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Some PPTs from Students, Structural Analysis Strain Energy
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HJD Institute Of Technical Education And Research - Kera
Presenting by,Shah Parth L. (120850106009)Vaghela Suyagn D. (120850106010)Patel Parth R. (120850106014)Saradhara Divyesh R. (120850106016)Patel Jay H. (120850106029)Bhatti Bhishma J. (120850106046)
STRAIN ENERGY
1. Elastic strain energy
2. Strain energy due to gradual loading
3. Strain energy due to sudden loading
4. Strain energy due to impact loading
5. Strain energy due to shock loading
6. Strain energy due to shear loading
7. Strain energy due to bending (flexure)
8. Strain energy due to torsion
9. Examples
When a body is subjected to gradual, sudden or impact load, the body deforms and work is done upon it. If the elastic limit is not exceed, this work is stored in the body. This work done or energy stored in the body is called strain energy.
Energy is stored in the body during deformation process and this energy is called “Strain Energy”.
What is Strain Energy ?
Strain energy = Work done
Resilience :
Total strain energy stored in a body is called resilience.
Proof Resilience :
Maximum strain energy which can be stored in a body is called proof resilience.
∴𝐮= 𝛔𝟐
𝟐𝐄×𝐕
p
Where, V = volume of the body
Where,
Modulus of Resilience :
Maximum strain energy which can be stored in a body per unit volume, at elastic limit is called modulus of resilience.
m
Strain Energy due to Gradual Loading :
• Consider a bar of length L placed vertically and one end of it is attached at the ceiling.
Let P =Gradually applied loadL =length of barA =Cross-sectional area of the barδl =Deflection produced in the barσ =Axial stress induced in the bar. It may
be tensile or compressive, depending upon if the bar under consideration is under tensile or compressive load
E =Modulus of elasticity of bar material
L
δl
P
Work done on the bar = Area of the load – deformation diagram
… (1)
¿12× 𝑃×𝛿 𝑙
Work Stored in the bar = Area of the resistance –
Deformation diagram
=
= … (2)Now,Work done = Work stored
P × l = × A ×l
P = × A
….. stress due to gradual load.
Strain Energy = l
= × A ×l R = × A = × A = × A E = = ×A
u = × v… strain energy due to gradual load.
Strain Energy due to sudden loading :
When the load is applied suddenly the value of the load is P throughout the deformation.
But, Resistance R increase from O to R
Work done on the bar =P× ... (1)
Work stored in the bar = ×R×
=×A× ...(2) Now,
Work done = Work stored =×A× 𝛿𝑙
P = ×A ∴
= ∴
Hence , the Maximum Stress intensity due to a suddenly applied load is Twice the stress intensity produced by the load of the same magnitude applied gradually.
L
δl
PCollar
h
Strain Energy due to Impact loading :
Load P is dropped through a height h, before it commences to load the bar.
Work done on the bar = Force × Deformation =P( h +)
=P( h + ) … (1) Work stored in the bar = ×× R =Strain Energy
= … (2)
Now,Work done = Work
stored
P( h +)=
𝛿𝑙=𝜎 ∙ 𝑙𝐸
∴
P( h + ) = P × h + P × h × × =
+ =
=
+
( +
(
… Stresses due to impact load∴σ= P
A+√ 2 EPhAl +
p2
A2
If load is applied suddenly, h = 0
=
When is very small as compered to h , then
Work done = P h
= P h
𝜎=√ 2 h𝐸𝑃𝐴𝑙
Let, Work done on the bar by shock = u
Work stored in the bar =
... max instantaneous stress
Strain Energy due to shock loading :
=
¿ σ2
2 E× A ×l
∴U= σ 2
2 E× A ×l
∴σ2=2UEAl
∴σ=√ 2UEAl
Strain Energy due to shear loading :
If t is the uniform shear stress produce in the material by external forces applied within elastic limit, the energy Stored due to shear Loading is given by,
𝐮= 𝛕𝟐
𝟐𝐆×𝐕
Where, t = shear stressG = Modulus of rigidity
Consider a square block ABCD of length l , Faces BC and AD are subjected to shear stress , Let face AD is fixed.
The section ABCD will deform to AD through the angle
= Shear strain
= is very small =
= ….. Shear Strain
Force P on face BCP =
When P in applied gradually In case of gradual load.
u = average force = = = = = = G = = BC = A
u =
The elastic energy stored due to shear loading is known as shear resilience
Strain Energy due to bending (flexure) : Consider two transverse section 1-1 and 2-2 of a
beam distant dx apart as shown in fig.
Consider a small strip of area da at distant y from the neutral
axis. B.M. in small portion dx will be constant.
… (1)
Strain energy stored in small strip of area da.u = v
=
=
= …(2)
= = second
moment of = area.
…(3)
Now, for strain energy in entire beam, integrate between limits 0 to l.
... Strain energy due to bending.
∴u=∫0
lM 2
2 EI∙dx
Strain Energy due to torsion :
We have seen that, when a member is subjected to a uniform shear stress , the strain energy stored in the member is .
Consider a small elemental ring of thickness dr, at radius r.
Strain energy due to torsion for uniform shear stress, in the ring.
… strain energy for
one ring.
Total strain energy for whole section, is obtained by integrating over a range from r = 0 to r = D/2 for a solid shaft.
∴R =
∴A =
u= τ 2
4G×V
… Strain energy due to torsion
examples Ex-1 :
An axial pull of 50 kN is suddenly applied to a steel bar 2m long and 1000 in cross section. If modulus of elasticity of steel is 200 kN/ .Find, (i) maximum instantaneous stress
(ii) maximum instantaneous extension(iii) Strain energy(iv) modulus of resilience.
Solution : here, P = 50 kN (Sudden load)
A = 1000 l = 2m = 2000 mmE = 200 kN/ = 200 N/
(i) Maximum instantaneous stress :
(ii) Maximum instantaneous extension :
=
¿1mm
(iii) Strain energy (u) :
(iv) Modulus of resilience () :
¿50,000N ∙mm
¿0.025N ∙mm /mm3
EX –2 :A 1500 mm long wire of 25 cross sectional
area is hanged vertically. It receives a sliding collar of 100 N weight and stopper at bottom end. The collar is allowed to full on stopper through 200 mm height. Determine the instantaneous stress induced in the wire and corresponding elongation. Also determine the strain energy stored in the wire. Take modulus of elasticity of wire as 200 GPa.Solution :
here, P = 100 N A = 25 l = 1500 mm h = 200 mm E = 200 GPa = 200
= 4 + 461.89
= 465.89 N/
Strain energy,
= 3.49 mm
= 20,348.76 N.mm
THANK YOU