Contributed by:
Elastic constants, Longitudinal strain, Lateral strain, Volumetric strain, Bulk modulus, Superposition principle, Thermal stresses
1.
ELASTIC CONSTANTS
There are three types of elastic constants. They are
1. Modulus of elasticity or Young’s Modulus (E)
2. Bulk modulus (K)
3. Shear Modulus or Modulus of Rigidity (G or C)
We know that the length of the a bar increases by axial tension. But the same time, there is a
reduction or decrease of the lateral dimension of the bar.
When a body is loaded axially, it deforms longitudinally or lengthwise as well as transversely
or crosswise and the following strain are developed.
i) Longitudinal strain
ii) Lateral strain
LONGITUDINAL STRAIN
The ratio of axial deformation to the original length of the body is known as longitudinal or
linear strain
longitudinal strain (el) = δL / L
where
δL = Change in length
L = Original length
2.
LATERAL STRAIN
The length of the bar will increase while the diameter of the bar will decrease.
consider a circular bar due to an axial load shown in fig.
Where
δL = Increase in length
δd = Decrease in diameter
3.
Lateral Stain (et) = δd / d
Where,
δd = Change in diameter
d = Original diameter
In Case of Rectangular Bar,
Lateral strain (et) = δb / b or δt / t
Where,
δb = Increase in breadth
δt = Decrease in thickness
POISSON’S RATIO
When a body is stressed, within its elastic limit, the ratio of lateral strain to the longitudinal
strain is constant for a given material.
Poisson’s ratio (µ or 1/m) = lateral strain / longitudinal strain
1/m = et / el
4.
VOLUMETRIC STRAIN
The ratio of change in volume to the original volume is known as volumetric strain.
volumetric strain = change in volume / original volume
ee = dV / V
Volumetric Strain of a Rectangular Bar
ee = dV / V
= δL / L [ 1- (2/m)]
where,
δL = change in length
L = original length
1/m = poison's ratio
5.
Volumetric Strain of a Cylindrical Rod
ee = dV / V
= δL / L – [2 δd /d ]
where,
δL = change in length
L = original length
δd = Change in diameter
d = Original diameter
Young’s Modulus or Modulus of Elasticity (E)
When a body is stressed , within its elastic limit, the ratio of tensile stress to the
corresponding tensile strain is constant.
E = Tensile stress / tensile strain
or
E = Tensile stress / longitudinal strain
E = σ / et
6.
Modulus of Rigidity or Shear Modulus (G)
When a body is stressed, within its elastic limit, the ratio of shearing stress to the
corresponding shearing strain is constant.
Its is denoted by G or C or N
Modulus of Rigidity (G) = Shear stress/ Shear strain
= τ/ø
Where
τ = Shear stress
ø = Shear strain
I. Relationship between Modulus of Elasticity (E) and Modulus of Rigidity (G)
i,e., E = 2G [1+(1/m)]
Where
E = Young’s modulus
G or C or N = Modulus of Rigidity
1/m = Poisson’s ratio
7.
II. Relation between Bulk modulus (K) and Young’s modulus (E)
i.e., E = 3k [1- (2/m)]
Formula Used
1. Stress (σ) = Load/Area
= P/A N/mm2
2. Longitudinal strain (el) = Change in length / Original length
= δL / L
3. Lateral strain (et) = δb / b or δt / t or δd / d
4. Volumetric strain = change in volume / original volume
ev = dV / V
5. Poisson’s ratio (µ or 1/m) = lateral strain / longitudinal strain
1/m = et / el
8.
6. Young’s modulus ( E ) = Tensile stress / longitudinal strain
= σ / et
7. Young’s modulus E = 2G [1+(1/m)]
8. Young’s modulus E = 3k [1- (2/m)]
9. Modulus of Rigidity (G) = Shear stress/ Shear strain
= τ/ø
10. Volumetric Strain = change in volume / original volume
ee = dV / V
11. Volumetric Strain of a Rectangular Bar
= δL / L [ 1- (2/m)]
12. Volumetric Strain of a Cylindrical Rod
= δL / L – [2 δd /d ]
9.
