Strength of Materials all formulas in pdf it subject iosd also klnown as mechanics of Soilid. in this pdf there are formulas of stress strain springs - closed coil helical spring , open coil helical Springs etc.

1. [Strength of Materials]
[SOM]
Anuj Singh 12/25/20 Formulas

2. 1 | P a g e W r i t t e n b y A n u j S i n g h
Unit 1
Compound Stresses and Strain
Formulas
1. Stress -
𝜎 =
𝑃
𝐴
Where, P = Load and A = Area, Unit = N/m2
2. Strain –
Strain =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛
𝑂𝑟𝑖𝑔𝑖𝑛𝑜𝑙 𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛
𝜖 =
𝛿𝑙
𝑙
=
𝑃
𝐴𝐸
Dimension Less
3. Volumetric Strain -
𝜖 =
𝛿𝑉
𝑉
Unit Less
4. True Stress –
𝜎𝑡𝑢𝑟𝑒 = 𝜎(1 + 𝜖) Where 𝜎 & 𝜖 are Engineering Stress
& Engineering Strain
5. True Strain –
𝜖 𝑇 = ∫
ⅆ𝑙
𝑙
𝐿
𝐿0
= 𝑙𝑛 (
𝐿
𝐿0
) = 𝑙𝑛(1 + 𝜖) = ln (
𝐴0
𝐴
) = 2 ln(
ⅆ0
ⅆ
)
6. Engineering Strain –
𝜖 = ⅇ 𝜖 𝑇 − 1
7. Hooks Law –
𝜎 ∝ 𝜖
𝜎 = 𝐸𝜖 Where E is called Young’s Modules
8. Young’s Modules of Elasticity or Modules of Elasticity –
𝐸 =
𝜎
𝜖
Unit = N/m2
9. Modules of Rigidity –
𝜏 ∝ 𝜙
𝜏 = 𝐺𝜙
𝐺 =
𝜏
𝜙
Where, G is the Modules of Rigidity
10.PoissonRatio –
𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜(𝜇) =
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑆𝑡𝑟𝑎𝑖𝑛
𝐿𝑜𝑛𝑔𝑖𝑡𝑢ⅆ𝑖𝑛𝑎𝑙 𝑆𝑡𝑟𝑎𝑖𝑛

3. 2 | P a g e W r i t t e n b y A n u j S i n g h
11.Expression for the Elastic Content –
 𝐾 =
𝐸
3(1−2𝜇)
 𝐸 =
9𝐾𝐺
3𝐾+𝐺
 𝐸 = 2𝐺(1 + 𝜇) Where, K = Bulk Modules
12.Elongation in Bar –
𝛿 =
𝑃𝐿
𝐴𝐸
Elongation of a uniform bar due to its self-weight,
𝛿 =
𝑊𝐿
2𝐴𝐸
Elongation of a conical bar due to its self-weight,
𝛿 =
𝑊𝐿
6𝐴𝐸
13.Elongation in CompositeBar –
𝛿 = 𝛿1 + 𝛿2 + 𝛿3
14.Thermal Stresses –
𝜎 𝑇ℎ = 𝑎𝑇𝐸
Load or thrust on rod = Stress × Area
𝑃 = 𝛼𝑇𝐸 × 𝐴
Stress and Strain when supported Yields –
Actual Expansion = Expansion due to rise in temp. −𝛿
Actual Expansion = 𝑎𝑇𝐿 − 𝛿
Actual Strain =
𝛼𝑇𝐿−𝛿
𝐿
Actual Stress = 𝐸 ×
𝛼𝑇𝐿−𝛿
𝐿
15.Principle Planes –
The planes which have no Shear Stress.
 When A Member is subjected to a direct stress in one plane –
Normal Stress (𝜎𝑛 ) = 𝜎 𝑐𝑜𝑠2
𝜃 =
𝑃
𝐴
𝑐𝑜𝑠2
𝜃
Tangential Stress or Shear Stress (𝜏) =
𝜎
2
𝑠𝑖𝑛 2𝜃 =
𝑃
2𝐴
𝑠𝑖𝑛 2𝜃
 Two mutually Perpendicular direct Stress –
Normal Stress (𝜎𝑛 ) =
𝜎 𝑥+𝜎 𝑦
2
+
𝜎 𝑥−𝜎 𝑦
2
cos 2𝜃
Tangential Stress or Shear Stress (𝜏) =
𝜎 𝑥−𝜎 𝑦
2
𝑠𝑖𝑛 2𝜃
Maximum Shear Stress (𝜏max) =
𝜎 𝑥−𝜎 𝑦
2

