PROPERTIES OF MATERIAL
Nominal/engineering/yield/avg stress = load/Original area
Actual/True stress = load/Actual Area
Actual area = Original Area ± ∆A
Actual σ = σ o (1 ± εo)
shaft subjected to Torsion have zero normal stress
Strain rosettes → measure linear strain
Strain is fundamental behaviour & Stress is derived behaviour
Stress is internal property while Pressure is external property.
Compressive stress: Acts into the Area
Tensile stress: Acts away from Area
Bearing stress : due to load transfer from one surface to another.
Tangential or shearing stress: Force acts tangentially to surface of the body
Normal or hydrostatic stress → Subjected to uniform force from all sides
Jacketing → Strengthening weak Beam or column
At N.A. → Bending stress(normal stress) = 0 & Shear stress(tangential) = maximum
Lift rope σ = w(1 + a/g)/A
Beam having P at e → δ = L²Pe/8EI = L²/8R
Extensometer → Normal strain measured
Margin of safety = FOS - 1
FOS = yield/allowable stress
For Ductile → Yield stress & Brittle →
Yield Stress > working stress
UTM(Universal testing machine) → Load & Elongation measured
Tempering → to steel in hardening process for improving Characteristics like Ductility, strength, roughness .etc
When the nut is tightened on the bolt → tensile stresses are induced in the bolt
Prying force → additional tension force developed in bolts
Bauschinger effect (Strain softening) → mild steel specimens subjected to tensile test cycles, the elastic limit in tension is raised and elastic limit in compression is lowered
Modular ratio = Ratio of E of two materials
Isotropic hardening → Dilatation of yield surface
Tenacity → Ultimate strength in tension, Resist fracture under tensile load
Red short iron cracks when bent are due to the presence of sulphur
The rate of increase of stress is large in case of bonded beams
Homologous temp → At which creep uncontrollable
Presence of free graphite → material is cast iron, Steel has no graphite
Carbon Percentage
C% ↑es → Yield stress ↑es, but fracture strain ↓es
Ductility ∝ 1/C%
Strength, Brittleness, Hardness ∝ C %
Pig iron = 3.8 - 4.7% → max carbon content
Cast iron = 2 - 4%
Wrought iron < 0.1% → Purest form of iron
Mild steel = 0.05 - 0.25%
Structural steel < 0.6 %.
Stress-Strain Curve
A (Limit of proportionality) → depends only on type of material, hooke's law is valid
Limit of proportionality ≤ Elastic limit
B: Elastic limit → Regain shape
Yielding point: extension takes place more quickly than increase in load or stress, material undergoes plastic deformation.
Yield strength → Stress require to produce certain arbitrary plastic deformation
C (upper yield point) → Can we avoided by hot working, depends on c/s area, shape, equipment and not depend on length
D (lower yield point) → Actual yielding starts here, Extension increased quickly as compare to stress/load
E → Strain hardening starts
F: Ultimate point
G: Fracture point
EF (Strain Hardening) → material undergoes changes in atomic and crystalline str, +ve slope, increased resistance to further deformation.
Necking region FG: between ultimate & rupture point.
Endurance limit → Max stress that can be applied to a material for an infinity number of cycles of repeated stress without causing failure
Yield zone is not considered for steel with high carbon content
Semi-compact section → can attain a yield moment but not the plastic moment before failure by plate buckling.
Strain Energy (U)
Strain Energy = Work done = Force x distance = ½ stress x strain x vol
U = ½ σ ε = ½ P ∆ = σ ²/2E = P²L/2AE → Quadratic eqn
Resilience = (σ ²/2E) x Volume
Point load U = ∫P²dx / 2AE = P²L / 2AE
SF = ∫S²dx / 2AG
Moment = ∫M²dx / 2EI
Torsion = ∫T²ds / 2GIp
Due to shear stress = (τ ²/2G) x Vol
In beam → U ∝ 1/I
Strain energy density → J/m^3 or kJ/m^3
AE → Axial rigidity, EI → Flexural rigidity, GIp → Torsional rigidity
AE/L → Axial Stiffness, EI/L → Flexural Stiffness, GIp/L → Torsional Stiffness
Proof stress
Stress required to caused a permanent extension equal to defined % of gauge length
0.2 % proof stress = stress at which if unloading is made there will be 0.2% permanent strain
Resilience
Resilience → Area under load-deformation curve within elastic limit, or energy stored/absorbed within elastic limit.
