T dition CHTR MCHNICS OF MTRIS Ferdinand. Beer. Russell Johnston, Jr. John T. DeWolf ecture Notes: J. Walt Oler Texas Tech University Stress and Strain xial oading
- Contents Stress & Strain: xial oading Normal Strain Stress-Strain Test Stress-Strain Diagram: Ductile Materials Stress-Strain Diagram: Brittle Materials Hooke s aw: Modulus of lasticity lastic vs. lastic Behavior Fatigue Deformations Under xial oading xample.01 Sample roblem.1 Static Indeterminacy xample.04 Thermal Stresses oisson s Ratio Generalized Hooke s aw Dilatation: Bulk Modulus Shearing Strain xample. Relation mong, ν, and G Sample roblem.5 Composite Materials Saint-Venant s rinciple Stress Concentration: Hole Stress Concentration: Fillet xample.1 lastoplastic Materials lastic Deformations Residual Stresses xample.14,.15,.16
- Stress & Strain: xial oading Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced under loading. Statics analyses alone are not sufficient. Considering structures as deformable allows determination of member forces and reactions which are statically indeterminate. Determination of the stress distribution within a member also requires consideration of deformations in the member. Chapter is concerned with deformation of a structural member under axial loading. ater chapters will deal with torsional and pure bending loads.
- 4 Normal Strain σ stress δ ε normal strain σ δ ε σ δ δ ε
- 5 Stress-Strain Test
- 6 Stress-Strain Diagram: Ductile Materials
- 7 Stress-Strain Diagram: Brittle Materials
- 8 Hooke s aw: Modulus of lasticity Below the yield stress σ ε Youngs Modulus or Modulus of lasticity Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of lasticity) is not.
- 9 lastic vs. lastic Behavior If the strain disappears when the stress is removed, the material is said to behave elastically. The largest stress for which this occurs is called the elastic limit. When the strain does not return to zero after the stress is removed, the material is said to behave plastically.
- Fatigue Fatigue properties are shown on S-N diagrams. member may fail due to fatigue at stress levels significantly below the ultimate strength if subjected to many loading cycles. When the stress is reduced below the endurance limit, fatigue failures do not occur for any number of cycles.
- 11 Deformations Under xial oading From Hooke s aw: σ σ ε ε From the definition of strain: δ ε quating and solving for the deformation, δ With variations in loading, cross-section or material properties, i δ i i i i
- 1 xample.01 9 6 psi D 1.07 in. d 0.618 in. SOUTION: Divide the rod into components at the load application points. pply a free-body analysis on each component to determine the internal force Determine the deformation of the steel rod shown under the given loads. valuate the total of the component deflections.
- 1 SOUTION: Divide the rod into three components: pply free-body analysis to each component to determine internal forces, 1 60 15 0 lb lb lb valuate total deflection, δ i i i i i 1 9 6 75.9 1 1 1 ( 60 ) 1 ( 15 ) 1 ( 0 ) + + 0.9 0.9 0. in. 1 + + 16 1 1 1 in. 0.9 in 16 in. 0.in δ 75.9 in.
- 14 Sample roblem.1 SOUTION: The rigid bar BD is supported by two links B and CD. ink B is made of aluminum ( 70 Ga) and has a cross-sectional area of 500 mm. ink CD is made of steel ( 00 Ga) and has a cross-sectional area of (600 mm ). For the 0-kN force shown, determine the deflection a) of B, b) of D, and c) of. pply a free-body analysis to the bar BD to find the forces exerted by links B and DC. valuate the deformation of links B and DC or the displacements of B and D. Work out the geometry to find the deflection at given the deflections at B and D.
- 15 Sample roblem.1 SOUTION: Free body: Bar BD Displacement of B: δ B ( 60 N)( 0.m) -6 9 ( 500 m )( 70 a) 514 6 m M F F B 0 M CD D 0 B 0 ( 0kN 0.6m) 0 ( 0kN 0.4m) + F + 90kN tension F CD B 60kN compression 0.m 0.m Displacement of D: δ D δ B 0.514 mm ( 90 N)( 0.4m) -6 9 ( 600 m )( 00 a) 00 6 m δ D 0.00 mm
- 16 Sample roblem.1 Displacement of D: BB DD BH HD 0.514 mm 0.00 mm x 7.7 mm ( 00 mm) x x DD δ 0.00 mm δ H HD 1.98 mm ( 400 + 7.7) mm 7.7 mm δ 1.98 mm
- 17 Static Indeterminacy Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate. structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium. Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations. Deformations due to actual loads and redundant reactions are determined separately and then added or superposed. δ δ + δ R 0
- 18 xample.04 Determine the reactions at and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied. SOUTION: Consider the reaction at B as redundant, release the bar from that support, and solve for the displacement at B due to the applied loads. Solve for the displacement at B due to the redundant reaction at B. Require that the displacements due to the loads and due to the redundant reaction be compatible, i.e., require that their sum be zero. Solve for the reaction at due to applied loads and the reaction found at B.
