# Strength of Materials

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Published on December 15, 2013

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Strength of Material complete stress strain torsion etc etc

1. Structures, loads and stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013

This course is concerned with structures: A structure is a solid object or assembly. A structure connects components, carries loads, provides form and integrity.

F22, 2002 Wright Flyer, 1903

Burj al Arab Qutab

• Distance between any two points does not change when a force is applied on it.

An upward reaction force at support Thus, an unbalanced moment results. a moment at But equilibrium requires a force and the support. Where does it come from?

Bar Resisting Force As force increases, elongation increases till the equilibrium is restored again. This implies that there is a force resisting the deformation, and that force increases with deformation.

Stress, ζ = P/A Strain, ε = δ/L Area A P Robert Hooke in 1678 showed l E is the elastic modulus, or simply the Elasticity δ P P

 Dimensions of stress: F/Area = F/L2 Units of stress = N/m2 = (Pa)scal, same as that of pressure. A very small unit. Standard atmospheric pressure = 1.03×105 Pa MPa and GPa (106 Pa and 109 Pa, respectively) are commonly used

 Strain δ/L is dimensionless, hence NO UNITS.

  E = Stress / strain, and therefore, has dimensions of stress, i.e., F/L2. Units of E are, accordingly, Pa(scal).

Material Aluminium 2024-T3 Aluminium 6061-T6 Aluminium 7075-T6 Concrete Copper Glass fibre Cast iron Steel, High strength Steel, Structural Titanium Wood Value of E in GPa 70 70 70 20 – 35 100 65 100 200 200 100 10-15

The external forces acting on a structure result in strains. The strains so produced result in stresses within the material of the members. The stresses, for the most part, are proportional to the strains. The constant of proportionality is termed as the modulus of elasticity.

Stresses due to various loads Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013

Bar F L δ

Tension in the belt

P P P P

Stress: Force Intensity 20 mm 5 mm ζ = F/A = 300N 20×5×10− 6 m2 = 3 MPa 300 N

Uniform stress is an approximation. Valid only in simple loadings. Away from ends.

105 Steel Concrete Soil N Bearing strengths Steel >> concrete >> soil Required areas Steel << concrete << soil

105 N Permissible compressive stress in steel is about 400 MPa. Steel Concrete Soil So the area of steel required is 105 N/400 MPa, or 2.5 10−4, or about 16 mm 16 mm

105 N Permissible compressive stress concrete is about 60 MPa. Steel Concrete Soil So the area of concrete required is 105 N/ 60 MPa, or 1.67 10−3, or about 41 mm 41 mm

Permissible bearing strength 105 N of soils varies widely. For good cohesive soil, it could be between 100 to 400 kPa, if it is above the water table. Steel Concrete Soil So the area of the footing required is 105 N/ 200 kPa (say), or 0.5 m2, or about 710 mm 710 mm

Shear Stresses

Bearing (Compressive) Shear

Blanking force = shear strength Shear area = perimeter shear area sheet thickness

Compression Shear Compression Shear

P P P Shear Shear area

Shear stresses on the back face of the shaft The stresses result in a moment that balances the twisting moment

Compression near top Extension near bottom

Bending of Beams Net tensile force is zero!

• Forces that tend to reduce the size of a structural member produce compressive strains which, in turn, produce compressive stresses.  Forces that tend to distort the shape of a member produce shear strains which in turn produce shear stresses.

• A twisting moment applied to a shaft produces shear strains. These shear strains give rise to shear stresses which result in a moment that balances the external twisting moment.  A bending moment produces both tensile and compressive strains and stresses. These give rise to a resisting moment which balances the bending moment.

Tensile members and trusses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013

10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN

A member on which external forces act only at two distinct points (and there is no external torque acting on it) is termed as a two-force member. The forces acting on a two-force member are equal and opposed. But is it enough?

The forces acting on a two-force member are equal, opposed and collinear.

Two-force members 60o 30o F1cos30o – F2cos60o = 0 500 N F1sin30o + F2sin60o – 500 = 0 F1 = 125 N; F2 = 433 N

Simply supported: No moment. Only reaction force. • Cannot translate, • Can rotate.

