Chapter 8 Structural Design and Analysis 1 Strength and stiffness 5 types of load: Tension Compression Shear Bending Torsion
Normal Stress Stress is a state when a material is loaded. For normal forces (tension and compression): Stress Force Area σ P A (1) Normal stress: tensile stress and compressive stress. Stress has the same unit as pressure, N/m, Pascal (Pa), or N/mm (MPa) Yield strength of steel, aluminum, and titanium.
Normal Strain 3 Strain: elongation per unit length. δ ε L () Strain is dimensionless. Relation between stress and strain: Hooke s Law σ Eε (3) Young s modulus, modulus of elasticity, steel, aluminum. Stress is the state when a material is strained. Combining Equation (1), (), and (3), we can obtain δ FL EA ()
Boundary conditions and reaction forces of a cantilever beam and a simply supported beam. F F V M F V F V M F V M F V F V M Cantilever Beam and Simply Supported Beam F F P F P P F P F P P
Bending Stress 5 When a beam is in bending, the inner layer of the material is in compression, and the outer layer of the material is in tension. F P P Bending moment creates normal stresses in cantilever beams or simply supported beams, and can be calculated by σ σ max My I Mc I (5) (7) neutral 中性軸 axis
Moment of Inertia and Cross Sectional Modulus 6 I x y da, x da (6) I y D d D h I ( D d ) I πd 6 π bh I 6 1 b 3 stress externalload cross-sectionalproperty σ max M z (8) z I c (9)
Example 1. Calculation of Bending Stress (I) A 0 cm-long stick with radius of 5 mm free-body diagram are supported in the form of a simply supported beam. A 10 N force is exerted in the middle of the stick. Find the maximum bending stress produced. 5N shear force diagram 5N 10N 5N 7 0cm -5N 0.5Nm 10cm 0cm bending moment diagram
Simply Supported Beam, Concentrated Force 8 w a b wb wa l wa wb l l l free-body diagram wb l 0 l wa l shear force diagram bending moment diagram y max Wa b 3EIl
Simply Supported Beam, Uniform Pressure 9 w w l w w 0 w 1 free-body diagram shear force diagram bending moment diagram y max 5 38 3 Wl EI
Cantilever Beam, Concentrated Force 10 w w l 0 1 free-body diagram shear force diagram bending moment diagram y max 3 Wl 3EI
Example 1. Calculation of Bending Stress (II) 11 I πd 11.968 10 m 6 c.5 10-3 m σ max.1 10 7 Pa.1MPa If the 10-N force is in axial direction, the normal stress is A F σ 0.51MPa In this example, when the beam is in bending, the stress is 80 times larger than when the beam is in tension or compression by the same force. Maximum deflection: y max 5.61 10 7 E -1 m. If the load is in axial direction, the deflection is FL δ 1.0 10 E EA 5 1 m In this example, when the beam is in bending, the deflection is 500 times larger than when the beam is in tension or compression by the same force.
Summary 1 When a structure is loaded, if stress exceeds the yield strength or ultimate strength of the material, failure will occur. If the stiffness of a structure is too low, the deflection of the structure when loaded is large, which will affect the precision of motion of the mechanical system, and problems in vibration and noise will occur. For structures, the ability to sustain tension and compression is much higher than the ability to sustain bending, for example, breaking egg shells, dome structures. In mechanical design, we should try to avoid structures under bending, and we should be very sensitive to the regions under bending because they will be the source of failure.