Problems
1. Determine the change in length, breadth and thickness of a steel bar which is 5m long,
20mm wide and 15mm thick subjected to an axial pull of 100 kn in the direction of its
length. Take E = 200 Gpa and Poisson's ratio = 0.3
2. A bar of 20mm diameter is tested in tension. It is observed that when a load of 40 kn is
applied, the extension measured over a gauge length of 200mm and 0.12 mm and
contraction in diameter is 0.0036 mm .find the Poisson's ratio and elastic constants E, G
and K
3. A steel rod of 300 mm diameter, 40 mm wide and 25 mm thick is subjected to a pull of
180 kn. Determine the change in volume of the bar. Take E = 2 × 105 N/mm2 and
Poisson’s = 0.3
10.
STRESSES IN BARS OF VARYING CROSS SECTIONS
Consider the following non-uniform cross section of member AB, BC and CD having cross
sectional areas of A 1, A2 and A3 with lengths of L1, L2 and L3 as shown in fig.1.1
11.
Tensile stress in portion , AB, σ = Load/ Area = P/A1
Elongation of AB, δL1 = PL1/ A1E …………..(1)
Tensile stress in portion , BC, σ = Load/ Area = P/A2
Elongation of BC, δL2 = PL2/ A2E ……………(2)
Tensile stress in portion , CD, σ = Load/ Area = P/A3
Elongation of CD, δL3 = PL3/ A3E ……………..(3)
Total Elongation = δL1 + δL2 + δL3
δL = (PL1/ A 1E ) + (PL1/ A1E ) + (PL1/ A1E )
δL = P/E [(L1/A1) + (L2/A2) + (L3/A3) ] ……(4)
12.
1. An axial load of 40 kn is acting on a bar consisting of three sections of length 300 mm,
259 mm, and 200 mm and diameter 20mm, 40mm,and 50mm respectively. Find the stress
in each section and total extension of the bar. Take E = 2 ×105N/mm2.
2. A road bar as shown in fig.1.6 is subjected to an axial tensile load of 100kn. Determine the
diameter(D1) of the first part if the shear stress in the first part is 100 MN/m 2. find also
total elongation. Take E = 290 Gpa.
3. The bar shown in fig.1.7 is subjected to a tensile load of 150 kn. Find the diameter of the
middle portion if the stress is limited to 160 N/mm 2. find also the length of the middle
portion if the total elongation of the bar is 0.25 mm. Take E = 200 GN/m 2.
4. A steel bar 900 mm long. Its two ends are 40 mm and 30 mm in diameter and the length of
each rod is 200 mm. the middle portion of the bar 15 mm in diameter and 500 mm long. If
the bar is subjected an axial tensile load of 15 kn. Determine 1) Stress in each section and
2) Total extension.
13.
Assignment Problems:( Varying Cross Section)
1. An axial pull of 40 KN is acting on a bar consisting of three section of length 300 mm,
250 mm and 200 mm and of diameter 20mm, 40mm, and 50mm respectively. Find the
stress in each section and total extension of the bar. Take E = 2 ×105N/mm2.
2. The bar shown in fig1.1 is subjected to a tensile load of 150 KN. Find the diameter of the
middle portion if the stress is limited to 160 N/mm2. . Find also the length of the middle
portion if the total elongation of the bar is 0.25mm, Take E = 2 GN/m 2.
3. A member formed by connecting a steel bar to an aluminum bar is shown in fig.1.2.
assuming that the bars are prevented from buckling sideways, calculate the magnitude of
force P that will cause the total length of the member to decrease 0.25 mm, the values of
elastic modulus for steel and aluminum are 2 ×105N/mm2 and 7 ×104N/mm2 respectively.
14.
PRINCIPLE OF SUPER POSITION
When a number of loads are acting on a body, the forces or (loads) are split up and their
effects or (deformation) are consider on individual sections, according to the principle of
superposition.
While using this principle for an elastic body which is subjected to a number of direct force
(tensile or compressive ) at different sections along the length of the body, first the free body
diagram of individual section is drawn.
Then the deformation of the each section is obtained.
The total deformation is equal to the algebraic sum of the deformation of the individual
section.
1. A member ABCD is subjected to loading as shown in fig.1.1.Determine the total elongation.
Take E = 2 ×105N/mm2.
2. A member ABCD is subjected to part loads as shown in fig.1.2. Calculate the force P 2
necessary for equilibrium and total change in length of the bar. Take E = 210 KN / mm 2
3. Calculate the force P3 and change in length for the following fig.1.3. Take E = 200GN/mm 2.
P1 = 120 KN ; P2 = 220 KN ; and P4 = 160 KN.
20.