- 19 xample.04 SOUTION: Solve for the displacement at B due to the applied loads with the redundant constraint released, δ 1 1 1 0 i i i i i 400 600 4 6 m 1.15 9 N 0.150 m 4 4 900 50 Solve for the displacement at B due to the redundant constraint, δ 1 1 1 R 400 i R ii 6 0.00 m i i B m ( 1.95 ) 50 R B 6 m N 6 m
- 0 xample.04 Require that the displacements due to the loads and due to the redundant reaction be compatible, δ δ + δ 1.15 δ R B R 577 0 9 ( 1.95 ) N 577 kn R B 0 Find the reaction at due to the loads and the reaction at B F 0 R 00 kn 600kN + 577 kn R y kn R R B kn 577 kn
- 1 Thermal Stresses temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports. Treat the additional support as redundant and apply the principle of superposition. δ T α ( T ) α thermal expansion coef. δ δ α T + δ 0 ( T ) + 0 δ The thermal deformation and the deformation from the redundant support must be compatible. δ δt + δ 0 α( T ) σ α ( T )
- oisson s Ratio For a slender bar subjected to axial loading: σ ε x x σ y σ z 0 The elongation in the x-direction is accompanied by a contraction in the other directions. ssuming that the material is isotropic (no directional dependence), ε y ε z 0 oisson s ratio is defined as lateral strain ε y ν axial strain ε x ε z ε x
- Generalized Hooke s aw For an element subjected to multi-axial loading, the normal strain components resulting from the stress components may be determined from the principle of superposition. This requires: 1) strain is linearly related to stress ) deformations are small With these restrictions: ε ε ε x y z σ νσ x y νσ + z νσ σ x y νσ + z νσ νσ x y σ + z
Dilatation: Bulk Modulus Relative to the unstressed state, the change in volume is e 1 ( 1+ ε )( 1+ ε )( 1+ ε ) 1 1+ ε + ε + ε ε x 1 ν [ ] [ ] + ε y x + ε ( σ + σ + σ ) x z y y z dilatation (change in volume per unit volume) z For element subjected to uniform hydrostatic pressure, ( 1 ν ) p e p k 1 ( ν ) k bulk modulus Subjected to uniform pressure, dilatation must be negative, therefore 0 <ν < 1 x y z
- 5 Shearing Strain cubic element subjected to a shear stress will deform into a rhomboid. The corresponding shear strain is quantified in terms of the change in angle between the sides, τ xy ( ) f γ xy plot of shear stress vs. shear strain is similar the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains, τ Gγ τ Gγ τ Gγ xy xy yz yz where G is the modulus of rigidity or shear modulus. zx zx
- 6 xample. rectangular block of material with modulus of rigidity G 90 ksi is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force. Knowing that the upper plate moves through 0.04 in. under the action of the force, determine a) the average shearing strain in the material, and b) the force exerted on the plate. SOUTION: Determine the average angular deformation or shearing strain of the block. pply Hooke s law for shearing stress and strain to find the corresponding shearing stress. Use the definition of shearing stress to find the force.
- 7 Determine the average angular deformation or shearing strain of the block. γ xy 0.04in. tan γ xy γ xy 0.00 rad in. pply Hooke s law for shearing stress and strain to find the corresponding shearing stress. τ xy Gγ xy ( 90 psi)( 0.00 rad) 1800psi Use the definition of shearing stress to find the force. τ ( 1800 psi)( 8in. )(.5in. ) 6 lb xy 6.0kips
- 8 Relation mong, ν, and G n axially loaded slender bar will elongate in the axial direction and contract in the transverse directions. n initially cubic element oriented as in top figure will deform into a rectangular parallelepiped. The axial load produces a normal strain. If the cubic element is oriented as in the bottom figure, it will deform into a rhombus. xial load also results in a shear strain. Components of normal and shear strain are related, G ( 1+ν )
- 9 Sample roblem.5 circle of diameter d 9 in. is scribed on an unstressed aluminum plate of thickness t /4 in. Forces acting in the plane of the plate later cause normal stresses σ x 1 ksi and σ z 0 ksi. For x 6 psi and ν 1/, determine the change in: a) the length of diameter B, b) the length of diameter CD, c) the thickness of the plate, and d) the volume of the plate.
- 0 SOUTION: pply the generalized Hooke s aw to find the three components of normal strain. ε ε ε x y z σ + x 1 6 psi + 0.5 νσ x 1.067 νσ x νσ y σ + y νσ + 1.600 νσ z ( 1ksi) 0 ( 0ksi) in./in. νσ z y in./in. σ + z in./in. 1 valuate the deformation components. δ B ε d ( + 0.5 in./in. )( 9in. ) x δ C ε d D t δ t ε y z δ B + 4.8 in. ( + 1.600 in./in. )( 9in. ) δ C D + 14.4 in. ( 1.067 in./in. )( 0.75in. ) δ t 0.800 Find the change in volume e ε V x + ε ev y + ε z 1.067 1.067 V in /in in. ( ) 15 15 0.75 +0.187in in
- 1 Composite Materials Fiber-reinforced composite materials are formed from lamina of fibers of graphite, glass, or polymers embedded in a resin matrix. Normal stresses and strains are related by Hooke s aw but with directionally dependent moduli of elasticity, x ν σ x ε xy x ε y ε x y ν σ ε xz y y ε z ε x z σ z ε Transverse contractions are related by directionally dependent values of oisson s ratio, e.g., Materials with directionally dependent mechanical properties are anisotropic. z
- Saint-Venant s rinciple oads transmitted through rigid plates result in uniform distribution of stress and strain. Concentrated loads result in large stresses in the vicinity of the load application point. Stress and strain distributions become uniform at a relatively short distance from the load application points. Saint-Venant s rinciple: Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points.
