some moments of inertia:

Circular Shaft:

Tubular Shaft:

Co and Ci are the outer and inner radiuses respectively.

Power Transmission: If a torque is causing a shaft to rotate, then we can find the power transmitted by the formula

OR

Note: If the frequency and power transmission of shaft are known, torque can be found which

lets us calculate the area of cross-section.

Angle of Twist:

For very long pipes like thousands of feets of bore

pipe, it is important to keep angle of twist dφ in

mind since it increases axially.

Shear Strain γ increases with distance from

circular center while Angle of twist increases

with axial length


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The formula can be derived by equating s=rθ for both angles γ and ρ

Now, replacing using hooke’s law and then replacing shear stress by our formula, we get therelation

Special Case: Constant cross-sectional Area

Note: Multiple Torques:

In case of multiple torques, twist is found from each torque

individually and then summation is taken.

Caution: Shafts must not yield for these formulas to be

valid

Statically Indeterminate Torque Loaded Members:

As previously done, we get one equation due to summation

equation,

but we need a compatibility equation which is obtained by setting the net angle of twist due to

moments at A and B equal to zero

Caution: This is only valid within elastic limit.

Recall:

Note the similarities (compare)

Sign Convention

Use right hand rule for torque. If

thumb points outwards, thentorque is positive else negative.

Anti-Clockwise=positive


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Plastic deformation and residual stresses phenomenon is very similar to the force case. They

will be explained if in assignment.

Stress Concentration:Just like before, in sudden change areas, stress reaches a max value that is K times the average

stress. We will only be showing graph for one type of change.

Chapter 6: Bending

Beams:Members that are slender and support loadings that are applied perpendicular to their

longitudinal axis are known as beams.

- Simply Supported beam: Pinned at one end and roller at other end.

- Cantilever beam: Fixed at one and free at other end.


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Ex 6.2, 6.6 - Overhanging beam: one or both its ends freely extended over the supports

Shear and Moment Diagrams:In beams shear forces and moments are generated which may vary along the length of the

beam. This shear force and moment when plotted along the length of the beam is called a

shear or moment diagram.Choice of origin:

To plot the length, one end of beam must be taken as origin and a positive direction must be

assumed. Usually it’s the left end with left direction as positive.

Note: Shear and moment diagrams cannot be drawn just by looking at the diagrams but they

can be roughly guessed by looking at a few things of the freebody. (coming later)

Beam Sign Convention:Case: External distributed load, upward positive

Case: Internal Shear, clockwise positive

Case: Internal Moment, Compressive on upward surface istaken positive

Caution: These are only Internal Shear and Internal Moment

conventions.

Before we take a plunge into shear and moment diagrams,

you should have a grip on the derivative and integral graph

making.

Derivative Integral Graphical relationship: Derivative shows only the change in a function

If function is horizontal at any point, derivative graph is at zero on that point.

If function is increasing (moving upwards), derivative graph will be in positive region.

If function is decreasing (moving downwards), derivative graph will be in negative region.

If function change is constant (straight line) (changing constantly), then derivative is a

horizontal line.

- If function change is increasing and function is increasing, the derivative graph will become more

vertical upwards.

- If function change is increasing and function is decreasing, the derivative graph will become more

vertical downwards.

viceversa...

REVERSE: We will need a starting point to plot function from the "change graph".

practice the reverse. Practice doesn't make you perfect but it certainly increases you self-esteem.


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Shear and Moment Diagrams:

After analyzing a generalized beam under distributed loading, we reach the following

conclusions

Thus in order to master the diagrams, you must have good practice of making integral graphs.

Some of the common

curves will be shown here.

If a point load acts on a beam, shear force receives a sharp change which is equal to the

magnitude of that force, while moment receives a sharp edge.

If a point Moment acts on a beam, Internal Moment receives a sharp change which is

equal to the magnitude of that moment.

Note: The equations are solved by all conventions but the diagrams are drawn by the new

conventions. How this is done? While drawing fbds, draw the shear force and internal moment

as positive by convention. Then solve by any convention. Thus if they are actually negative

beam convention, we will get negative values.


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Bending Deformation of a Straight Member:

From the figure, we can see that in upper portion, compression takes place while in lower

portion, elongation takes place. Thus there must exist a neutral axis where no change will be

observed longitudinally.

We will base our calculations on some assumptions:

Assumptions:

1) The neutral axis does not undergo any change in length. (it will curve though)

2) All cross-sections of beam remain plain and perpendicular to the neutral axis.

Calculations:

By making use of s=rθ, we will

calculate the longitudinal strain

Substituting the values, we get


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Note: This neutral axis is not necessarily in the center.