Modulus of Resilience(MOR) → Area under Stress-Strain curve within elastic limit
Proof Resilience → Max strain energy stored at Elastic limit without undergoing permanent Deformation
U per unit Vol = σ ²/2E = MOR = Proof Resilience/Vol
Toughness
Ability to Absorb mechanical energy up to failure or ability to resist fracture.
Area under stress-strain curve represent toughness
Bend test → To check toughness
More failure strain → More though
Ductile materials → Though & Brittle materials → Hard
Toughness → Area under load-deformation curve up to fracture
Modulus of Toughness/Fracture → Area under stress - strain curve up to fracture
Charpy test
Specimen supported as → A Simply supported beam
Use → Relative toughness or impact toughness of material
Brittleness
Fracture & ultimate point are same
brittle material → No plastic zone
Post elastic strain < 5%
Ex. Rubber, glass, cast iron
Ordinary glass is nearly ideal brittle material
Ductility
Drawn out into wires without necking down.
Has long plastic elongation range and large deformation at Failure.
Depends on → Temperature of structure, Size of the structure
Std measure of ductility → % elongation in Length
Post elastic strain > 5%
Ex. Lead, mild steel, Copper
Gauge length → lo=5.65Ao
Failure of material
Ductile → Weak in shear
Brittle → Weak in tension
Brittle: Tension → Right angle to axis, Compression → Oblique plane, shear fail(45°), Torsion → 45°
Ductile: Tension → 45°(cup & cone shear), Compression→ 90°(bulzing failure) , Torsion → 90°
Generally 5% fracture strain is adopted as border between brittle and ductile material
Malleability
Hammered into sheets without Rupture
Plastic response of a material to compressive force is malleability.
Durability
Perform it's intend function throughout its design life without Deterioration
Creep
Deform continuously at slow rate without any further increase in stress
Relaxation
Loss of stress with the time at constant strain
Fatigue
Repeated cycle of Stress → Phenomenon of decreased resistance of a material to reverse of stress
Under stressing → improves the fatigue strength of a metallic material
Endurance limit
Stress level below which even a large no of stress cycles can't produce fatigue failure or stress below which material has a high probability of not failing under reversal of stress
Endurance limit = ½ of ultimate strength
Hardness
Resist scratch or abrasion
Scratch hardness by mohr's method
Brinell hardness test uses a steel ball of 10mm dia as indenter
Brinell hardness No P / (D/2)(D-D2-d2
D → steel ball dia, d → indentation ball dia
Thrust
Tension → +ve, Compression → -ve
Elasticity
Return to its original shape after removal of load
Diamond > Mild steel > Rubber
Perfectly elastic → Regains its original shapes on removal of the load
Shear stress τ = σ/2
Poisson ratio (μ)
μ = 1/m = - Lateral strain/Longitudinal strain
Range = 0 - 0.5 → for engineering material
Limiting value/General range = -1 to 0.5
Cork or rigid body = 0 ← Lowest
Concrete = 0.15 - 0.25
Cast iron = 0.21 - 0.26
Steel = 0.27 - 0.30
Wrought iron = 0.3
Aluminium = 0.334
Copper = 0.35
Rubber = 0.5 ← Highest
μ↑es → Elasticity ↓es
μ is constant for linear elastic, homogeneous and isotropic materials
Young's modulus of elasticity (E)
E = σ/ε
Steel E = 2 x 10⁵ Mpa.
E → Copper > Aluminium > Glass > wood.
Esteel/Etimber = 1
Material heated up → Elastic modulus decreases.