1. A circular rod 2.5 m long, tapers uniformly from 25 mm diameter to 12 mm diameter.
Determine the extension of the rod under a pull of 30 kn. Assume modulus of elasticity of
the rod is 200 kn/mm2..
2. A steel flat plate ab of 1 cm thickness tapers uniformly .from 10 cm to 5 cm width in a
length of 40 cm. Determine the elongation of the plate, if an axial tensile force of 5000 N
acts on it. Take E =2 ×105N/mm2.
21.
STRESSES IN COMPOSITE BARS
A bar, made up of two or more bars of equal length but of different materials rigidly fixed.
So that the system is extension or compression as a single unit.
In such case, the following two governing principles are to be followed.
1. Extension or Compression in each bar is equal. So , the strain induced in those bars
are
also equal.
22.
Change in length of bar (1) = Change in length of bar (2)
P1L1 / A1E1 = P2L2 / A2E2 ………………………….(A)
2. The sum of loads carried by individual materials of a composite member is equal
load
applied on the member.
Total load P = Load carried by bar (1) + Load carried by bar (2)
P = P1 + P2 …………………………( B)
23.
Practice Problems:
1. A compound bar of length 500 mm consists of strip of aluminum 40 mm wide × 15 mm
thick and a strip of steel 40 mm wide × 10 mm thick rigidly joined at ends. If the bar is
subjected to a load of 50 kn, find the stress developed in each material and the extension
of the bar. Take modulus of elasticity of aluminum and steel as 1.1 × 10 5 N/mm2 and
2.1 × 105 N/mm2 .
2. A steel rod of 25 mm diameter is enclosed centrally in a copper hollow tube of external
diameter 40 mm and internal diameter 30 mm. the composite bar is then subjected to an
axial pull of 4500 N. if the length of each bar is equal to 130 mm, determine :
1. The stress in the rod and tube
2. Load carried by each bar
Take Es = 2.1 × 105 N/mm2 , Ec = 1.1 × 105 N/mm2
3. A reinforced concrete column 50 cm × 50 cm is section is reinforced with 4 steel bars of
2.5 cm diameter, one in each corner. The column is carrying a load of 2 MN. Find the
stresses in the concrete and steel bars. Take Es = 2.1 × 10 5 N/mm2 , Ec = 1.4 × 104
N/mm2
4. Two vertical rods one of steel and the other of copper are each rigidly fixed at the top and
50 cm apart. Diameter and lengths of each rod 2 cm and 4m respectively. A cross bar
fixed to the rods at the lower ends carries load of 5000 N such that the cross bar remains
horizontal even after loading. Find the stress in each rod and the position of the load on the
bar. Take Es = 2 × 105 N/mm2 , Ec = 1× 105 N/mm2
24.
TEMPERATURE or THERMAL STRESS AND STRAIN
When a material is free to expand or contract due to change in temperature, no stress and
strain will be developed in the material.
But when the material is rigidly fixed at both the ends, the change in length is prevented.
Due to change in temperature, stress will be developed in the material.
Such stress is known as thermal stress and the corresponding strain is known as thermal
strain.
In other words, thermal stress is the stress developed in material due to change in
temperature.
DETERMINATION OF THERMAL STRESS AND STRAIN
Consider a rod AB of length L, fixed at both ends A and B as shown in fig.1.1.
25.
If the rod is free to expand due to temperature, then extension of the rod is given by
dL = αTL
Where,
α = Coefficient of linear expansion
T = Temperature
L = Length
When the fixture at the end B is removed, the rod is freely expand by dL. From that we came
to known BB’ = αTL . Shown fig .1.2
26.
Compressive load P is applied at B’, the rod is decreased in its length L + αTL to L as shown
in fig. 1.3
27.
We know that
Compressive strain or thermal strain = Decrease in length / Original length
= αTL / (L + αTL)
= αTL / L
Thermal strain = αT ………………….(1)
We know that
E = Stress / Strain
σ= e×E
Thermal stress σ = αT E ……………………( 2)
We know that ,
Stress = Load / Area
Load = Stress × Area
P = αT E ×A …………………( 3)
28.
Suppose a rod of length, when subjected to a rise of temperature is permitted to expand only by δ,
then
Thermal strain = (Actual expansion allowed) / Original length
= (αTL – δ)
L
Thermal Stress = Thermal strain × E
= (αTL – δ ) × E …………………………(4)
L
Practice Problems:
1. A rod of 3 m long is initially at a temperature of 15 ˚ C and it is raised to 90˚ C . Find the
expansion of the rod and if the expansion is prevented, find the stress in the material.
