STREMA

Question 1

It deals with analyzing stresses and deflections in materials under load

Answer 1

Strength of Materials or
Mechanics of Deformable Bodies

Question 2

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

Answer 2

Stress

Question 3

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

Answer 3

Axial Stress/ Normal stress

Question 4

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

Answer 4

Shearing Stress

Question 5

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

Answer 5

Single shear

Question 6

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

Answer 6

Double shear

Question 7

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

Answer 7

Punching shear

Question 8

Axial stress formula

Answer 8

σ = P/A

Question 9

Double shear formula

Answer 9

σₛ = P/2Aₛ

Question 10

Single shear formula

Answer 10

σₛ = P/Aₛ

Question 11

Punching shear formula

Answer 11

σₚ = P/πdt

Question 12

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.

Answer 12

Thin-walled Pressured Vessels

Question 13

also known as Circumferential Stress/ Hoop Stress/ Girth Stress

Answer 13

Tangential Stress

Question 14

Tangential stress formula

Answer 14

σₜ = PᵢD/ 2t

Question 15

Longitudinal Stress

Answer 15

σₗ = PᵢD/ 4t

Question 16

Spherical Shell formula

Answer 16

σ = PᵢD/ 4t

Question 17

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

Answer 17

a. σₜ = 2σₗ

Question 18

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

Answer 18

proportional limit

Question 19

the stress is directly proportional to strain

Answer 19

Hooke’s Law

Question 20

Hooke’s law formula

Answer 20

σ = Eε

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

Question 21

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.

Answer 21

Elastic limit

Question 22

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

Answer 22

Yield point

Question 23

The maximum ordinate in the stress-strain diagram.

Answer 23

Maximum strength/ stress

Question 24

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

Answer 24

Rupture point

Question 25

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

Answer 25

Modulus of Resilience

Question 26

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

Answer 26

resilience

Question 27

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).

Answer 27

Modulus of Toughness

Question 28

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

Answer 28

toughness

Question 29

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

Answer 29

Working stress

Question 30

The maximum safe stress that a material can carry.

Answer 30

Allowable stress

Question 31

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

Answer 31

Factor of safety

Question 32

Elongation due to the applied load formula

Answer 32

δ = PL/ EA

Question 33

Elongation due to its weight

Answer 33

δ = ρgL² / 2E δ = mgL/ 2AE

Question 34

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

Answer 34

Shear strain

Question 35

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

Answer 35

Poisson's Ratio

Question 36

Poisson's Ratio

Answer 36

ν = - εₗₐₜ/ εₒₙ

Question 37

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

Answer 37

G = E/ 2(1+ν)

Question 38

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

Answer 38

K = E/ 3(1 − 2ν)

Question 39

Sidewise deformation/ lateral strain formula

Answer 39

εₗₐₜ = -Δd/ d Δd → change in diameter d → original diameter

Question 40

Longitudinal deformation/ strain formula

Answer 40

εₗₒₙ = ΔL/ L ΔL → change in length L → original length

Question 41

Thermal Stress formula

Answer 41

σₜ = EαΔT E → Young's modulus α → coefficient of linear expansion

Question 42

Yield stress formula

Answer 42

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

Question 43

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.

Answer 43

Torsion

Question 44

Torsional Shearing Stress formula

Answer 44

𝜏ₘₐₓ = Tr/ J 𝜏ₘₐₓ = 16T/ 𝝿D³ T → torque J → Polar moment of inertia D → diameter r → radius

Question 45

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

Answer 45

Solid shaft: J = 𝝿D⁴/32 Hollow shaft: J = 𝝿/32 (D⁴ − d⁴)

Question 46

Angle of twist formula

Answer 46

θ = TL/ JG θ → angle of twist (rad) G → modulus of rigidity

Question 47

Power through shaft

Answer 47

P = Tω = 2𝝿fT ω → angular velocity (rad/s)

Question 48

Maximum shearing stress of Helical Spring (Approximation Method)

Answer 48

𝜏ₘₐₓ = 16PR/𝝿d³ (1 + d/4R) R → mean radius P → load d → diameter of helical spring

Question 49

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

Answer 49

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

Question 50

AM Wahls factor

Answer 50

[(4m−1)/(4m−4) + (0.615/m)]

Question 51

Spring deflection

Answer 51

δ = 64PR³n/ Gd⁴ n → number of turns

Question 52

Spring constant in series

Answer 52

1/ kₜ = 1/k₁ + 1/k₂ + ... + 1/kₙ

Question 53

Spring constant in parallel

Answer 53

kₜ = k₁ + k₂ + ... + kₙ

Question 54

Helical spring connected in series

Answer 54

P₁ = P₂

Question 55

Helical spring connected in parallel

Answer 55

δ₁ = δ₂

Question 56

Cables

Answer 56

Parabolic and Catenary

Question 57

Tension at the Supports (Parabolic Cable) formula

Answer 57

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

Question 58

Tension at the lowest point (Parabolic cable) formula

Answer 58

H = ωL²/8d d→ sag H →tension at the lowest point L → span of the supports ω → weight per horizontal length

Question 59

Approximate length of cable (Parabolic cable)

Answer 59

S = L + (8d²/3L) + (32d⁴/5L³) S → length of cable d → sag L → span of the supports

Question 60

ω given is "weight per unit length"

Answer 60

Catenary cable

Question 61

ω given is "weight per horizontal length"

Answer 61

parabolic cable

Question 62

Half length of cable (Catenary cable) formula

Answer 62

S = c sinh (x/c) S → half length of cable c → clearance x → half of the span of the supports

Question 63

height of support formula (Catenary cable)

Answer 63

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

Question 64

Tension at the supports (Catenary cable) formula

Answer 64

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

Question 65

Tension at the lowest point (catenary cable) formula

Answer 65

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