Load Pinned Support: Reaction could be inclined. (1 DoF) Roller Support: Normal reaction only. (2 DoF)

FRICTION must act

RH2X4 – 600X0.5 = 0 or RH2 = 75 N RH2 4m Or, N RV1 600 N RH1 1m y x z

While in the pinned support, the member is restrained from translating, in the clamped support, the member cannot even rotate.

• Cannot translate, • Cannot rotate. No DoF at all. Also called built-in support

P Fy Fx Mz L y Statically determinate x z

Resisting moment Tension will build up faster than the moments due to bending, and therefore, can treat the joint as pinned

Two-force members

3m θ cos = 4/5, sin = 3/5

Method of Joints Symmetry: FGB FGC FGF FGA 15 kN 2FGC sinθ – 15 kN = 0 FGC = (15 X 5)/(2 X 3) = 12.5 kN

Method of Joints -FCGsinθ - FCFsin θ = 0 FCF = - FCG = -12.5 kN FCB FCG FCF - FCB – FCGcosθ + FCFcosθ = 0 FCB = –2FCGcosθ = 10 kN

FFC ∑FV = RB,V +(3/5)FFC = 0 or, RB,V = - (3/5)FFC = + 7.5 kN ∑FH = -(4/5)FFC - FFG = 0 or, FFG = - (4 /5)FFC = 10 kN FFG RB,V

B D C 20 kN

B B A two-force member D D C 20 kN FBD,D

F2 B F1 D 20 kN C D 20 kN ∑Fy = 0

• Co-planar • Concurrent

Notation for stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013

First index: normal to the plane on which acting. z x Second index: direction of y the stress component itself

Stress Intensity of Force

The stress vector t depends upon the location as well as the direction of the surface.

z x y The sign of a stress component depends on the direction of normal and the direction of force: If both have same sign then the stress component is positive, if the two have different signs, then the stress component is negative.

ζyy is negative ηyx is positive ηyz is negative z x y

ζxx is positive ηxy is negative ηxz is negative z x y

Face Area Force xy(direction of (assume outward unit depth) compone compone nt nt normal) x δy•1 σxx• δy τxy• δy −x δy•1 −σxx• δy −τxy•δy y δx•1 σyy• δx τyx• δx −y δx•1 −σyy• δx −τyx•δx Consider the moment balance about the mid-point:

Thin-walled cylinders are used extensively in industry and homes because they are very efficient structures. • Oil storage tanks are cylindrical • So are oxygen bottles, cooking gas cylinders • Deodorant bottles are pressurized cylinders. • So are beer cans.

  Cylindrical and spherical pressure vessels are commonly used for storing gas and liquids under pressure. A thin cylinder is normally defined as one in which the thickness of the metal is less than 1/20 of the diameter of the cylinder.

  In thin cylinders, it can be assumed that the variation of stress within the metal is negligible, and that the mean diameter, Dm is approximately equal to the internal diameter, D. At mid-length, the walls are subjected to hoop or circumferential stress, and a longitudinal stress, .

  The internal pressure, p tends to increase the diameter of the cylinder and this produces a hoop or circumferential stress (tensile). If the stress becomes excessive, failure in the form of a longitudinal burst would occur.

C o n sid e r th e h a lf cylin d e r sh o w n . F o rce d u e to in te rn a l p re ssu re , p is b a la n ce d b y th e fo rce d u e to h o o p stre ss,  h . i.e . h o o p stre ss x a re a = p re ssu re x p ro je cte d a re a h x 2 L t = P x d L h = (P d ) / 2 t W h e re : d is th e in te rn a l d ia m e te r o f cylin d e r; t is th e th ickn e ss o f w a ll o f cylin d e r.