Perfectly rigid body → E = ∞, Strain = 0
Modulus of Rigidity (G) or Shear modulus
G = Shear stress / Shear strain = τ/ϕ
Diagonal strain = ϕ/2
Pressure meter test → G determination
Bulk modulus (K)
K = σ/volumetric strain =σ/V/V
if σx = σy = σz → K = E/3(1 - 2μ)
Strain
Dilation = Sum of strain in all direction → Volume change due to hydrostatic stress
Perfectly plastic material → V=0, v=0, =0.5
Volumetric strain(Dilation of material)(εv ) = ΔV/V = (σx + σy + σz)(1 - 2μ) / E
Volumetric strain = 3 x Linear Strain → if σx = σy = σz
Cylinder → εv = εL + 2εd
Sphere → εv = 3 x εd
Vol strain of sphere varying dia = 3(d2-d1)/d1
Strain = P(1 - 2μ)/AE
x=x/E-y/E-z/E → Similarly for y and z
Relationship b/w constant
E > K > G
E = 2G(1 + μ) = 3K (1 - 2μ) = 9KG/(3K + G)
μ = (3K-2G)/(6K+2G) → 1/3 ≤ G/E ≤ 1/2
Hooke's law
Stress ∝ strain → σ = Eε
Valid up to the limit of proportionality
Thermal Stress & Strain
σ = E α ∆T
∆L = L α ∆T
Strain = α ∆T
α → Al > brass > copper > steel (ABCS)
T ↑es & Restrained → Expand → Compressive stress
T↓es & Restrained → Shrinkage → Tensile stress
∆ Due to combined σ & T
Temp fall = - L α T + σL/E
Temp rise = + L α T - σL/E
Deformation of bar Due to axial load P
Prismatic bar → ∆ = PL/AE
Bar in series → ∆ = ∆1 + ∆2 + ...= (PL/AE)1 + (PL/A)2+...
Bar with varying Width → ∆ = PL2.303 log(B/b) / E.t(B-b)
Cone Frustum → ∆ = 4PL/πD1D2E
11.11 %error → if d is taken avg of D1 & D2
Deformation of bar Due to self
∆ ∝ L², σ ∝ L → Not depends on c/s area
W = γAL
Cone = WL/2AE = γL²/6E = ⅓ x Prismatic bar
Prismatic bar = WL/2AE = γL²/2E = 3 x Cone
Lifting W load → Cable ∆ = WL/2AE
Composite Bar
Composed of more than one material rigidly connected together so as to behave as one piece.
α → Al > brass > copper > steel (ABCS)
P = P1 + P2, ∆1 = ∆2 → Use these eqⁿ to solve qtns.
∆1 = ∆2 = P1L/A1E1 = P2L/A2E2 = PL/(A1E1+A2E2)
P1 = PA1E1/(A1E1+A2E2) & P2 = PA2E2/(A1E1+A2E2)
Equivalent E = (A1E1 + A2E2)/(A1 + A2)
σ1/σ2 = E1/E2
P = P1 σ1 + P2 σ2
Independent & Total elastic constants
Homogeneous, isotropic, elastic material obeying hooke's law = 2(E,μ) & 4
Orthographic (wood) = 9 & 12
Anisotropic = 21 & infinity
Isotropic → Elastic properties same in each and every direction (steel)
Homogenous → Material having Uniform composition throughout or properties same throughout its volume.
Anisotropic → Elastic properties are not same in any direction (wood)
Orthotropic → Elastic properties are same in all direction other than that in perpendicular direction(wood, ply)
Visco-elastic material → Time dependent stress-strain curve
SHEAR FORCE & BENDING MOMENT
Shafts → Torque
Tie → Tension
Strut → Compression
Beams → Transverse loading only i.e, BM & SF
Helical Spring → Twisting
Thrust diagram → Variation of axial load along the span
Compatibility eqn → Extra eqn to analyse str
Arching of Beam → To reduce BM
Max Free bending moment over fixed beam = Sum of fixed end moment
At point of application of a concentrated load on a beam there is → Maximum BM
Share Force
Resultant of all transverse forces to the right or left of sectⁿ
At point of symmetry → SF = 0
Bending Moment
Resultant moment at a section due to all the transverse forces either to left or right of the sectⁿ
Sagging = +ve BM,
Hogging = -ve BM.
