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DAY 3 > STREMA > Flashcards

STREMA Flashcards

1
Q

It deals with analyzing stresses and deflections in materials under load

A

Strength of Materials or
Mechanics of Deformable Bodies

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2
Q

is defined as the internal force which is resisting the external force per unit area.

A

Stress

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3
Q

is a stress that occurs when a member is loaded by an axial force.

A

Axial Stress/ Normal stress

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4
Q

is a force that causes layers or parts to slide upon each other in opposite directions

A

Shearing Stress

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5
Q

is load applied in one plane that would result in the fastener being cut into two pieces

A

Single shear

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6
Q

is load applied in one plane that would result in the fastener being cut into three pieces

A

Double shear

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7
Q

is a failure mechanism in structural members like slabs and foundation by shear under the action of concentrated loads

A

Punching shear

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8
Q

Axial stress formula

A

σ = P/A

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9
Q

Double shear formula

A

σₛ = P/2Aₛ

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10
Q

Single shear formula

A

σₛ = P/Aₛ

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11
Q

Punching shear formula

A

σₚ = P/πdt

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12
Q

A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which resist bursting, developed across longitudinal and transverse sections.

A

Thin-walled Pressured Vessels

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13
Q

also known as Circumferential Stress/ Hoop Stress/ Girth Stress

A

Tangential Stress

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14
Q

Tangential stress formula

A

σₜ = PᵢD/ 2t

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15
Q

Longitudinal Stress

A

σₗ = PᵢD/ 4t

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16
Q

Spherical Shell formula

A

σ = PᵢD/ 4t

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17
Q

Which is true among the choices
a. σₜ = 2σₗ
b. 2σₜ = σₗ
c. σₜ = σₗ
d. σₜσₗ = 2

A

a. σₜ = 2σₗ

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18
Q

the endpoint of the stress-strain curve that is a straight line.

A

proportional limit

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19
Q

the stress is directly proportional to strain

A

Hooke’s Law

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20
Q

Hooke’s law formula

A

σ = Eε

σ → stress
E → Young’s Modulus/ Modulus of Elasticity
ε → strain

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21
Q

is the limit beyond which the material will no longer go back to its original shape when the load is removed, or it is the maximum stress that may be developed such that there is no permanent or residual deformation when the load is entirely removed.

A

Elastic limit

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22
Q

is the point at which the material will have an appreciable elongation or yielding without any increase in load.

A

Yield point

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23
Q

The maximum ordinate in the stress-strain diagram.

A

Maximum strength/ stress

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24
Q

is the strength of the material at rupture. This is also known as the breaking strength.

A

Rupture point

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25
Q

is the work done on a unit volume of material as the force is gradually increased from the linear range, in N·m/m3. This may be calculated as the area under the stress-strain curve from the origin O to up to the elastic limit E

A

Modulus of Resilience

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26
Q

The ______ of the material is its ability to absorb energy without creating a permanent distortion.

A

resilience

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27
Q

is the work done on a unit volume of material as the force is gradually increased from O to R, in N·m/m3. This may be calculated as the area under the entire stress-strain curve (from O to R).

A

Modulus of Toughness

28
Q

The ______ of a material is its ability to absorb energy without causing it to break.

A

toughness

29
Q

is defined as the actual stress of a material under a given loading.

A

Working stress

30
Q

The maximum safe stress that a material can carry.

A

Allowable stress

31
Q

The ratio of this strength (ultimate or yield strength) to allowable strength.

A

Factor of safety

32
Q

Elongation due to the applied load formula

A

δ = PL/ EA

33
Q

Elongation due to its weight

A

δ = ρgL² / 2E

δ = mgL/ 2AE

34
Q

The ratio of the shear stress τ and the shear strain γ

A

Shear strain

35
Q

The ratio of the sidewise deformation (or strain) to the longitudinal deformation (or strain)

A

Poisson’s Ratio

36
Q

Poisson’s Ratio

A

ν = - εₗₐₜ/ εₒₙ

37
Q

Relationship of E(modulus of Elasticity), v(Poisson’s ratio), G(modulus of rigidity)

A

G = E/ 2(1+ν)

