STREMA

Question 1

Stress

Answer 1

σ = P / A

P = tensile or compressive load
A = cross-sectional area,

Question 2

Doulble shear

Answer 2

σ = P / nA

n =2 (clevis)

P = tensile or compressive load
A = cross-sectional area

Question 3

Punching Shear

Answer 3

σ = P / πDt

πDt = punching area

P = tensile or compressive load
A = cross-sectional area,

Question 4

Thin walled pressured vessel

Tangetial Stress

Answer 4

σₜ = PᵢD / 2t = (Pᵢ - Pₒ)D/2t

σt = 2σL

Question 5

Thin walled pressured vessel

Longitudinal/Spherical Stress

Answer 5

σₜ = PᵢD/4t = (Pᵢ - Pₒ)D/2t

σt = 2σL

Question 6

Hooke’s Law

Answer 6

σ = Ε ε

Ε = Young’s Modulus/ Modulus of Elasticity
ε = strain

Question 7

Factor of Safety

Answer 7

FoS = σᵤ/σₐ

σₐ = σt

Question 8

Elongation

Answer 8

δL = PL/EA

Question 9

Elongation due to its weight

Answer 9

δW= ρgL²/2E = mgL/2AE

Question 10

Total Elongation

Answer 10

δT = δL + δW

Question 11

Shear Modulus

Answer 11

G = τ / γ
G= shear stress/ shear strain

Question 12

Poisson’s Ratio

Answer 12

v = - ε lat / ε long
Δd/d / ΔL/L

Question 13

Modulus of Rigidity

Answer 13

G = E / 2(1+ ⱴ)

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Question 14

Bulk Modulus

Answer 14

k = E / 3(1-2ⱴ)

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Question 15

Thermal Stess

Answer 15

σᴛ = αEΔT

σₐ = σᴛ + σ,yield

α = coefficient of linear expansion

Question 16

Torsion

Torsional Shearing Stress

Answer 16

τ, max = Tr/J

where:
T = torsion
r = radius
J = Polar moment of inertia

Question 17

Torsion

Polar Moment of Inertia

Solid Shaft

Answer 17

J = πD⁴/32

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Question 18

Torsion

Polar Moment of Inertia

Hollow Shaft

Answer 18

J = π(Dₒ⁴-dᵢ⁴)/32

Question 19

Torsion

Angle of Twist

Answer 19

θ = TL/JG x 360°/2π

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Question 20

Torsion

Power through shaft

Answer 20

P = 2πƒT

ƒ = Hz, cps, rps

Question 21

Helical Spring

Approximation Method

Answer 21

τ, max = 16PR/πD³ · [ 1 + d/4R]

Question 22

Helical Spring

AM Wahl’s Formula
Exact Method

Answer 22

τ, max = 16PR/πD³ · [ 4m-1/4m-4 + 0.615/m]

m = 2R/d
R = mean radius

Question 23

Helical Spring

Spring Deflection/Deformation

Answer 23

δ = 64PR³n/Gd⁴

where:
n = no. of twists of spring

Question 24

Spring Constant

Answer 24

k = Pᴛ/δᴛ

Question 25

Spring Constant

Series

Answer 25

P1 = P2
δᴛ = δ1+δ2
1/k = 1/k1 + 1/k2 + … + 1/kn

Question 26

Spring Constant

Parallel

Answer 26

Pᴛ = P1 + P2
δ1 = δ2
k = k1 + k2 + … + kn

Question 27

Cables: Parabolic Cables

Tension at the Supports

Answer 27

T = √(ωL/2)² + H²

ω = weigh of horiz. length
L = spand of supports

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Question 28

Cables: Parabolic Cables

Tension at the lowest point

Answer 28

H = ωL²/8d

ω = weigh of horiz. length
L = spand of supports
d = sag

Question 29

Cables: Parabolic Cables

Approx. Length of Cable

Answer 29

S = L + 8d²/3L - 32d⁴/5L³

L = spand of supports
d = sag

Question 30

Cables: Caternary Cables

1. Half Length
2. Support Weight

Answer 30

1. S = C sinh (x/C)
2. y = C cosh (x/C)

y² = S² + C²

c = clearance

Question 31

Cables: Caternary Cables

Max Tension

Answer 31

T = ωy

y = height

Question 32

Cables: Caternary Cables

Tension at Lowest Point

Answer 32

H = ωC

c = clearance

Question 33

Cables: Caternary Cables

W

Answer 33

W = ωS

c = clearance