T h e in te rn a l p re s s u re , P a ls o p ro d u c e s a te n s ile s tre s s in lo n g itu d in a l d ire c tio n a s s h o w n a b o v e .  d F o rc e b y P a c tin g o n a n a re a lo n g itu d in a l s tre s s ,   d t  L  d 4 L is b a la n c e d b y a c tin g o v e r a n a p p ro x im a te a re a , (m e a n d ia m e te r s h o u ld s tric tly b e u s e d ). T h a t is : x d t  P x  L 4 2  P d 4t 2

1. Since hoop stress is twice longitudinal stress, the cylinder would fail by tearing along a line parallel to the axis, rather than on a section perpendicular to the axis.  The equation for hoop stress is therefore used to determine the cylinder thickness.  Allowance is made for this by dividing the thickness obtained in hoop stress equation by efficiency (i.e. tearing and shearing efficiency) of the joint. 

Take section of a pressurized cylinder And the upper half FBD of the lower half σθθ p σθθ

FT FR p We can show by symmetry arguments that: (a) Both shear should be inwards or outwards (b) Shear should be ZERO

FT FBD of the ‘contents’ p Net forced on the curved surface = p×2r×δl Equilibrium: FT = ζ 2δl t = p×2r×δl This gives: Hoop stress

Forces on the rim Pressure on ζ the back cap Axial stresses are lone-half of hoop stresses

Forces on the rim Pressure on the ‘content’ p Maximum stress in a spherical vessel is one half that of a cylindrical vessel of same radius and thickness

Shaped structures

Arch Keystone All stones are subjected to compressive forces only.

Towers Load bearing cables Deck of bridge Cables support the bridge through tension. Towers carry compression,

The main span of the Golden Gate suspension bridge is 1.287 km long. The sag in the cables is 140 m. The design loading is 400 kN/m.

Tension in the cable at the lowest point is; To = 2.96×108 N Max tension = 3.23×108 N Each cable consists of 27,572 strands of 4.88-mm diameter wires bundled parallel. Cross-sectional area of the cable = 27,572×[π×0.004882/4] = 0.516 m2 So stress = 625.5 MPa

z x y

Take section of a pressurized cylinder And the upper half FBD of the lower half σθθ p σθθ

FT FR p We can show by symmetry arguments that: (a) Both shear should be inwards or outwards (b) Shear should be ZERO

p

2. Deformations, strains and material properties Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013

P/A L δ/L Stress = E×Strain P

Load Equilibrium Stress Macro Geometry Micro Strain Micro Geometry Material Property Deformation Macro

Cross-section: 6 cm2 Section Tension, N Stress, kPa Strain Length, m Elongation, m AB BC CD 250 500 800 416.7 4.16×10- 3 833.3 8.35×10- 3 1333.3 13.33×10- 3 1.5 2.0 1.5 6.24×10- 3 16.70×10- 3 20.02×10- 3 DE EF 550 300 916.7 9.20×10- 3 500.0 5.00×10- 3 1.5 2.0 13.78×10- 3 10.00×10- 3

W = ρAxg = T ζ = T/A = ρxg 72 m ε = ζ/E = ρxg/E dδ = εdx = ρgxdx/E T dx x W(x)

or For a Nylon wire: density, ρ ~ 0.8X103 kg/m3, and E ~ 400 MPa. We get δ ~ 52 mm For steel: density is 7.6X103 kg/m3, and E is 200 GPa. We get δ ~ 1 mm

RA,y RA,x A B RB,y Member Force kN Lengt h m Area m2 AC 28.3 1.41 1.77×10−4 BC − 20 1 1.77×10−4 RA,y = 20 kN RA,x = − 20 kN RB,y = 20 kN C 20kN Stress MPa TAC = 28.8 kN TBC = − 20 kN Strain Elongation m 7.6×10−4 1.07×10−3 −113.2 −5.4×10−4 −0.54×10−3 160.1

A A y x 45o B E C D C1 45o B E C D 45o G F C1 X-displacement of C ~ shortening of BC =−0.54 mm y-displacement of C ~ EF + FC1 = CD/cos45o + FG(=EC) ~ 1.25 mm

Statically-indeterminate structures Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013

PP P R1 R2 R3 Reaction at middle support (and hence, at all supports depends on the bending of plank.