At hinge → BM = 0
Max BM in beam occurs where SF changes sign
Max BM due to moving load on a fixed ended beam → At a Support
Pure bending → BM = constant, SF = 0
Flexural Shear
Shear associated with change of bending moment along the span
Point of Contraflexure
POC → BM changes sign & BM = 0
Propped Cantilever beam subjected to UDL/P/UVL → Contraflexure point = 01
Two Span continuous beam with both end fixed → Points of contraflexure = 04
For a fixed beam having UDL = L/2√3 = 0.289L ← From centre, 0.211 L ← from Support
Focal length → Distance b/w adjacent contraflexure
Point of inflection
Deflected shape of beam changes
Points to remember
At hinge BM = 0
At the point of symmetry SF = 0.
dV/dx = W, dM/dx = V
Variation if loading = n → SF = n+1, BM = n + 2.
∆M = M2 - M1 = Area under SF diagram.
SF = 0 → M is constant at that particular sectⁿ & vice-versa.
SF = 0 → BM = max for SSB.
Sf = 0 → BM is max or min.
SF Changes sign → BM is max or min but not vice-versa.
Locus of reactⁿ of 2H semicircular arch → a straight line.
intermediate support sinks than -ve BM ↓es & +ve BM ↑es.
Jacketing → When Beam/columns become weak or insufficient
Shear Span → Zone where SF = Constant.
Non-yielding support → Has zero slope, Can take any amount of reactⁿ.
SSB (UVL loading) → max BM = wl²/9√3 ← at x = L/√3 = 0.519L
Types of support
Free or roller or Rocker, Built in or fixed, Hinged or pinned, Link, Slider support
Types of BM
Cantilever, Propped Cantilever, Simply supported Beam, Fixed End/ Encastre Beam, Continuous Beam,Overhanging Beam
For SSB → BM at support = 0
Fixed Beam is an example of 2D structure
Continuous Beam
CB may or may not be an OB
Sagging moment → middle region of span
Hogging moment → intermediate support
SHEAR STRESS
Shear stress → q = VAy̅/Ib = fQ/It=SAy̅/Ib
Q = Ay → 1st moment of area
Normal stress → by BM
Shear stress → by SF
Shear stress → Max at centre or N.A. & 0 at extreme fibres
Bending stress → Zero at centre or N.A. & max at extreme fibres
Equal strength form → When stress in each c/s is just equal to working stress
To avoid Shear failure → Shear strength = 2 x Shear strength
Shear stress distribution
Parabolic distⁿ
0 at extreme fibre.
τ avg = SF/Area
qavg=SF/Area=τavg
Rectangle q=6S(d2/4-y2)/bd3 → y from NA
Circle q=4S(R2-y2)/3R4 → y from NA
Shear centre or Centre of flexure
Point through which if transverse bending load passes , the beam will have no twisting, only Bending
Always lies on the axis of symmetry
Semi circle SC = 4R/π
For no torsion → Plane of bending should pass through shear centre of section
BENDING STRESS
√2 = 1.414 , √3 = 1.732
Bending section mod = I/y
Torsional sectⁿ mod = J/r
Compression → Direct stress > Bending stress
Beam stiffness = δ max/Span
Uniform strength → Same bending stress at all sectⁿ
Pure Bending
SF = axial = torsional force = 0
Bending moment = constant
Prismatic bar Shape → Arc of circle
Assumption Theory of simple Bending
Material of beam homogeneous & isotropic
E tension = E compression
Plane Sectⁿ before bending remain plane after bending
Isotropic → Elastic constant are same in all the direction
Flexure formula/Bending equation
M/I = σ/y = E/R
Bending stress σ = My/I = M/Z.
MOR = (σ max) x (I/y)
Curvature 1/R = M/EI
Radius of curvature R = EI/M.