Take E = 2.1 × 105 N/mm2 ; α = 12 × 10-6 / ˚ C
2. A steel rod 4 m long and 30 mm diameter is connected to two grips and the rod is maintained
at a temperature of 70 ˚ C . Find out the force exerted by the rod after it has been cooled to 25˚ C ,
(a) The ends do not yield, and (b) The ends yield by 1.5 mm.
Take Es = 2.1 × 105 N/mm2 ; α = 12 ×10-6 / ˚ C
29.
TEMPERATURE STRESSES IN BARS OF VARYING CROSS SECTION
Consider a bar of varying cross section as shown in fig .1.1
This bar is fixed at A and C and subjected to temperature variation.
When the temperature varies, the bar will tend to expand or contract.
but the same is prevented as it is fixed at both ends, so that the temperature stress will be
produced in that bar.
30.
The total force or pull exerted in the bar (P 1 = P2) = σ 1 A1 = σ 2 A2 ………….(1)
Change in length , δL = (σ 1 L1 / E ) + (σ 2 L2 / E)
δL = 1/E(σ 1 L1 + σ 2 L2 ) …………...(2)
When the materials having different Modulus of elasticity,
δL = (σ 1 L1 / E1 ) + (σ 2 L2 / E2) ……………(3)
Problems:
1. A rod made of brass and steel is held between two right supports A and B as shown in fig. find the
stresses in each material if the temperature rises by 50 ˚ C .
Take Eb = 1.1 × 105 N/mm2
Es = 2.1 × 105 N/mm2
αb = 18 × 10-6 / ˚ C ; αs = 12 ×10-6 / ˚ C
2. Calculate the values of the stress and strain in portions AC and CB of the steel bar shown in fig. A close
fit exists at both of the rigid supports at room temperature and the temperature is raised by 75 ˚ C . Take E
= 200 Gpa and α = 12 ×10-6 / ˚ C for steel. Area of cross section of AC is 400 mm2 and of BC is 800
mm2.
31.
TEMPERATURE STRESSES IN COMPOSITE BARS
A composite member is composed of two or more different materials which are joined
together.
Fig 1.2 shows the composite bars consisting of brass and steel which are subjected to
temperature variation.
32.
Due to different co-efficient of linear expansion, the two materials ( brass and steel)
expand or contract by different amount.
When the ends of bars are rigidly fixed, then the composite section as a whole will expand
or contract.
Since the linear expansion of brass is more than that of steel.
So, the actual expansion of the composite bars will be less than that of brass.
Therefore, brass will be subjected to compressive stress, whereas steel will be subjected to
tensile stress.
We know that,
stress in brass = (load on the brass) / (Area on the brass)
σb = Pb / Ab
Pb = σb × Ab
similarly
load on the steel , Ps = σs × As
under equilibrium condition, compression in the brass bar is equal to tension in the steel
bar,
load on the brass = load on the steel
σb × Ab = σs × As. …………………….( 1)
..
33.
We know that
Actual expansion of steel = Actual expansion of brass …………………..(A)
Actual expansion of steel, dLs = (free expansion of steel) + (Expansion due to tensile stress in
steel)
dLs = α sT Ls + (PsLs /As Es) ………..(2)
Actual expansion of brass, dLb = (free expansion of brass) +(compression due to tensile stress in
brass)
dLb = α bT Lb - (PbLb / AbEb) ……………………..(3)
Sub equation (2) and (3) in equation (A)
α sT Ls + (PsLs / AsEs) = α bT Lb - (PbLb / AbEb) (Ls = Lb )
α sT + (Ps/AsEs) = α bT - (Pb/AbEb) (σs = Ps/As )
α sT + (σs / Es) = α sT - (σs / Es) ……………………….(4)
34.
Practice Problems:
1. A steel rod of 30 mm diameter passes centrally through a copper tube of 60 mm external
diameter and 50 internal diameter. The tube is closed at each end by rigid plates of
negligible thickness. Calculate the stress developed in copper and steel when the temperature
of the assembly is raised by 60 ˚ C
Take Ec = 1 × 105 N/mm2
Es = 2 × 105 N/mm2
αc = 18 × 10-6 / ˚ C ; αs = 12 ×10-6 / ˚ C
2. A gun metal rod 25 mm diameter screwed at the end passes through a steel tube 30 mm and
35 mm internal and external diameters. The temperature of the whole assembly is raised to
125 ˚ C and the nuts of the rod are then screwed lightly home on the ends of the tube.
