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Strength of Materials (GATE Aerospace and GATE Mechanical)
By Mr Dinesh Kumar
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Axially Loaded Members
Simplest example of axially loaded member is rope which can undergo only tension loads.
Next example is spring which can undergo both tensile and compressive loads.
o Springs subjected to an axial load.
If 훿 be the change in length,
P = k훿
훿 = f.P
→Where k and f are constants of proportionality,
o 훿= Deflection in the spring o k = stiffness o f = flexibility
→k =
Next is Bar/Rod which can also undergo both tensile and compressive axial load.
→What is the difference b/w spring and Bar/Rod?
Ans. - spring has high flexibility and Bar/Rod has high stiffness.
Spring can only bear axial load but Bar/Rod can undergo axial as well as transverse or torsion load, then same prismatic bar/rod is classified as different structure like beam or shaft.
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o Prismatic Bar
A prismatic bar is a structural member having a straight longitudinal axis and constant c/s section throughout its length and made of isotropic elastic material.
Stress developed in the bar
Stress 휎 =
Strain developed in the bar
Strain ∈=
Hook’s law:-
Stress ∝ strain
휎 ∝∈
휎 = E∈
=
→ 휹 = 푷푳푨푬
o P is applied end load o A is area of c/s o L is the initial length o 휹 is the deflection in bar/rod o = k = stiffness of axial bar/rod o AE = axial rigidity o = f = flexibility of bar
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o Governing differential equation
If q is the load intensity (load per unit length) acting on the prismatic bar Stress developed in the bar
Stress 휎 =
Strain developed in the bar
Strain ∈=
Here u is the displacement field
Hook’s law:-
Stress ∝ strain
휎 ∝∈
휎 = E
= E
P= AE
qL= AE
q= 푨푬푳
흏풖흏풙
Above equation is known as governing differential equation for bar/rod under axial loading
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o Bar with Intermediate Axial loads
Consider a Bar subjected to intermittent point loads as shown in the figure [A] This Bar can be analyzed using free body equilibrium as shown in Figure [B] to [D]
Equilibrium
For [B]: → 푁 + 푃 + 푃 - 푃 = 0
→ 푁 = 푃 - 푃 - 푃
For [C]: → 푁 + 푃 - 푃 = 0
→ 푁 = 푃 - 푃
For [D]: → 푁 = 푃
Stress
o σ =
o σ =
o σ =
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Change in length
o δ =
o δ =
o δ =
Total change in length → 훿 = 훿 + 훿 + 훿
→ 휹 = ∑푵풊푳풊푨풊푬풊
풏풊 ퟏ
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o Bar with varying loads or dimensions or material properties
Consider a bar having varying c/s or having varying material properties or undergoing a varying load as shown
If d훿 be the elongation of differential element,
푑훿 = ( )( ) ( )
Total elongation of bar 훿 = ∫ 푑훿 = ∫ ( )( ) ( )
Here Load N(x), Area A(x) and Modulus of elasticity E(x) are function of x.
Limitation:
o Linear elastic materials. o The stress distribution is uniform across sections (using this we derived 훿 = ).
So, the above formula is applicable only for very small angle of taper less than 20°.
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o Statically indeterminate Bar/Rod
A structure is statically indeterminate if member of unknown reactions are more than number of equilibrium equations.
∵ a + b = L
→ Equilibrium equation,
푅 - P + 푅 = 0 ....... (1)
→ From compatibility equation,
훿 = 0
훿 + 훿 = 0 ......... (2)
훿 = ..
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훿 = - ..
From equation (2),
..
- ..
= 0
If 퐴 = 퐴 = A (for prismatic shaft), then
푅 = .
From equation (1),
∴ 푅 =
∴ 푅 =
Displacement of point c → 훿 = 훿 = ..
= .
. =
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o Composite Bar
Consider a composite bar made of two different materials as shown in figure.