Flexural rigidity = EI
Sectⁿ modulus
Z = I/y ← in mm³
Strength of beam is measured by Z
Elastic Strength ∝ z ∝ I/y
Z↑es → Strength ↑es.
0.011d portion should be removed from top and bottom of a circular c/s of dia d in order to obtain maximum section modulus
Strongest rectangular sectⁿ from a circular log → width b = D/√3 = 0.577D & depth d = √2D/√3 = 0.816D → b/d = 0.707 = 1/√2, d/b = 1.414 = √2
Same bending stress → Uniform strength.
Economical → Sqr > Rec > circle (When wt or area is equal)
Z sqr = 1.81 Z circle.
Weight → Circle > Square > Rectangular (Same strength,stress)
Z rectangular/ Z diamond = √2 & I will be the same for both = bd³/12.
Two prismatic beam of Same material, length, flexural strength → Weight circular/Square = 1.118
I Beam
Flange → Bending & Web → Shear.
Most efficient & economical
More Bending stress & lateral stability is higher
Z & MOI is high.
80% BM resistance by Flange hence preferred over rectangular sections & MOI is high.
Beam of constant strength or fully stressed Beam.
Max stress at every X-sectⁿ of beam is equal to the max allowable bending stress in the beam.
Use eqn → σ = My/I = M/Z = constant
MOI & CENTROID
1st moment of area = A y̅ = 0 about CG for all sectⁿ.
Section modulus → 1st moa about axis of Bending.
2nd moment of area = Moment of inertia ≠0 ≠-ve.
(ΣA) y̅ = A1y1 + A2y2 + A3y3 +...
Moi is a concept applicable in the case of a rotating body.
MOI → Resistance against Rotation
Orthogonal axis Σ MOI = Constant
Eclipse → Locus of moi about inclined axis to principal axis
Principal axis→ Product of MOI = 0.
Locus of MOI : Ellipse about inclined axis to principal axis.
Ix = Ixx + A y̅ ²
Iy = Iyy + A x̅ ²
Polar MOI
MOI about z-axis
Iz = Ip = Ixx + Iyy
Rectangular sectⁿ
at centre Ixx = bh³/12 , Iyy = b³h/12
about diagonal = b³h³/6(b²+h²)
Triangular section:
at centre Ixx = bh³/36 , Iyy = b³h/48
at base = bh³/12
equilateral Triangle C.G. = a/2√3
Semicircular
x̅ = d/2
y̅ = 4r/3π
Ixx = 0.11r⁴
Circular
x̅ = y̅ = d/2
Ixx = Iyy = πD⁴/64
Ip = πD⁴/32
Ring
I = πR³t
A = 2πRt
Quadrant of circle:
x̅ = y̅ = 4r/3π = 0.636R
Trapezoidal:
y̅ = (2a+b)h/(a+b)3. a<b
Solid cone:
y̅ = h/4
Hollow cone:
y̅ = h/3
Solid half sphere (hemisphere):
y̅ = 3r/8
DEFLECTION
For Beams & Frames major deflection → is due to Bending
For Trusses deflection is caused → by internal Axial Forces
EI is Flexural rigidity
δ ∝ 1/EI
Max or minimum deflection of a beam → Zero slope location of beam
Beam → Carries transverse loading only
Beams of uniform strength are preferred to those of uniform section bcz → Economical for large span
Strain energy = (1/2 )P∆
Deflection Depends on
y ∝ Load(P)
y ∝ BM
y ∝ Span(L)
y ∝ 1/A (Cross section)
y ∝ 1/EI
Deflection decreases By
Stronger material (E↑es)
increase MOI (I), Depth increases more I than width
↓es length of Beam
↓es Load on Beam
1). Cantilever
At Fixed end → Deflection(∆) = 0, Slope = 0
At Free end → ∆ = maximum, Slope = maximum
2). SSB
At support → ∆ = 0, Slope = maximum
Deflection(∆) = maximum → at a point where slope is Zero
3). Fixed/Builtin/Encastre Beam
At Ends → Slope = 0
Methods to Determine θ & ∆
Area moment theorem / Moment Area method / Mohr's method
Only if the deflected shape is Continuous.