Example. Principle of Superposition in Stress Calculation (I) 13 Calculate the stresses at point A and B. A A 30 o 5mm B 60mm 700N 5mm tension bending
1 Example. Principle of Superposition in Stress Calculation (II) + stress due to bending + stress due to tension total stress Stress due to bending: Mc σ I 1000.5 5 1,010MPa Stress due to tension: A F σ.5mpa Maximum stress (tension) at point A: 1010+.5103.5MPa Minimum stress (compression) at point B: 1010-.5985.75MPa
Example 3. Deflection Calculation of Gear Shafts (I) 15 For the three axes, I 1 6 10 3 mm, I 10 mm, I 3 10 5 mm, E 07GPa. Find the deformation of axis 1 and axis at the contact point of gears A and B. input 輸入軸 axis 1 輸出軸 output axis 3 3 輸入軸 1 輸出軸 3 A D B C axis 軸
Example 3. Deflection Calculation of Gear Shafts (II) 16 input 輸入軸 axis 1 輸出軸 output axis 3 3 輸入軸 1 輸出軸 3 A D B C axis 軸 Radial forces from the gears 150mm y A F A 1000N y A 3 FAL 8EI 1000N 8 07 10 9 ( 150mm) N/m 3 6 10 3 mm 0.0566mm 75mm
Example 3. Deflection Calculation of Gear Shafts (III) 17 F B 1000N F C 1500 N F B 1000N F C 1500N y B y B1 + y B 75mm 75mm 75mm 75mm 350mm 350mm 350mm y y FBa b 3EIL 3 07 10 ( 75mm) [( 350 75) mm] 1000N B1 9 B Fc ab 6EIL ( L a b ) N/m ( 350 75) 10 mm 350mm 0.0978mm [ ( 350mm) ( 75mm) ( 75mm) ] 0.1mm 1500N 75mm mm 6 07 10 10 350 9
Example 3. Deflection Calculation of Gear Shafts (IV) 18 Use superposition to calculate the total deflection at point B: y B yb B 1 + y 0.0978mm + 0.1mm 0.378mm Total separation of axis 1 and axis at the contact point of gears A and B y A + y B 0.0566 + 0.378 0.9mm
Direct Shear 19 When the beam is very short, the main effect of a shear force becomes direct shear. Direct shear stress is often the major cause of failure in connectors ctors and fasteners such as pins, bolts, screws, rivets, and welded joints. Shear stress: τ F A s (11) Example. Direct Shear Stress in a Pin 15mm P500N 10mm 0mm A s τ πd P As π 500 176.7 ( 15) 176.7mm N.8 mm.8mpa P P
Example. Direct Shear Stress in bolts 0 When torque T 0 100N m is applied to the shaft, the bolts are under direct shear. T 0 Area of a bolt: π d A Total shear force on bolts: D 50 T P 100N m P mm P Shear stress: π 10 mm 78.56mm 0 P τ A 000 N 78.56 mm 1.73MPa 000N D φ10 T 0
Shear Stress and Shear Strain 1 When material is under shear stress, the material is not compressed or stretched, instead, the perpendicular plane is shifted. This change in angle γ is called shear strain. Hooke s Law again: τ Gγ (1) G is called modulus of elasticity in shear. Three ways to create shear stresses: direct shear, vertical shear, and torsional shear. Vertical shear stress is often neglected, compared to the bending stress generated by the vertical shear force. Torsional shear stress is very important in transmission shaft design and analysis.
Torsional Shear Stress When a torque is applied to a shaft, the outer surface of the shaft experiences the greatest shearing strain and therefore the largest torsional shear stress. Tr Tc τ (13) τ max (17) J J Shear stress becomes zero at the center of the shaft. J is called polar moment of inertia, defined as: J r da (1) πd For a solid round shaft, J (15). For a hollow shaft, 3 Polar section modulus J T Z p τ max (18) c D d π 3 ( ) J (16) Z p
Example 5. Torsional Shear Stress in Transmission Shafts 3 Power of the motor is 750W, working at 1750rpm. Diameter of the shaft is 10mm, find the maximum torsional shear stress of the shaft. π Power(Watt)Torque(N m) rotational speed(rpm) (0) 60 1750 rpm 183 rad/s T 750 183 πd J 3 τ Tc J.098 π 3 ( N m),098( N mm) ( 10) 98( mm ) ( 098)( 5) max 98 0.87( MPa)
Example 6. Combine Torsion and Bending (I) Find the deflection at point A and B, and the maximum stress. Displacement due to bending: I πd 6 π ( 0.05) 8 6 10 m 00mm y max 3 WL 3 EI 1000N 9 3 07 10 N/m 6. 10 m ( 0.m) 3 10 8 m A 100mm φ5mm 1000N B
Example 6. Combine Torsion and Bending (II) 5 Displacement due to torsion: TL θ (1) GJ J πd 3 π ( 0.05m) 8 3 3.8 10 m A 00mm 100mm φ5mm 1000N T 1000N 0.1m 100N m θ TL GJ 80 10 100N m 0.m 9 8 N/m 3.8 10 m 0.0066(rad) B Total displacement at point B: 6. 10 m + 0.1m 0.0066 6. 10 m + 6.6 10 m 1.3 10 3 m
Example 6. Combine Torsion and Bending (III) 6 Stress due to bending: σ MC I 1000N 0.m 0.05 m 8 1.5 10 Pa 8 10 m max Stress due to torsion: 15MPa τ TC J 100N m 0.05 m 6 3.9 10 Pa 8 3.8 10 m max 3.9MPa 00mm How to add these two stresses? Mohr s circle can be used to describe the state of combined stress. A 100mm φ5mm 1000N B
-D Stress Element 7 y σ y To visualize the general case of combined stress, it is helpful to consider a small stress element of the load-carrying member. σ x τ xy τ yx σ x x For equilibrium, both normal stresses and shear stresses should exist in pairs, and τ xy τ yx For normal stress: tension +, compression σ y τ yx τ xy τ xy indicates the shear stress acting on the element face that is perpendicular to the x- axis and parallel to the y-axis. A positive shear stress is one that tends to rotate the stress element clockwise.