Actual Deformation:

We made some assumptions about no

strain in cross-sectional areas. In anactual case, the cross-sectional area

below neutral axis will become smaller

while the area above will become larger (if load is applied downwards on the beam). This can

be calculated by Poisson’s ratio.

Common Conventions:Remember:

Normal force is positive when directed out of body. Compressive is negative.

Strain is positive when expansion takes place. Negative in case of compression.

Stress is positive when directed normal to surface.Make sure the calculated stress and strain are in accordance with these conventions.

The Flexure Formula:

By Hook’s law, we can deduce that,

Also, we can see that for y positive, we get strain negative and stress negative. Moment is

positive for y pos or neg. The stress strain formulas are in accordance with these.

Finding the neutral Axis:

We set the condition that the net force

Due to stress distribution is equal to zero.

Since the stress max or c cannot be zero, so First Moment must be equal to zero.

This condition (first moment zero) is only satisfied at centroidal axis.


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Finding the stress:

We set the condition that the resultant internal moment must

be equal to the moment about the neutral axis produced due

to stress distribution.

Now, we can see that we have moment of inertia formed. So wecan deduce the following equations. These are flexure formulas

Drawing Shear/Moment Diagrams: While making diagrams, draw the shear force and internal moment assumed direction such that they

are positive by beam convention.

Make the beam to be in sections. Different formulas govern different sections.

Moments also includes all moments due to forces of freebody diagram. Forces do not include forces

due to moments.

While calculating internal moment, take the moment about the sectioned point since, shear force

might not be known.

It is derived for beams that are

symmetrical across an axis

perpendicular to neutral axis

Moment of Inertia for a rectangular

beam is given by I=1/12 bh3


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Unsymmetric Bending:

Now we imply some conditions for beams that are not symmetrical across an axis perpendicular

to the neutral axis.

Imagine a couple moment to be applied on a beam. Thus we can say a few things about it

Summation forces is zero

Moment across y axis is zero.

Moment across z axis=Internal

Moment

So we get some results

This term is called Product of Inertia and it is equal to zero in this case.

Moment Arbitrarily Applied:

Moment might not be applied just across z axis. It might be

in the middle of z and y axis. In this case, we add the flexure

formulas of the moment components

Orientation of neutral axis:


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Stress Concentrations ?

Chapter 7: Transverse Shear

This chapter is explains the shear force more in detail and how it is applied in crossections.

Shear in Straight members:Due to complimentary property of shear, longitudinal shear

stresses are formed and thus the member undergoes cross-section warp.

Consequence:

As a result, flexure formula which assumed “All cross-section of

beam remain plain and perpendicular to the neutral axis.” That

doesn’t hold true. BUT we can generally assume that for

members that have small depth compared to their length

(Slender Beams), this warping is so little that we can neglect it.

The Shear Formula:

We need to make a formula for the complementary shear force applied in the beam.

So let’s assume a general beam


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Now we take an arbitrary cross-section from it

- NA is the neutral axis

- We ignored the vertical forces (distr load and shear force since they don’t affect

horizontal complimentary shear)

- One side of the crossection has moment M while the other has M+dM

- One side has dF’’ while other has dF’. Important things is that above and bottom forces

cancel each other out.

Now to calculate complimentary shear, we take a section from this cross-section(the shaded

one) (above figure) (we will refer this as section from now on(self-made name))

- Now before we took this section, above and bottom forces satisfied ∑Fx=0

(btw since beam isn’t moving, the condition must satisfy)


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- But now we see that moment on both sides differ by dM and thus on the right side, the

forces are greater. (positive moment on right side suggest forces in direction of forces

that are already on the right side)(the internal moment is taken about neutral axis)

- So to balance x directed forces, there must be a shear force underneath the section

which is the complimentary shear

Now we balance the x directed forces

(Note that Area A’ on both sides of our section is equal)(the above picture is the guide for this)

- Here, dM/dx is shear force V,- Integral of ydA’ is the first moment of area A’ about neutral axis(since y is distance from

neutral axis).

- Also we can call this integral as Q and since , we can say that

(This y’ bar is the distance of the centroid of our section from the neutral axis)

Now we put in the values and get the “shear formula”

The shear formula gives us

the longitudinal shearstress, where is our

transverse shear stress???

(First diagram of this chap tells the

difference between them)

According to the

complementary property of

shear, these two are equal

Didn’t we say that shear stress is simply shear force divided by Area???

Yes, that was average shear But actually shear force also causes longitudinal shear

which due to complimentary produces extra transverse shear stress. Thus stress by

shear formula is reater It is actual


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Limitation on the use of Shear Formula:

1) Since Flexure formula was used in its calculation, the material must remain elastic.