Equation in Slope deflection methods → Derived using Moment area theorems
Mohr Theorem 1 → Slope = θ2 - θ1 = Area of (M/EI) diagram
Mohr Theorem 2 → ∆ = ∆1 - ∆2 = Moment of (M/EI) diagram
Not suitable for Continuous beam
Double integration method
Gives deflection only due to Bending
y = deflection
dy/dx = Slope = Deformation corresponding to Moment
d²y/dx² = M/EI = dθ/dx = 1/R = Curvature .
d³y/dx³ = V/EI
d⁴y/dx⁴ = W/EI
M = EId²y/dx² → SF = dM/dx = EId³y/dx³
Flexural rigidity = EI
Flexural Stiffness = Flexural rigidity/Length = EI/L.
Conjugate Beam thᵐ
Thᵐ 1 → Slope in Real beam = SF in Conjugate Beam
Thᵐ 2 → Displacement = BM in Conjugate Beam
Loading in CB = M/EI diagram of actual beam
Fix ⇄ free
Roller ⇄ roller
Slider ⇄ slider
Pin ⇄ pin
internal pin/roller → Hinge
internal hinge → internal roller
PRINCIPAL STRESS & STRAIN
Plane stress
When two faces of cubic elements are free from any stress, the stress condition is called plane stress condition → σ = τzx = τ zy = 0
Plane stress components → σ x , σ y , τ xy.
Principal plane
Only σ exists & τ = 0.
Product of moi = 0 i.e. Ixy = 0 → Principal axes of sectⁿ
MOI = max or minimum.
τ xy = 0 → σ x' = σ1 cos²θ + σ2 sin²θ
Angle of principal plane → tan2p=2xy/(x-y)
Transformation of Plane Stress
θ should be taken from the major axis.
θ major = 90 + θ minor or 180 - θ minor
Cos2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ
σ x + σ y = σ x' + σ y' = σ 1 + σ 2 = constant
σ r = √(σ n² + τ max²)
At obliquity of 45° → σ n = τ = σ/2 → normal = tangential stress
Plane of max obliquity = π/4 + Ømax/2 ← inclined to major principal plane
Principal Stress & maximum shear stress
Extreme values of normal stresses are called principal stresses.
Greatest Normal stress ( σ 1) = 2 τ max & σ 2 = 0.
τ max = max of [½(σ 1- σ 2) or ½(σ 2- σ 3) or ½(σ 3- σ 1)]
Angle b/w max τ & max σ = 45° or 135°
Mohr's Circle
Centre always lie on x-axis symmetrical
C = [½ ( σ x + σ y), 0] = [½ (σ 1 + σ 2), 0]
R = τ max
σ 1, σ 2 = C ± R
Planes of max Shear stress are 45° or 135° to principal planes.
at planes of max shear stress → σ x/σ y = ± ½(σ 1+σ 2) = 0, min, max.
tan(2θ) = 2τxy/(σ x-σ y)
Diff Cases/Conditions
Pure shear stress
σ1 = -σ2 → σ 1 = +τ xy, σ 2 = - τ xy
R = τ, C = (0,0) at origin
σ x = σ y = 0
Hydrostatic loading or stress
Mohr circle → Reduce to a point.
C = (σ,0), R = 0
σ x = σ y = σ
Uniaxial tension
Analysis of Strain
1=1/E-2/E
1=E(1+2)/(1-2)
2=E(2+1)/(1-2)
Total strain energy per unit volume = (11+22+33)/2
SE/Vol = (12+22+32-2(12+23+31))/2E → 3D case
THEORIES OF FAILURE
1>>>>>> 2,3 → All theory gives same results
COLUMN'S
in case of eccentrically loaded struts Composite sectⁿ is preferred.
to determine allowable stress in axial compression ISI adapted Secant formula(1984)
i. Short Column
λ < 32
Fails in crushing (yielding)
ii. Medium size column
32 < λ < 120
Combined failure
iii. Long Column
λ > 120
Fails in Buckling (elastic instability)
IS 456 → λ = leff/r
Crippling or Buckling or Critical load (Pcr)
Max axial load which is sufficient to keep a column in a small deflected shape.