4. 3 | P a g e W r i t t e n b y A n u j S i n g h
Resultant Stress 𝜎 𝑅 = √ 𝜎 𝑛
2 + 𝜏2
Obliquity –
𝑡𝑎𝑛 𝜙 =
𝜏
𝜎 𝑛
 Two Dimensional Stresses (General) –
 Normal Stress (𝜎𝑛 ) =
𝜎 𝑥+𝜎 𝑦
2
+
𝜎 𝑥−𝜎 𝑦
2
cos 2𝜃 + 𝜏 𝑥𝑦 sin 2𝜃
 Tangential Stress or Shear Stress (𝜏) =
𝜎 𝑥−𝜎 𝑦
2
𝑠𝑖𝑛 2𝜃 − 𝜏 𝑥𝑦 cos 2𝜃
 Position of Principle Plane is Given by-
𝑡𝑎𝑛 2𝜃 =
2𝜏 𝑥𝑦
𝜎 𝑥−𝜎 𝑦
16.Principle Stresses-
𝜎 =
𝜎 𝑥+𝜎 𝑦
2
± √[
𝜎 𝑥−𝜎 𝑦
2
]
2
+ 𝜏 𝑥𝑦
2
 Major Principal Stress (𝜎1 ) =
𝜎 𝑥+𝜎 𝑦
2
+ √[
𝜎 𝑥−𝜎 𝑦
2
]
2
+ 𝜏 𝑥𝑦
2
 Minor Principal Stress( 𝜎2 ) =
𝜎 𝑥+𝜎 𝑦
2
− √[
𝜎 𝑥−𝜎 𝑦
2
]
2
+ 𝜏 𝑥𝑦
2
 Maximum Shear Stress(𝜏max) =
𝜎1−𝜎2
2
Or
 Maximum Shear Stress(𝜏max) =√[
𝜎 𝑥−𝜎 𝑦
2
]
2
+ 𝜏 𝑥𝑦
2
 Radius of Mohr’s Circle (r) = √[
𝜎 𝑥−𝜎 𝑦
2
]
2
+ 𝜏 𝑥𝑦
2
 Centre of Mohr’s Circle (a) =
𝜎1+𝜎2
2
 𝑡𝑎𝑛2𝜃 =
2𝜏 𝑥𝑦
𝜎 𝑥−𝜎 𝑦
17.Principle Strain-
𝜖 =
𝜖 𝑥+𝜖 𝑦
2
± √[
𝜖 𝑥−𝜖 𝑦
2
]
2
+ (
𝛾 𝑥𝑦
2
)
2
 Maximum Principal Strain (𝜖1) =
𝜎1
𝐸
−
𝜎2
𝜖𝐸
=
1
𝐸
( 𝜎1 − 𝜇𝜎2 )
 Minor Principal Strain(𝜖2) =
𝜎2
𝐸
−
𝜎1
𝜖𝐸
=
1
𝐸
( 𝜎2 − 𝜇𝜎1)
 Maximum Shear Strain ( 𝛾) = 2√[
𝜖 𝑥−𝜖 𝑦
2
]
2
+ (
𝛾 𝑥𝑦
2
)
2

5. 4 | P a g e W r i t t e n b y A n u j S i n g h
UNIT – 2
Beams and Torsion
Formulas
1. Bending Equation –
𝑀
𝐼
=
𝜎
𝑦
=
𝐸
𝑅
2. Section Modules –
𝑍 =
𝐼
Ymax
or, 𝑀 = 𝜎 𝑚𝑎𝑥 × 𝑧
3. Section Modules for Different Shapes of Beam Section –
 Rectangular Section –
𝑍 =
𝑏 ⅆ3
6
 Hollow Rectangular Section –
𝑍 =
1
6𝐷
[ 𝐵𝐷3
− 𝑏𝑑3]
 Circular Section –
𝑍 =
𝛱
32
𝑑3
 Hollow circular Section –
𝑍 =
𝜋
32𝐷
[ 𝐷4
− 𝑑4]
4. Torsion Equation –
𝑇
𝐼𝑝
=
𝜏
𝑅
=
𝐺𝜃
𝐿
Where, T = Torsion, Ip = Polar
moment of inertia
 Polar Moment of Inertia of the Shafts-
 For a Solid Shaft –
𝐼𝑝 =
𝜋
32
𝐷4
 For a Hollow Shaft –
𝐼𝑝 =
𝜋
32
( 𝐷4
− 𝑑4)
 Torque transmitted by the Shaft –
 For a solid Shaft –
𝑇 = 𝜏 ⋅
𝜋
16
𝐷3
 For a Hollow Shaft –
𝑇 = 𝜏 ⋅
𝜋
16
(
𝐷4−ⅆ4
𝐷
)
 Power Transmitted by the Shaft –
𝑃 =
2𝜋𝑁𝑇
60×1000
𝑘𝑊