1. Consideration of static equilibrium and determination of loads 2. Consideration of relations between loads and deformations, (first converting loads to stresses, then transforming stresses to strain using the properties of the material, and then converting strains to deformations), 3. Considerations of the conditions of geometric compatibility

Indeterminate because x is not known! 1.3 m 2.6 m F2 F1 150 kN 1m

R1 Taking moments about the pivot point, 2R1 + 2R2 – P = 0 P Indeterminate because 4 unknown forces and only three equations to determine them. R2 Geom. Comp. δ1 = δ2 R1L1/E1A1 = R2L2/E2A2

R2 - F - R1 = 0; R1L – Fx = 0 Geom. Comp. h + δ = 2(h - δ ) 1 2 F h R1 = kδ1 R2 = kδ2 L x

P P = R1+ R2 R2 R1 R1 = (E1A1/L1)δ1 R2 = (E2A2/L2)δ2 Geom. Comp. δ1 = δ2

(a) Tendon being stressed during casting. Tension in tendon, no stress in concrete. (b) After casting, the force is released and the structure shrinks. (c) FBD of tendon. The concrete does not let the tendon shrink as much as it would on its own. This results in residual tension in the tendon. (d) FBD of concrete. The residual force in the tendon is trying to compress the concrete..

A concrete beam of cross-sectional area 5 cm 5 cm and length 2 m be cast with a 10 mm dia mild steel rod under a tension of 20 kN. The external tension in steel released after the concrete is set. What is the residual compressive stress in the concrete? Calculate the extension of steel under the tension of 20 kN T = 20 kN →σ = 255 Mpa →ε = 1.21 10- 3 →δ = 2.42 mm

2.42 mm δs δc δs + δc = 2.42 mm

Lateral strains: Poisson ratio, ν Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013

σxx εxx = σxx/E εyy = - ν εxx ν is Poisson ratio σxx Another material property

Let us consider εxx. σxx produces an εxx = σxx /E σyy produces an εyy = σyy /E, which through Poisson ratio gives εxx = -νεyy = - νσyy/E. Similarly for σzz .

Shear stresses do not cause any normal strain Therefore, εxx = ζxx/E – νζyy/E - ν ζzz/E = [ζxx – ν(ζyy + σzz)]/E Similarly for εyy and εzz

F ζyy = −F/A, ζzz = 0 What is ζxx and εyy y x Geometric compatibility: εxx = 0 εxx = [ζxx – ν(ζyy + ζzz)]/E 0 = [ζxx – ν(ζyy + 0)]/E, → ζxx =νζyy = − ν F/A εyy = [ζyy – ν(ζxx + ζzz)]/E = [−F/A + ν F/A]/E = −(1− ν)F/AE

σyy Steel: εx = 0.6×10−4 εy = 0.3×10−4 Find σxx and σyy : Plug in: εxx = [σxx – ν(σyy + σzz)]/E εyy = [σyy – ν(σzz + σxx)]/E E = 200 GPa, ν = 0.3 σxx

Shear strains and stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013

Apply shear stresses to a block: Shear strain γ is π/2 − θ Shear strain is also seen as: θ1 + θ2 θ2 θ θ1 Since angles are measured positive counter-clockwise, the angle θ2 above is a negative angle. In general terms, then, γ = θ1 − θ2 with θs measured positive when counter-clockwise

A square blocks 0.2 mm × 0.2 mm deforms under shear ystress Coordinates after C D deformation (in mm) are: θ2 θ1 A A(0,0), B(0.194, 0.013), and D(−0.012, 0.196). B x θ1 = 0.013/0.2 = 0. 065 θ2 = 0.012/0.2 = 0. 06 γxy = 0.65 − 0.60 = 0.05 radians

Shear strain γ is related to shear stress τ by γxy = τxy/G, where G is shear modulus It can be shown that γxy does not depend on other components of stress.