- Stress Concentration: Hole Discontinuities of cross section may result in high localized or concentrated stresses. K σ σ max ave
- 4 Stress Concentration: Fillet
- 5 xample.1 SOUTION: Determine the largest axial load that can be safely supported by a flat steel bar consisting of two portions, both mm thick, and respectively 40 and 60 mm wide, connected by fillets of radius r 8 mm. ssume an allowable normal stress of 165 Ma. Determine the geometric ratios and find the stress concentration factor from Fig..64b. Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor. pply the definition of normal stress to find the allowable load.
- 6 Determine the geometric ratios and find the stress concentration factor from Fig..64b. D d 60mm 1.50 40mm K 1.8 r d 8mm 40mm 0.0 Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor. σ σ K 165 Ma 1.8 max ave 90.7 Ma pply the definition of normal stress to find the allowable load. σ ( 40mm)( mm)( 90.7 Ma) ave 6. N 6.kN
- 7 lastoplastic Materials revious analyses based on assumption of linear stress-strain relationship, i.e., stresses below the yield stress ssumption is good for brittle material which rupture without yielding If the yield stress of ductile materials is exceeded, then plastic deformations occur nalysis of plastic deformations is simplified by assuming an idealized elastoplastic material Deformations of an elastoplastic material are divided into elastic and plastic ranges ermanent deformations result from loading beyond the yield stress
- 8 lastic Deformations σ ave σ max K σ Y Y K lastic deformation while maximum stress is less than yield stress Maximum stress is equal to the yield stress at the maximum elastic loading U σ Y K Y t loadings above the maximum elastic load, a region of plastic deformations develop near the hole s the loading increases, the plastic region expands until the section is at a uniform stress equal to the yield stress
- 9 Residual Stresses When a single structural element is loaded uniformly beyond its yield stress and then unloaded, it is permanently deformed but all stresses disappear. This is not the general result. Residual stresses will remain in a structure after loading and unloading if - only part of the structure undergoes plastic deformation - different parts of the structure undergo different plastic deformations Residual stresses also result from the uneven heating or cooling of structures or structural elements
- 40 xample.14,.15,.16 cylindrical rod is placed inside a tube of the same length. The ends of the rod and tube are attached to a rigid support on one side and a rigid plate on the other. The load on the rod-tube assembly is increased from zero to 5.7 kips and decreased back to zero. a) draw a load-deflection diagram for the rod-tube assembly b) determine the maximum elongation c) determine the permanent set d) calculate the residual stresses in the rod and tube. σ r r Y, r 0.075in. 0 6 6ksi psi σ t t Y, t 0.0in. 15 6 45ksi psi
- 41 xample.14,.15,.16 a) draw a load-deflection diagram for the rodtube assembly Y, r σ Y, r r ( 6ksi)( 0.075in ).7 kips δ Y,r ε Y, r σ Y, r Y, r 6 0 6 psi 0in. 6 psi - in. Y, t σ Y, t t ( 45ksi)( 0.0in ) 4.5 kips δ Y,t ε Y, t σ Y, t Y, t 45 15 6 psi 0in. 90 psi - in. r + t δ δ δ r t
- 4 xample.14,.15,.16 b,c) determine the maximum elongation and permanent set at a load of 5.7 kips, the rod has reached the plastic range while the tube is still in the elastic range r t σ t Y, r t t.7 kips r ( 5.7.7).0 kips 0ksi 0.1in kips.0 kips δ t ε t σ t t 0 15 6 psi 0in. psi δ max δ t 60 in. the rod-tube assembly unloads along a line parallel to 0Y r 4.5kips m 15kips in. slope - 6 in. δ m max 5.7 kips 45.6 15kips in. in. δ p δ max + δ ( 60 45.6) in. δ p 14.4 in.
- 4 xample.14,.15,.16 calculate the residual stresses in the rod and tube. calculate the reverse stresses in the rod and tube caused by unloading and add them to the maximum stresses. δ ε 45.6 0in. in. 1.5 in. in. σ r σ t ε ε t r 6 ( 1.5 )( 0 psi) 6 ( 1.5 )( 15 psi) 45.6 ksi.8ksi σ σ residual, r residual, t σ + σ σ + σ t r t r ( 6 45.6) ksi 9. 6ksi ( 0.8) ksi 7.ksi