6. 5 | P a g e W r i t t e n b y A n u j S i n g h
5. Combined Bending and Torsion –
 Equivalent Bending Moment –
𝑀𝑒 =
𝑀+√𝑀2+𝑇2
2
 Equivalent Torque–
𝑇𝑒 = √ 𝑀2 + 𝑇2
 Maximum Stress –
𝜎 𝑚𝑎𝑥 =
𝜎 𝑑
2
+ √(
𝜎 𝑑
2
)
2
+ 𝜏2
Where 𝜎ⅆ is Caused by Bending
 𝜏 𝑚𝑎𝑥 = √(
𝜎 𝑑
2
)
2
+ 𝜏2

7. 6 | P a g e W r i t t e n b y A n u j S i n g h
UNIT – 3
Springs Column and Struts
Formulas
I. Springs –
I. The Closed Coil Helical Springs-
 Shear Stress –
𝜏 =
16𝑤𝑅
𝜋ⅆ3
 Deflection of Springs –
𝛿 =
64𝑤𝑅3 𝑛
𝐺ⅆ4
 Stiffness of Springs –
𝑘 =
𝑤
𝛿
𝑘 =
𝐺 ⅆ4
64𝑅3 𝑛
 Strain Energy Stored by Springs –
𝑈 =
1
2
× 𝑤𝛿 =
1
2
𝑇𝜃
 Whal’s Correction Factor –
𝜏 =
16𝑤𝑅
𝜋ⅆ3
× 𝑘
𝑘 =
4𝑠−1
4𝑠−4
+
0.615
𝑠
Where, Spring Index (s) =
D
d
II. The Closed Coil Helical Springs with axial twist –
 Angle of Twist–
𝜙 =
𝑀𝑙
𝐸𝐼
𝜙 =
128𝑀𝑅𝑛
𝐸ⅆ4

8. 7 | P a g e W r i t t e n b y A n u j S i n g h
 Bending Equation –
𝜎𝑏 =
32𝑀
𝜋ⅆ3
𝑀 =
𝜎 𝑏 𝜋 ⅆ3
32
 Strain Energy –
𝑈 =
𝜎 𝑏2
8𝐸
× 𝑉𝑜𝑙𝑢𝑚ⅇ 𝑜𝑓 𝑆𝑝𝑟𝑖𝑛𝑔
III. Open Coil Helical Spring with Axial Load –
 Torsion –
𝑇 = 𝑊𝑅 𝑐𝑜𝑠 𝛼
 Bending –
𝑀 = 𝑊𝑅 𝑠𝑖𝑛 𝛼
 Deflection –
𝛿 = 2𝑊𝑅3
𝑛𝜋 𝑠ⅇ𝑐 𝛼 [
𝑐𝑜𝑠2
𝛼
𝐺 𝐼 𝑃
+
𝑠𝑖𝑛2
𝛼
𝐸𝐼
]
 Angular Rotation –
𝜓 = 2𝑤𝑅2
𝑛𝜋sin 𝑎 [
𝑐𝑜𝑠2
𝑎
𝐺𝐼 𝑝
−
1
𝐸𝐼
]
Where,
W = Applied Load
R = Mean Coil Diameter
n = No. of Coils
α = Helix Angle
G = Modulus of Rigidity
E = Young’s Modulus

9. 8 | P a g e W r i t t e n b y A n u j S i n g h
IV. Open Coil Helical Spring with Axial Load –
 Angular Rotation –
𝜙 = 2𝑤𝑅2
𝑛𝜋 sec 𝑎 [
𝑠𝑖𝑛2
𝑎
𝐺𝐼 𝑝
+
𝑐𝑜𝑠2
𝑎
𝐸𝐼
]
 Deflection –
𝛿 = 2𝑇𝑅2
𝑛𝜋 sin 𝑎 [
1
𝐺𝐼 𝑝
−
1
𝐸𝐼
]
V. Springs in Series –
1
𝑘
=
1
𝑘1
+
1
𝑘2
VI. Springs in Parallel –
𝑘 = 𝑘1 + 𝑘2
VII. Laminated Spring –
 Semi Elliptical Spring (Leaf Spring) –
o Bending Stress –
𝜎𝑏 =
3𝑊𝑙
2𝑁𝑏𝑡2
o Deflection –
𝛿 =
3𝑊𝑙3
8𝐸𝑁𝑏𝑡3
o Strain Energy –
𝑈 =
𝜎 𝑏2
6𝐸
× 𝑉𝑜𝑙𝑢𝑚ⅇ 𝑜𝑓 𝑆𝑝𝑟𝑖𝑛𝑔
 Quarter Elliptical Spring –
o Bending Stress –
𝜎𝑏 =
6𝑊𝑙
𝑁𝑏𝑡2
o Deflection –
𝛿 =
6𝑊𝑙3
𝐸𝑁𝑏𝑡3