2) Assumption is that the longitudinal shear stress is uniformly distributed across beam but

it is actually arc shaped ( larger on the corners ).

This has little affect unless width to height b/h is much greater like this

Beam on right

And also the shear applied is actually average shear stress

3) Shear formula does not give accurate results at sharp corners since stress concentration

is in effect there.

4) Another important limitation is that in members where the boundary isn’t rectangular

Fig (a), at the outer areas, we can make two two components of shear stress. τ’ and τ’’

which act on parallel and perpendicular planes to the boundary. Fig (c). Since boundary

is stress free, so τ’ is zero. And shear direction is on perpendicular to the boundary. Fig

(d). Thus we cannot find the shear simply. We can only find shear along the coloured

lines of Fig (a) in this case.

To Summarize, the shear formula does not give accurate results when applied tomembers having cross sections that are short or flat, or at points where the cross-

section suddenly changes. Nor should it be applied across a section that intersects the

boundary of the member at an angle other than 90°.


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Shear Flow in built-up members:Often members are joined with each other and the adhesive material (nails, glue) has to resist a

certain force. This force per unit length of the beam is called Shear Flow.

Now just like we previously formulated, summation of x directed forces will give us this

equation,

Just like before, we introduce Q, divide both sides by dx thus getting shear force V and dF/dx

becomes q i.e shear flow

Note: The concept of Q will only be clear after practicing said numericals.

Clarification of Q: In it, Q=ybar x A’. This y is the distance of the centroid of A’ from the neutral

axis. Carefully note what is A’.

Now after calculating shear flow q, this q will be resisted by the number of nails given. Etc.


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Chapter 8: Combined Loadings

Thin Walled Pressure Vessels:

- Provided that the vessel wall is “thin”, the stressdistribution in it will not vary greatly and can be

assumed constant.

The pressure in the vessel will always be the gauge pressure

since the atmospheric pressure is assumed to be present

outside the cylinder before the cylinder is pressurized.

Now we perform some stress analysis,

Cylindrical Vessels:

Now we take a section from this cylinder as specified.

- Stress pulls the section inwards

- Pressure pushes it outwards. Pressure

Is taken at the middle of cylinder since

It is the same pressure that is applied to the circular region.

(no additional force acts between these two points)

- Self Explanatory

Thin Wall refers to a vessel

having an inner radius to wall

ratio of 10 or more r/t >= 10

Reason: If radius and thickness

ratio is 10, the stress calculations

of the wall give a stress which is

4% less than the actual max stress

in the wall.

Error will be even smaller in larger

rations

Caution: This is radius to thicknessratio, diameter to thickness rationmust be >= 20


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Spherical Vessels:

Some Results:

From the above calculations, we found that an element taken from pressure vessel (cylindrical

or spherical) is subjected to biaxial stress. i.e a Normal Stress existing in only two directions.

Actually a radial stress is also occurring which is equal to ‘P’ at the inside and reduces to zero at

the outer surface since the gauge pressure there is zero. However for thin walled vessels we

can ignore this component since the circumferential σ2 and longitudinal σ1 stresses are 5 and 10

times greater than the maximum radial stress ‘P’. (r is 10 times remember)

In Cylindrical Vessels, we can see

that circumferential stress is

twice as much as longitudinal

stress.

Thus circumferential joints must

be designed twice as strong as

longitudinal joints.


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Chapter 8: Combined Loadings

Plane Strain Transformation:We have studied that at any point in a body there are six normal stresses and other shear

stresses. But in actual case ‘Engineers’ make assumptions and simplifications to analyze thesestresses in a single plane. In this case, the material is said to be subjected to plane strain.

Example:

When there is no stress on outer surface of a material, than there is obviously no stress on the

opposite inner face either. Thus the material is said to be subjected to plane strain.

Seen in: Cylindrical and spherical thin walled pressure vessels.

Representation of Stresses:

In an xy plane, the element is arranged, we can represent the stresses by two normal stress

components and one shear stress component (Fig (a))

However we change its orientation, the

stresses will have different values (since

area also plays a role) (Fig (b))

Thus we can say that,

“The state of plane stress at the point isuniquely represented by two normal stress

components and one shear stress

component acting on an element that has a

specific orientation at the point”


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Ex 9.1 page

440

Stress Transformation:

We will learn how to transform stress components from one orientation (a) to another

orientation (b). This is just like finding x and y components of a resultant vector and a waste of

my time if you ask me.

Not so fast there buddy : Force can be transformed considering its magnitude and direction.

But stress must take care of its area as well.