Calculate the stresses developed in gun metal and steel tube when the temperature of the
assembly has fallen to 20 ˚ C.
Take Eg = 1 × 105 N/mm2
Es = 2.1 × 105 N/mm2
αg = 20 × 10-6 / ˚ C ; αs = 12 ×10-6 / ˚ C
35.
PRRINCIPLE STRESSES AND STRAIN
In many engineering problems both direct (tensile or compressive stress ) and shear stresses
are at the same time.
or
In many situations, machine components are subjected to two or more stresses on a given
plane.
These stresses were acting in a plane, at right angles to the line of action of the force.
In such situation, the resultant stress across any cross section will be neither normal or
tangential to the plane.
The stresses, acting on an inclined plane or (oblique section) will be analyses.
PRINCIPLE PLANES AND PRINCIPLE STRESSES
The planes, which have no shear stress, are known as principle planes.
Hence principle planes are the planes of zero shear stress.
These planes carry only normal stresses
The normal stresses acting on a principle plane are known as principle stresses.
MHEODS FOR DETERMINING STRESSES ON OBLIQUE SECTION
Analytical Method
Graphical Method or Mohr’s method
36.
Analytical method
1) A member subjected to a direct stress in one plane.
2) A member subjected to direct stresses in two mutually perpendicular direction.
3) A member subjected to a simple shear stress.
4) A member subjected to a simple stress in one plane accompanied by a simple shear stress
5) A member subjected to two direct stresses in mutually perpendicular direction
accompanied by a simple shear stress
Mohr,s circle method
1) A body subjected to two mutually perpendicular unequal and like principle stresses
2) A body subjected to two mutually perpendicular unequal and unlike principle stresses
3) A body subjected to two mutually perpendicular unequal and like stresses accompanied
by a simple shear stress.
37.
1. A MEMBER SUBJECTED TO A DIRECT STRESS IN ONE PLANE
Fig 1.1. shows a rectangular member of uniform cross section area (A) and unit thickness
subjected to a principle tensile stress (σ)
38.
The Direct stress along x – axis , σ = P/A
P = Axial force acting on the member.
A = Area C/S , which is perpendicular to the of action of the force ‘P’
Area of section AB = AB × 1
= A ( t = 1 for unit thickness)
The stress on the section AB is give by
σ = P/A
The stress on section AB is only normal stress and no shear is acting
Now consider a section which is at an angle Ø with the normal cross section as shown in
fig.1.2
39.
The stress on section AB is only normal stress and no shear is acting
Now consider a section which is at an angle Ø with the normal cross section as shown in
fig.1.2
40.
Area of section ,BC = BC × 1 ( t = 1 for unit thickness)
= ( AB/ cos Ø) × 1 [ In triangle ABC , cos Ø =
( AB/BC)
BC = ( AB / cos Ø
= ( A / cos Ø ) [ AB ×1 = A)
Stress on section , BC = (P/A)
= [P/(A/ cos Ø)]
= ( P/A) cos Ø
= σ × cos Ø [ σ = (P/A)]
41.
The stress may be resolved in two components
One components will be normal to the section BC whereas the second components will be
along the section BC (i.e, tangential to the section BC)
The normal stress and tangential stress on the section BC are calculated as follows ( ref fig
1.3 )
42.
Pn = the tensile force perpendicular to the plane BC
= P cos Ø
Pt = the tensile force parallel to the plane BC
= P sin Ø
Find the normal stress and tangential stress across section BC
Normal Stress σn = (force perpendicular to the plane BC/ Area of section BC)
= [ Pn / (A / cos Ø)]
= [(P cos Ø) /(A / cos Ø)]
= ( P/A) × (cos2 Ø)
σn = σ cos2 Ø …………………………………1
43.
Tangential stress or shear stress (σt) = (force parallel to the plane BC/ Area of section BC)
= [ Pt / (A / cos Ø)]
= [(P sin Ø) /(A / cos Ø)]
= ( P/A) × (sin Ø × cos Ø )
= (σ /2) × 2 sin Ø × cos Ø
σt = (σ /2) sin 2 Ø ……………………….2
[sin 2 Ø = 2 sin Ø ×
cos Ø
44.