38
Q

Relationship of E, v and K(Bulk’s Modulus)

A

K = E/ 3(1 − 2ν)

39
Q

Sidewise deformation/ lateral strain formula

A

εₗₐₜ = -Δd/ d

Δd → change in diameter
d → original diameter

40
Q

Longitudinal deformation/ strain formula

A

εₗₒₙ = ΔL/ L

ΔL → change in length
L → original length

41
Q

Thermal Stress formula

A

σₜ = EαΔT

E → Young’s modulus
α → coefficient of linear expansion

42
Q

Yield stress formula

A

σ = σₜ + σᵧᵢₑₗₔ

σ → axial/ mechanical stress
σₜ → thermal stress
σᵧᵢₑₗₔ → yield stress

43
Q

In solid mechanics, it is the twisting of an object due to an applied torque.
In circular sections, the resultant shearing stress is perpendicular to the radius.

A

Torsion

44
Q

Torsional Shearing Stress formula

A

𝜏ₘₐₓ = Tr/ J
𝜏ₘₐₓ = 16T/ 𝝿D³

T → torque
J → Polar moment of inertia
D → diameter
r → radius

45
Q

Polar moment of inertia formula (Solid shaft and hollow shaft)

A

Solid shaft: J = 𝝿D⁴/32

Hollow shaft: J = 𝝿/32 (D⁴ − d⁴)

46
Q

Angle of twist formula

A

θ = TL/ JG

θ → angle of twist (rad)
G → modulus of rigidity

47
Q

Power through shaft

A

P = Tω = 2𝝿fT

ω → angular velocity (rad/s)

48
Q

Maximum shearing stress of Helical Spring (Approximation Method)

A

𝜏ₘₐₓ = 16PR/𝝿d³ (1 + d/4R)

R → mean radius
P → load
d → diameter of helical spring

49
Q

Maximum shearing stress of Helical Spring (A.M. Wahls Formula)

A

𝜏ₘₐₓ = 16PR/𝝿d³ [(4m−1)/(4m−4) + (0.615/m)]

m = D/d = 2R/d

D → mean diameter
R → mean radius
P → load
d → diameter of helical spring

50
Q

AM Wahls factor

A

[(4m−1)/(4m−4) + (0.615/m)]

51
Q

Spring deflection

A

δ = 64PR³n/ Gd⁴

n → number of turns

52
Q

Spring constant in series

A

1/ kₜ = 1/k₁ + 1/k₂ + … + 1/kₙ

53
Q

Spring constant in parallel

A

kₜ = k₁ + k₂ + … + kₙ

54
Q

Helical spring connected in series

A

P₁ = P₂

55
Q

Helical spring connected in parallel

A

δ₁ = δ₂

56
Q

Cables

A

Parabolic and Catenary

57
Q

Tension at the Supports (Parabolic Cable) formula

A

T = √[(ωL/2)² + H²]

T →tension at the supports
H → tension at the lowest point
ω →weight per horizontal length
L → span of the supports

58
Q

Tension at the lowest point (Parabolic cable) formula

A

H = ωL²/8d

d→ sag
H →tension at the lowest point
L → span of the supports
ω → weight per horizontal length

59
Q

Approximate length of cable (Parabolic cable)

A

S = L + (8d²/3L) + (32d⁴/5L³)

S → length of cable
d → sag
L → span of the supports

60
Q

ω given is “weight per unit length”

A

Catenary cable

61
Q

ω given is “weight per horizontal length”

A

parabolic cable

62
Q

Half length of cable (Catenary cable) formula

A

S = c sinh (x/c)

S → half length of cable
c → clearance
x → half of the span of the supports

63
Q

height of support formula (Catenary cable)

A

y = c cosh (x/c)

y = c + d
y² = S² + c²

y → height of support
c → clearance
x → half of the span of the supports
d → sag
S → half length of cable

64
Q

Tension at the supports (Catenary cable) formula

A

T = ωy

T → tension at the supports
ω → weight per unit length
y → height of the support

65
Q

Tension at the lowest point (catenary cable) formula

A

H = ωc

H → tension at the lowest point
ω → weight per unit length
c → clearance

DAY 3 (3 decks)

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