Principle Stresses 8 In the same stress state, when the coordinate on a stress element is rotated, the magnitudes of normal stresses and shear stresses change. σ We are interested in the maximum normal stress, maximum shear stress, and the planes on which these stresses occur. σ 1 x There will be only normal stresses σ 1 and σ on the stress element at a specific orientation. If σ 1 > σ, σ 1 is called the maximum principle stress or first principle stress, and σ is called the minimum principle stress or second principle stress. σ x + σ y σ x σ y σ 1 + + τ φ σ 1 xy tan 1 [ τ ( σ σ )] xy x y σ σ x + σ y σ x σ y σ + τ xy σ 1 φ σ
Construction of Mohr s Circle 9 Given σ x, σ y (from tension, compression, or bending) and τ xy (from direct shear or torsion). τ ( σ avg, τ max σ avg, τ ) φ φ τ Construct a σ-τ plane. Locate two points (σ x, τ xy )and (σ y, τ yx ) with proper signs. Draw a line segment between these two points. This line establishes the center O crossing the σ -axes. y σ 0 σ, τ x ) ( τ x xy σ 1 x φ φ σ σ Draw a circle using the line segment as diameter. The coordinates of O is σ, τ y ) ( τ y xy yx σ x + σ y, 0
Reading Mohr s Circle 30 Mohr s circle intersects the σ axes at (σ 1, 0) and (σ, 0). τ ( σ avg, τ max σ avg, τ ) φ φ τ From Mohr s circle, the maximum shear stress occurs at (σ avg, τ max ). σ, τ x ) ( τ x xy x The vector connecting point O and (σ x, τ xy ) represents x-axis on the stress element. σ 0 σ 1 φ φ σ σ The vector connecting point O and (σ y, τ yx ) represents y-axis on the stress element. Angles on the Mohr s circle double the true angles on the stress element. All points on Mohr s s circle represent the same stress state, viewed from different coordinates. y σ, τ y ) ( τ y xy yx
Example 7. Sketch Mohr s Circle ( a) σ x 0, σ y - 0, τ xy 0 ( b) σ 30, σ 0, τ 0 x y xy 3 o x 31 (0,0) (30,0) 58 o σ ( 0,0) σ 1 ( c) σ 0, σ 0, τ 0 x y xy σ σ 1 x 5 o 135 o ( 0, 0) (0,0) y σ σ 1 ( 0, 0) y
Example 8, 9. Pure Uniaxial Tension and Pure Torsion 3 τ τ τ max φ τ φ τ σ σ x 0 σ 1 σ y σ σ τ xy 0 σ 1 τ xy σ τ max 1 σ y at 5 degrees σ max τ xy at 5 degrees
Example 10. Three-Dimensional Mohr s Circle (I) 33 The problems presented thus far are plane stress problems, no stress is given for the third direction. A cylinder rolled by thin steel plate in a 75 degree angle, is subjected to an internal pressure 3.5MPa, with its ends closed. The thickness of the plate is 1mm, and the diameter of the cylinder is 00mm. 75 o 75 o Determine the principle stresses, maximum shear stress, and the orientation of the maximum shear stress element.