Equation of compatibility: If both materials are glued together then deflection in both materials will be same
훿 = 훿 Force displacement relation:
δ =
δ =
→ =
⇒ 푃 = 푃 ( )
Axial force → P = 푃 + 푃
푃 = P. (
)
푃 = P. (
)
Axial stresses:
휎 = = P. ()
휎 = = P. ()
Change in length:
훿 =
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o Normal and shear stresses on any inclined plane
Stresses on inclined plane is given by below equations and these equations are derived from stress transformation
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o Maximum normal and shear stresses
Here figure (b) shows plane of principal stresses and figure (c) shows the plane of max shear stresses.
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The value of principal stresses and max shear stresses and normal stresses on the plane of max shear stresses are shown in the figures.
Ref: Mechanics of Materials by GERE
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o Thermal stresses
Changes in temperature produce expansion or contraction of structural material, resulting in thermal strains and if structures is restrained then result in thermal stresses. When the block of material is unrestrained and therefore free to expand, every element of the material undergoes thermal strain in all directions and consequently the dimensions of the block increases. For most structural material, thermal strain 휖 is proportional to the temperature change ∆T and independent of c/s properties of the structure.
휖 = 훼∆T
Here,
o 훼 = coefficient of thermal expansion o ∆T = Temperature loading
Sign Convention:
o expansion is positive o contraction is negative
Case (1) a bar with fixed & free end
o →Here ∆푇 is uniform and stress = 0
o →Strain 휖 = 훼∆푇 o = 훼∆푇
o 퐶ℎ푎푛푔푒푖푛푙푒푛푔푡ℎ, δT = 퐿 훼∆푇
If ∆푇 is not uniform in above figure
o Strain ≠ 0 o Stress ≠ 0
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Case (2) a bar with both ends are fixed
o Here ∆푇 is uniform or not uniform
o Strain = 0 o ∈ = 휖 + 휖 o 0 = 훼∆푇 +
o Stress,휎 = - 훼∆푇퐸 o Strain in lateral directions 휖 = (1 + 휈)훼∆푇
Case (3) a bar with one ends is fixed and other end has gap
Expansion due to thermal loading = δT = 퐿 훼∆푇
δT = δgap + RL/AE
δT = δgap + 휎
Thermal Stress 휎 = (δT - δgap ). (Compressive)
Thermal Stress 휎 = (퐿 훼∆푇 - δgap ). (Compressive)
δgap
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o Bar in series under thermal loading and supports are fixed
훿 = 퐿 훼 ∆푇
훿 = 퐿 훼 ∆푇
If R is the reaction in the bar then from boundary condition,
훿 = 0
훿 + + 훿 + = 0
퐿 훼 ∆푇 + 퐿 훼 ∆푇 + ( + ) = 0
Where R is the reaction force developed in the bars due to constraints
→ From above equation = 휎 can be found.
→ Movement of point c,
훿 = 퐿 훼 ∆푇 +
Or
= 퐿 훼 ∆푇 +
퐸
퐿
훼
퐸
퐿
훼
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o Bar is parallel under thermal loading and supports are moving
Here 푃 and 푃 are reaction forces in bar 1 and 2 as due to different thermal coefficient expansion of bar 1 and 2 will be different and to maintain compatibility of displacement both bar will have reaction forces.
→ Equilibrium equation,
푃 + 푃 = 0
→ Compatibility equation,
훼 ∆푇퐿 + = 훼 ∆푇퐿 +
From these two equations P1 and P2 can be found.
Thermal Stresses 휎 =
휎 =
Example: Bolted sleeve
Ref: Mechanics of Materials by GERE
퐴 ,퐸 ,훼
퐴 ,퐸 ,훼
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o Strain energy stored in the bar/rod
Area under load and deflection diagram is the strain energy stored in the bar/rod.