Euler's Theory
Applicable to long columns(λ > 120) only → buckling failure only
Material is isotropic, homogeneous & linear elastic
λ ≥ 80
Load → Pe = π²EI/Leff² = π²Er²A/Leff² = π²EA/λ²
Stress → σ cr =Pe/A = π²E/λ²
5 - 10% error assumption made not met is real life
Validity of euler's theory
σ crσ y → π²E/λ²σ y
λy/π²E
λc=90 Mild steel
Eff Length
Fix free = 2L
Hing Hing = L
Fixed fixed = L/2(.65L)
Fix Hinge = L/√2(.8L)
Electric pole = 2L
Fixed → eff held in position & restrained against rotation.
Hinged → eff held in position & Not restrained against rotation.
Rankine's Formula
All columns → Slenderness ratio has any value →Buckling and crushing both considered
1/Pr = 1/Pc = 1/Pe → Pr = (PePc)/(Pe+Pc)
Crushing load → Pc= σ c. A
Pr = σ c. A /(1+α2)
Rankine Constant
α = σ c / π²E ← depends on material
α → Cast iron > Timber > MS > WI.
Slenderness ratio (λ)
λ = Leff/rmin→ Short column ≤ 32, medium column = 32 - 120, long column ≥ 120.
λ = Leff/LLD→ Pedestal ≤ 3, short column = 3-12, long column ≥ 12
Cantilever column → λeff = 2L/r
λ = 0 → if its length is supported on all sides through its length, hence no Buckling
Failures of columns depend on the Slenderness ratio of the columns
Radius of Gyration:
measure of resistance against rotation or buckling.
The whole area of the body is considered from a given axis.
rmin = Imin/A
More r → more resistant to Buckling or rotation.
Column will buckle around min r.
Solid shaft r = D/4 = R/2
Hollow shaft = √((D^2 + d^2)/4)
Kern
No tension
Core area of the section in which if the load applied , tension will not be induced in the sectⁿ.
e = kern/2
TORSION OF CIRCULAR SHAFT
Assumption in Torsion Eqⁿ or Formula
Plane Sectⁿ remains plane after twisting → only for hollow or solid circular c/s
Circular sectⁿ remains circular after twisting
Twist along shaft is uniform
Shaft is straight & has a uniform c/s
Stress induced within elastic limit
Stress-strain variation is linear
Torsion eqⁿ → T/J = τ/r = Gθ/L → Valid for circular section
T = τJ/r = τ.Zp
Other imp Points
Polar modulus(Zp) = J/r, r = D/2
Circle → J = πD⁴/32
Hollow → J = π(D⁴-d⁴)/32
Tube → J = 2πR³t
Solid → τ max = 16T/πD³
Hollow → τ max = 16T/πD³(1-1/n4) → n= outer dia/inner dia
π = 180°→ 1° = π/180 radian
Torsional rigidity (GJ) ∝ strength
Torsional stiffness = T/θ
Hollow circular section is best in torsion.
Box type sectⁿ → Torsion ⭕, 🔲.
J hollow > J solid → τ develop in hollow < τ solid . ( Same wt.)
Position of τ max → Circular = Outermost fibres & Closed coil helical spring = innermost fibre.
Pure Torsion → Equal & opposite twisting moment at end.
At free end their will be max angle of Twisting
In rectangular shaft subjected to torsion max shear stress → At middle of longer side
Series Connection
T = T1 = T2 = ..
θ ad = θ ab + θ bc + θ cd
Parallel Connection
T = T1 + T2 + ..