MAXIMUM NORMAL STRESS
we know that
σn = σ cos2 Ø ……………………….A
here
cos2 Ø should be maximum
take
Ø = 0 , cos2 0 = 1
sub the above value in equation A
Maximum normal stress = σ …………………………..3
MAXIMUM SHEAR STRESS OR TANGENTIAL STRESS
we know that
σt = (σ /2) sin 2 Ø ……………………….B
here
cos2 Ø should be maximum
45.
2 Ø = 1 or 2 Ø = 90° or 270 °
Ø = 45° or 135 °
sub the above value in equation B
Maximum shear stress = (σ /2) sin 2 Ø
= (σ /2) sin 90 ° [2 Ø = 90° ]
[sin 90 °
=1]
= (σ /2) ………………….4
Note :
the maximum tangential stress is half of the maximum normal stress
THE RESULTANT STRESS (σ res)
(σ res) = (σn 2 + σt2) ½.............................................5
46.
Practice Problems:
1. A rectangular bar of cross – sectional area 10000 mm 2 is subjected to an axial load of 20
kn. Determine the normal stress and shear stress (i.e, tangential stress) on a section which
is inclined an angle of 30 ° with normal cross section of bar.
2. Find the diameter of a circular bar which is subjected to an axial pull of 160 kn, if the
maximum allowable shear stress on any section is 65 N/mm 2.
3. A member subjected to a pull P consists of two pieces of wooden frame of cross section
35 mm × 15 mm connected by glued joints as shown in fig. calculate the maximum
permissible value of P which can withstand, if the permissible normal and tangential stress
in glue are 13 N/mm2 and 8 N/ mm2
47.
MOHR,S CIRCLE METHOD
It is used to find out the normal ,tangential and resultant stress in oblique plane is Mohr's
method.
It is also graphical method
It is applicable in 3 cases such as
Case 1 : a body subjected to two mutually perpendicular unequal and like principal stress.
Case 2 : a body subjected to two mutually perpendicular unequal and unlike principal
stress.
Case 3 : a body subjected to two mutually perpendicular unequal like stresses
accompanied by a simple shear stress.
48.
PRACTICE PROBLEM FOR THEORY OF SIMPLE BENDING
1. An I section as shown in sketch span two supports 1.5m apart. Determine the total load
uniformly distributed on the entire span that the beam could carry in addition to a
concentrated load of 8 kn at its centre in order that extreme fiber stress is limited to 800
n/mm2
2. A simply supported timber beam of span 6m carries a UDL of 12 kn/m over the entire span &
a point load of 9 kn at 2.5m from the left support. If the bending stress in timber is not exceed
8 n/mm2, design a suitable section for the beam. The depth of the beam equals twice the
breadth.
3. A calculated the maximum bending stress & shear stress in a cantilever beam of span 6m.
Which carries a UDL of 5 kn/m over a distance of 4m from the free end. The cross section of
the beam is a rectangular of breadth 100mm & depth 150mm.
52.
UNIT – IV
PART –A
Deflection of beams
Slope & deflection of a beam subjected to Uniform Bending Moment
Relation between Slope , Deflection & Radius of curvature
Slope & Deflection of the beam may be determined analytically by following method
I. Double Integration Method
II. Macaulay's methods
III. Moment area methods
IV. Conjugate beam method
I. Double Integration Method for slope & deflection
1. Cantilever Beams
a) cantilever with a point load at the free end
b) cantilever with a point load at a distance of ‘a’ from A free end.
c) cantilever with UDL
d) cantilever with UDL from fixed end
e) cantilever with UDL from free end
f) cantilever with a UDL
53.
2. SIMPLY SUPPORTED BEAM
a) SSB with central point load
b) SSB with eccentric point load
c) SSB with UDL
II. MACAULAY'S METHODS FOR SLOPE & DEFLECTION
III. MOMENT AREA METHODS FOR SLOPE & DEFLECTION
1) cantilever beam
a) cantilever with point load at free end
b) cantilever with UDL
2) SSB
a) SSM with central point load
b) SSB with UDL
54.
IV Conjugate Beam method
1) cantilever beam
a) cantilever with point load at the free end
b) cantilever with UDL
c) cantilever with UVL
2) SSM
a) SSB with central point load
b) SSB with an eccentric point load.
c) SSB with UDL
PART –B
Shear stress distribution
Different cross section
1) rectangular sec
2) circular section
3) I- section
4) T- section
5) triangular section
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