Material Aluminium G, GPa 25 Steel 80 Glass 26-32 Soft Rubber 0.003- 0.001

8,000 N 4,000 N Shear stress τ = 4,000 N/ (0.1 m)(0.12 m) = 3.33×105 Pa Shear strain γ = τ/G 3.33×105 Pa/1 MPa = 0.33 Wall Wall Rubber blocks 10 cm × 10 cm with 12 cm height

Consider the rubber block on the left: 8,000 N γ =0.33 Wall And therefore, The vertical deflection of load = 0.33×0.10 m = 33 mm Wall Rubber blocks 10 cm × 10 cm with 12 cm height

We have so far introduced three elastic properties of materials. Material ν E, GPa G, GPa 70 25 0.33 Steel 200 80 0.27 Glass 50-80 26-32 0.21-.27 0.00080.004 0.0030.001 0.50 Aluminium Soft Rubber

Thermal strains and stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013

On heating, there is linear expansion: There is no thermal shear strain Material Steel Aluminium α (×10-6/oC) ~ 10 ~ 20

Putting Hooke law, Poisson effect and thermal strains all together, εxx = [σxx – ν(σyy + σzz)]/E + αΔT εyy = [σyy – ν(σzz + σxx)]/E+ αΔT εzz = [σzz – ν(σxx + σyy)]/E+ αΔT γxy = τxy/G, γyz = τyz/G, and γzx = τzx/G

Aluminium rod, rigid supports. Temperature raised by ΔT. What are the stresses? εxx = 0 = [σxx/E + αΔT] σxx = −αEΔT x

Tank is flush when empty. Find end forces when pressure is p Due to p: σzz = pr/ 2t, σθθ = pr/ t z If end forces F, axial stress due to it is F/2πrt εzz = [(pr/ 2t − F/2πrt) −νpr/t ]/E Equate it to 0 and determine F p

Determining stress-strain relations Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013

 A material property. Tensile Test Machine, UTM

σ (= F/Ao) Ductile Brittle ε (=∆L/Lo) Ductile Failure cup-and-cone Necking Brittle Failure

Y σ (= F/A0) Yield stress, σY 0.02% Permanent set ε (= ΔL/L0)

Y1 Ultimate stress σ (= F/A0) Y B ε (= ΔL/L0)

σ σ σ (a) Rigid ε ε (b) Perfectly elastic σ ε (c) Elastic-Plastic σ Increase in yield strength ε (d) Perfectly plastic ε (e) Elastic- Plastic (strain hardening)

= 9.82×10-6 m4

Let us check on the stresses: Quite safe

Φ 10 cm Φ 5 cm F Φ 2 cm Φ 6 cm 1m F 0.6 m 150 N.m −250 N.m 150 N.m

Φ 10 cm Φ 5 cm F Φ 2 cm Φ 6 cm 1m F 0.6 m 150 N.m

Φ 10 cm Φ 5 cm F Φ 2 cm Φ 6 cm 1m F 0.6 m 150 N.m Angle θ2 which represents the counter-clockwise movement of the smaller gear due to gearing alone is 10/6 of θ1 or 0.0085 rad counter-clockwise Rotation of the right end of second shaft wrt stationary wall is, therefore, 0.0085 rad + 0.12 rad = 0.1285 rad or 7.36 degree.

By geometry: γθz= rdΦ/dz Therefore, τθz = GrdΦ/dz r varies from R1 to R2, and θ varies from 0 to 2π

2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 Weight 0 0.2 0.4 0.6 0.8 1

2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 Weight Stiffness 1

2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 Strength to weight Stiffness Weight 0 0.2 0.4 0.6 0.8 1

M1 Mo M2 TMD Equilibrium Condition: - M1 + Mo – M2 = 0 Geometric Condition: Φ1 +Φ2 = 0

Shear flow on horizontal surfaces is same as on the vertical surfaces q1 = q2

Relating q to twisting moment T h o qds dT at O = qds×h = q×2×Grey area q = T/2A

T 100 Nm R 20 mm R 16 mm This gives τ = (49 N/m)/0.004 m = 12.25 MPa

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