θ1 = θ2 → TL/GJ = Constant
Torsional Strain Energy (U)
U = Tθ/2 = T²L/2GJ = (τ ²max/4G) x Vol
SE density = U/Vol of shaft
U solid = τ ²max/4G
U hollow = ( τ ²max/4G)x((Ro² + Ri²)/Ro²)
U hollow > U solid
U hollow/ U Solid = (D^2 + d^2)/D^2
Due to shear stress U = (τ ²/2G) x Vol
Shear Resilience → SE/Vol = τ ²/2G=/2
Resilience ∝ Elasticity → Regain
Power
P = T x ω = 2πNT/60 = 2πfT
ω = 2πf = 2πN/60
P = watt, T = N-m, ω = rad/sec, N = rpm
Watt = N-m/sec.
Same Dia → Power Solid > Hollow secⁿ
Same Weight/material → Power Solid < Hollow sectⁿ
Thin tube
T = πD²tτ/2 = 2πr²tτ
Ip = J = 2πr³t
Shear flow → q = τ t = T/2Am = constant
τ 1.t1 = τ 2.t2 = constant
COMBINED STRESS
Bending & Torsion
Equivalent Me = ½[M + √(M²+T²)]
Equivalent Te = √(M²+T²)
Max Bending stress = 32M/πD^3
Max Shear stress = 16T/πD^3
Max Bending/Shear stress = 2M/T
SPRING
Flexibility → Deformation of spring produced by unit load.
Spring constant or stiffness of spring (k) = P/δ.
Watch → Flat spiral spring
Spring or Axles are made up of Vanadium steel.
The pitch of the close coil spring is very small.
Proof load → The greatest load which a spring can carry without getting permanently distorted.
Proof Resilience → Strain energy stored when proof load is applied without being permanently distorted.
Proof Stress → Max stress in the spring when PL is applied
Leaf/Laminated Spring
Given an initial curvature → Bcz Spring becomes flat when it is subjected to design load
Subjected → To Bending stress
Supported → at Centre
Loaded → at Ends
Deflection ∝ 1/Stiffness
Carriage spring
Central deflection = 3WL^3/8Enbt^3
Closed Helical Spring
in closed helical springs material is subjected to the torsional moment & induce torsional shearing stress.
U = T²L/2GJ = kx²/2, k = stiffness & x = deformation
θ = TL/EI = 64TBn/Ed⁴
L = πDn
I = πd⁴/64
k = P/δ = GD⁴/64R³n
Proof load Pmax = π³ σ max / 16R.
Angle of Helix
Angle made by the coil with horizontal
Angle of helix ≤ 10° ← Closed coil
Angle of helix > 10° ← Open coil.
Parallel connection
Keq = K1 + K2 + K3…
Series connection
1/Keq = 1/K1 + 1/K2 + 1/K3..
THICK & THIN CYLINDER/SPHERE
Thin shell
Wall t < 1/15 to 1/20 its internal dia.
Thin Cylinder Subjected to internal pressure
Hoop or Circumferential stress σ h = pd/2t = 2 x σ L
Radial pressure = inside = P outside = 0
Longitudinal or Axial stress σ L = pd/4t = σ h/2
L/H Stress = 1/2
Longitudinal strain = pd(1-2μ)/4tE
Hoop strain = pd(2-μ)/4tE
L/H strain = (1-2μ)/(2-μ) = (m-2)/(2m-1)
τ max = (σ h - σ L)/2 = pd/8t.
Thin cylinder shell subjected to an internal pressure then → if σ h (tensile) → Dia↑es & σ L (tensile) → Length↑es.
Hoop stress induced in a thin cylinder or by winding it with wire under tension will be Compressive.
Thin Sphere Subjected to internal pressure
Hoop = longitudinal stress = pd/4t
Hoop strain = longitudinal strain = pd(1-μ)/4tE
Volumetric Strain = 3 x circumferential strain.
Max shear stress in plane = 0
Absolute Max shear stress = pd/8t.
Lame's theorem of Thick Shells
Based on max principal stress theory of failure
To find thickness of thick shells
Clavarious eqⁿ: t of Ductile + close end
Birnies eqⁿ : Ductile + open end.
Damping coefficient = Damping ratio x critical damping coefficient.
Relative stiffness = MOI/L
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