General Equations of Plane-Stress Transformation:

What’s our world if not automation :

The difficult procedure of the last topic can be automated by making general equations and

simplifying them. (You would have realized they were difficult if you had actually tried them)

So as always, forming conventions:

- Angle positive is taken from positive x axis to the positive x’ axis.- Positive normal stress is outward normal from the material surface.

- Positive Shear stress is along the positive y axis.

σ’ can be calculated by setting the angle to θ+90 in the

normal stress formula.(caution => there is a minus sign in the shear transformation formula)

Principal Stresses and Maximum In-Plane Shear Stresses:

From the above two topics we saw that the magnitude of the stresses dependon the orientation angle θ of the element. Since in Engineering, we are often

concerned with finding the orientation that causes the stress to be maximum

or minimum, this thing is a friggin’ goldmine for us.


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In-Plane Principal Stresses:Calming down…first of all, The maximum and minimum In-Plane Stress is called Principle Stress.

Now to calculate the maximum stress, we take the derivative of the stress transformation

formula and set it equal to zero.

Now to put it into the stress transformation formula,

we need cos and sin functions of this θp angle. We can

obtain those from the graph on the left side by plotting the

perpendicular and base of the triangle; for both positive and

both negative stresses.

Putting these values in the normal stress transformation

formula, we get

So maximum and minimum normal stresses can be found where σ1 >= σ2 This set is called Principal Stresses,

And their corresponding planes are called principal planes.

Furthermore if we put the trigonometric relations in the shear stress transformation formula,

we can find that the shear stress there is zero. So we can say that

“ No Shear Stress Acts On The Principal Planes.”

Maximum In-Plane Shear Stresses:Just like before, we take derivative, make tan, make triangles, put the values in shear

transformation formula. We get

At the Principal Plane, if

one normal stress is

maximum, then the other

component of normal

stress will be minimum.

Thus two principal planes

are possible in which 2

different components are

maximum.


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We observe two things in this case.

The maximum In-Plane shear stress occurs at an angle of 45 degree from the Principal

Plane.

At the Maximum shear plane, there is also normal stress present which is average of the

initial stresses.

Mohr’s Circle: It is used to analyze plane stress and understand the principle stress, transformed stress etc.

Recall the stress transformation equations (normal and

shear)

We can remove the angle by squaring and adding these

equations.

Now we the stresses on the right side are all known constants. So we replace them by R.

ThePositive Shear Stress is onthe downward axis of the Mohr’s

Circle.

A flower from the garden of Javed

Mohr’s Circle for Plane Stress

and Plane Strain is one of the

favorite tools of the interviewers


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This is an equation of Circle. From this we build a diagram called Mohr’s circle.

Keeping in mind the two circle equations, we can understand what point represents what.

1)

2) Let we have an element, it will have σx, σx, τxy. Remember that these are notnecessarily the maximum or minimum.

3) We can construct a circle of a radius R by these values. 4) For convenience, we take the origin of the circle to be at σavg.

5) Now plotting the constant values of circle of radius R, we get the P point where thedistance of P from the circle center is^ and from the y axis is σx . The distance from the xaxis is τxy.

6) Now imagine some scenarios where we transform the element and the values change.


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Figure (a)

σx’ = σx

τx’y’ = τxy Since there is no change in this case, this is point P (our reference point)

Figure (b) (rotating 90 degree anticlockwise)

σx’ = σy

τx’y’ = -τxy

On the Mohr’s circle , we

can see that plotting the

new σx’ τx’y’ we get thepoint G.

Thus we can see that byrotating the reference

line by 2θ, we obtain

transformed values of θ.

Values of normal stress in y direction can also be obtained by plotting mohr’s diagram

for it.

If instead the τ axis were established positive upwards, then the angle 2θ on the circle

would be measured in the opposite direction to the orientation θ of the plane. (what

the book said :/)

Absolute Maximum Shear Stress:

Previously:

On a body, if normal stress was acting in two directions, we wereable to find the orientations where one stress became max and the

other became min. We were assuming not considering z direction

stress.

If we consider z direction in a 3 dimensional body, then the

principal stresses will be max, intermediate, and minimum. This

condition is called triaxial stress

(Too complex for our scope. Too complex my ass )


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Finding Absolute Maximum Shear Stress:

We ignore z direction stresses and consider xy, xz

and yz plane.

Drawing Mohr’s circles for all these planes, we find

the maximum shear stress.

The above Diagram is pretty self-explanatory (look carefully at the subscribts, which plane’s

shear stress is which) along with these formulas. (circle center will always be at the averagenormal stress)


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Mohr’s circle

Absoulte maximum shear stress

mom summary points

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