CHATR Stress MCHANICS OF MATRIALS and Strain Axial Loading Stress & Strain: Axial Loading 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. Later chapters will deal with torsional and pure bending loads. - 1
- Normal Strain normal strain stress L A ε σ L A A ε σ L L A ε σ - 4 Stress-Strain Test
Stress-Strain Diagram: Ductile Materials 1 Ksi 6.89 Ma 1 lb 4.448 N 1in 5.4 mm - 5 Stress-Strain Diagram: Brittle Materials - 6
Hooke s Law: Modulus of lasticity Below the yield stress σ ε Youngs Modulus or Modulus of lasticity - 7 Hooke s Law: 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. - 8 4
lastic vs. lastic Behavior - 9 Fatigue - 10 5
Fatigue Fatigue properties are shown on S-N diagrams. A 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 Axial Loading Hooke s Law: σ σ ε ε strain: ε L A L A il i i Ai i - 1 6
Deformations Under Axial Loading Hooke s Law: σ σ ε ε strain: ε L quating, L A A With variations in loading, cross-section or material properties, il i i Ai i - 1 xample.01 6 9 10 psi D 1.07 in. d 0.618 in. Determine the deformation of the steel rod shown under the given loads. - 14 7
SOLUTION: Divide the rod into three components: Apply free-body analysis to each component to determine internal forces, 1 60 10 lb 15 10 lb 0 10 lb L1 L 1 in. A1 A 0.9 in L 16 in. A 0.in valuate total deflection, L 1 1 L1 L L i i + + i Ai i A1 A A 75.9 10 in. ( 60 10 ) 1 ( 15 10 ) 1 ( 0 10 ) 1 16 6 + + 9 10 0.9 0.9 0. 75.9 10 in. - 15 Sample roblem.1 The rigid bar BD is supported by two links AB and CD. Link AB is made of aluminum ( 70 Ga) and has a cross-sectional area of 500 mm. Link 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. - 16 8
Sample roblem.1 SOLUTION: Free body: Bar BD M B 0 0 0 ( 0kN 0.6m) FCD + 90kN tension MD 0 ( 0kN 0.4m) + FCD 0.m FAB 0.m FAB 60kN compression Displacement of B: B Displacement of D: D L A ( 60 10 N)( 0.m) -6 9 ( 500 10 m )( 70 10 a) 6 514 10 m L A B 0.514 mm ( 90 10 N)( 0.4m) -6 9 ( 600 10 m )( 00 10 a) 6 00 10 m D 0.00 mm - 17 Sample roblem.1 Displacement of D: BB BH DD HD 0.514 mm 0.00 mm x 7.7 mm H DD HD 0.00 mm 1.98 mm ( 00 mm) x x ( 400 + 7.7) mm 7.7 mm 1.98 mm - 18 9
Static Indeterminacy Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations. L + R 0-19 xample.04 Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied. - 0 10
xample.04 SOLUTION: Solve for the displacement at B due to the applied loads with the redundant constraint released, 1 0 600 10 N 4 900 10 N 6 6 A1 A 400 10 m A A4 50 10 m L1 L L L4 0.150 m 9 i Li 1.15 10 L i Ai i Solve for the displacement at B due to the redundant constraint, 1 RB 6 A1 400 10 m L1 L 0.00 m il i R i Ai i 6 A 50 10 m ( 1.95 10 ) RB - 1 xample.04 L + R 0 9 1.15 10 ( 1.95 10 ) RB 577 10 N 577 kn RB 0 Fy 0 RA 00 kn 600kN + 577 kn RA kn RA kn RB 577 kn - 11
Thermal Stresses A 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. T α( T ) L α thermal expansion coef. T + 0 L A α( T ) L + 0 L A The thermal deformation and the deformation from the redundant support must be compatible. T + 0 Aα( T ) σ α( T ) A - 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. Assuming that the material is isotropic (no directional dependence), ε y ε z 0 oisson s ratio is defined as lateral strain ε y ε ν z axial strain ε ε x x - 4 1
Generalized Hooke s Law 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: σ νσ x y νσ ε z x + νσ σ x y νσ ε z y + νσ νσ x y σ ε z z + - 5 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 1 ν [ ] [ ] x ε x + ε y + ε z ( σ + σ + σ ) x y y z dilatation (change in volume per unit volume) z For element subjected to uniform hydrostatic pressure, ( 1 ν ) p e p k k 1 ( ν ) bulk modulus Subjected to uniform pressure, dilatation must be negative, therefore 0 <ν < 1 x y z 1
Shearing Strain A 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 f γ xy ( ) xy A 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 - 7 xample.10 A 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. SOLUTION: Determine the average angular deformation or shearing strain of the block. Apply Hooke s law for shearing stress and strain to find the corresponding shearing stress. Use the definition of shearing stress to find the force. - 8 14
Determine the average angular deformation or shearing strain of the block. 0.04in. γ xy tan γ xy γ xy 0.00 rad in. Apply Hooke s law for shearing stress and strain to find the corresponding shearing stress. τ xy Gγ xy ( 90 10 psi)( 0.00 rad) 1800psi Use the definition of shearing stress to find the force. τ A ( 1800 psi)( 8in. )(.5in. ) 6 10 lb xy 6.0kips - 9 Relation Among, ν, and G An axially loaded slender bar will elongate in the axial direction and contract in the transverse directions. An 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. Axial load also results in a shear strain. Components of normal and shear strain are related, ( 1+ν ) G - 0 15
Sample roblem.5 A 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 10x10 6 psi and ν 1/, determine the change in: a) the length of diameter AB, b) the length of diameter CD, c) the thickness of the plate, and d) the volume of the plate. - 1 SOLUTION: Apply the generalized Hooke s Law to valuate the deformation components. find the three components of normal strain. B A ε x d ( + 0.5 10 in./in. )( 9in. ) σ νσ + x y νσ z B A + 4.8 10 in. ε x C D ε z d ( + 1.600 10 in./in. )( 9in. ) 1 1 ( 1ksi) 0 ( 0ksi) 6 10 10 psi C D + 14.4 10 in. + 0.5 10 in./in. νσ σ x y νσ ε + z y 1.067 10 in./in. νσ νσ x y σ ε + z z + 1.600 10 in./in. t t ε y ( 1.067 10 in./in. )( 0.75in. ) t 0.800 10 in. Find the change in volume e ε x + ε y + ε z 1.067 10 in /in V ev 1.067 10 ( ) 15 15 0.75 in V +0.187in - 16
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 Law but with directionally dependent moduli of elasticity, x σ x ε x ε y ν xy ε x y ν xz σ y ε y ε z ε x z σ ε Transverse contractions are related by directionally dependent values of oisson s ratio, e.g., Materials with directionally dependent mechanical properties are anisotropic. z z - Saint-Venant s rinciple Loads 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. - 4 17
Stress Concentration: Hole Discontinuities of cross section may result in high localized or concentrated stresses. σ K σ max ave - 5 Stress Concentration: Fillet - 6 18