Area under load deflection diagram = U →U = .P.훿
We know, 훿 =
U = (for prismatic bar)
= ∑ (for stepped bar)
= ∫ (for varying c/s or material properties or
varying load)
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o Strain energy density
Area under stress- strain diagram = strain energy density
u = .휎.휀
=
→Strain energy = strain energy density×volume
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Problems
1.
Sol. A
Stress is independent of material constant.
휎 =
H.W. - What is the deformation for this problem?
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2.
Sol. 4
Reaction force for BC section –
Free body diagram b/w B and C
푁 = F
Reaction force for AB section –
Free body diagram b/w A and B
푁 = F-P
Given, 훿 = 0
훿 = 훿 + 훿 = 0
+ = 0
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( )( )
+ ( )( )
= 0
= 4
3.
Sol. 2
We know, 훿 =
∆ =
∆ = . .( ) .
= = ∆
∆∆
= 2
4.
Sol. A
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Change in length 훿 = ∫
Here 퐴 = 푑
푑 = 푑 – ( ) 푥 (take tan휃 for different section and then equal them)
Change in length 훿 = ∫–( )
= × –( ) ×
( )×
푑1–(푑1−푑2)
퐿 푥
= (푑1−푑2)
푑2−
푑1
= (푑1−푑2)
푑1−푑2
푑1푑2
= 푑1푑2
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5.
Sol. 15 mm
Given E = 2× 10 MPa
p = 30휋kN
푑 = 20mm
푑 = 10mm
퐿 = 1.5 m (∵ mark B at mid section and C at root of bar)
퐿 = 2 m
Total deflection at A, 훿 = 훿 + 훿
= +
Putting all values in above equation we get,
훿 = 15 mm
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6.
Sol. 1.718 mm
Given,
E = 100푒 GPa
A = 100 푚푚
We know, 훿 = ∫
= ×× × ×
∫
∵ Here dimension must be converted to meter.
= (푒 ) m
= (푒 − 1) mm
= 1.718 mm
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7.
Sol. 35N-mm
Strain energy stored in stepped bar,
U = ∑
= ×( + )
= ( ) ×(× ×
+× ×
)
= 35N-mm
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8.
Sol. 0.16mm
E = 100GPa
A= 100푚푚
Here BD is a rigid link and to solve such a problem, draw free body diagram for rigid link BD.
∑푀 = 0
푅 ×300 = 6× 10 × 200
푅 = 4 kN
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Elongation in AB 훿 =
= × ×× ×
= 0.16mm
9.
Ans. (C)
Thermal stresses are developed only if bar is fixed on both the end for uniform thermal loading but strain is non zero when it is allowed free to expand.
10.
Ans. (A)
Thermal stress is independent of length of the bar
11.
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Sol. 25°
Given, T = 40°
σ = 60MPa
E = 200GPa, 훼 = 20× 10 /°c
σ = E훼∆T
σ = E훼 (T − T )
× × × = (T − T )
T = 40-15 = 25°
At T temperature rod will just lose contact with wall.
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12.
Sol. -239.8MPa
Given, length of the rod 퐿 = 250 mm
훿 = 0.2 mm
∆T = 200 °C
E = 200GPa
훼 = 10 /°C
휎 = - (훿 − 훿 )
= - (퐿 훼∆T-훿 )
= - (250× 10 × 200- 0.2) ×.
= - 239.8MPa
Here Negative has to be added in answer as in question, they didn’t ask magnitude or compressive stress
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13.
Sol. 500MPa
Given, ∆푇 = 250°C
E = 200GPa
훼 = 1 × 10 /°C
휎 = - E훼∆푇
= - 200× 10 × 250 × 10
= 500MPa (Here magnitude is asked)
14.
Sol. 222MPa
Given, ∆푇 = 100°C
E = 200GPa
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훼 = 11 × 10 /°C
휎 = - E훼∆푇
= - 200× 10 × 11 × 10 × 100
= 222MPa (Here magnitude is asked)
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