Example 10. Three-Dimensional Mohr s Circle (II) 3 For the pressure vessel problem, we have to consider the longitudinal stress and hoop stress. PD σ (Tension) t 3.75MPa y σ x PD t 87.5MPa (Tension) τ max Consider the Mohr s Circle for x-y plane: 87.5 3.75 τ 1.875 Consider the Mohr s Circle for x-z plane: 87.5 0 τ max 3.75 If the two principle stresses are of the same sign, we must consider the three dimensional Mohr s Circle in order to find the true maximum shear stress. (3.75,0) (87.5,0)
35 How Would a Structure Fail? What is the purpose for stress analysis? To predict whether the structure will fail under certain loading. How would a structure fail? Stress failure Maximum normal stress theory of failure Maximum shear stress theory of failure Maximum strain energy theory of failure All three theories compare the structure with the specimen in the e standard tensile test. Buckling Fatigue
Maximum Normal Stress Theory of Failure 36 The structure will fail if the maximum normal stress in the structure is larger than the allowable stress. For brittle materials, consider ultimate strength S u. For ductile material, consider yielding strength S y. Allowable stress S u /N or S y /N, N is called design factor or safety factor. Design factors are used to compensate the uncertainties in material properties, material quality, and loading conditions. Stress analysis is often simplified, too. Design factors are often determined according to designers confidence and understanding in the structural design problem. N3 in general cases; N for static loads, ductile materials, and if the designer has good understanding in the problem; N for impact loads, brittle materials, and if the designer does not have good understanding in the problem.
Maximum Shear Stress Theory of Failure 37 The structure will fail if the maximum shear stress in the structure is larger than the allowable stress. For ductile material, consider yielding strength in shear S sy, which is the maximum shear stress when the specimen in standard tensile test yields. 1 From the Mohr s Circle of pure uniaxial tension, τ max σ y at 5 degrees, therefore, 1 S sy S y Example 11. Maximum Shear Stress Failure in the Pressure Vessel Problem (I) 1 Ssy 15 Ssy S y 15MPa τ max 3.75MPa MPa 8.33MPa N 3 No shear stress failure
Maximum Strain Energy Theory of Failure 38 In the case of combined normal stress and shear stress, we should use the maximum strain energy theory of failure for failure prediction of the structure. 1 F 1 Strain energy per unit volume: U kx σ u Eε (5) k E If the strain energy per unit volume is larger than that of the specimen in standard tensile test when yielding, the structure will fail. If the maximum equivalent stress (or von Mises stress) is larger than the allowable stress S sy /N, the structure will fail. σ ( σ σ ) + ( σ σ ) + ( σ ) 1 3 1 σ 3 (6) Example 11. Maximum Strain energy Failure in the Pressure Vessel Problem (II) σ ( 87.5 3.8) + ( ) ( ) 87.5 0 + 3.8 0 75.70MPa No failure will occur σ S y N 90 96.67MPa 3
Stress Concentration 39 In many typical machine design situations, inherent geometric discontinuities are necessary for the parts to perform their desired functions. For example, shafts carrying gears, chain sprockets, or belt sheaves usually have several diameters that creates a series of shoulders that seat the power transmission members and support bearings. σ max Geometric discontinuities will cause the actual maximum stress in the part to be higher than the simple formulas predict. This is called stress concentration. Stress concentration phenomenon can be viewed as stress flow passing through geometric discontinuities. σ max Stress concentration factor: K t (7), σ σ 0 is called nominal stress. 0
0 Stress Concentration Factor Flat Plate with a Central Hole w d
Stress Concentration Factor Stepped Round Shaft, Tension 1 3.5 3 F F Kt.5 D r d 1.5 D/d.0 1.1 1. 1.01 1 0 0.05 0.1 0.15 0. 0.5 0.3 r/d
Stress Concentration Factor Stepped Round Shaft, Torsion.1 1.9 Kt 1.8 1.7 1.6 T T 1.5 1. D/d.0 1.3 1.11 1. 1.5 1.1 0 0.05 0.1 0.15 0. 0.5 0.3 r/d D r d
Example 13. Stress Concentration 3 r 1.5mm F 1mm 10mm F D d 1mm 10mm 1.0 r d 1.5mm 10mm 0.15 K t 1.60 πd F9800N A σ Ktσ K F A [( π )( 10mm) ] ( 9800N) 1.60 78.5mm t max 0 78.5mm 199.6MPa
Alternating Stress A point on the rotating shafts is subject to tensile stress compressive stress tensile stress. In alternating stress condition, we should consider fatigue failure, that is, structure fails when alternating stress repeats for a given number of times. input 輸入軸 axis 1 輸出軸 output axis 3 3 輸入軸 1 輸出軸 3 A D B C axis 軸
Mean Stress and Stress Amplitude 5 Mean Stress: Static part of the alternating stress. σ m ( + ) σ max σ min (8) Stress Amplitude: Dynamic part of the alternating stress. σ a ( ) σ max σ min (9) stress σ max σ a σ m 0 time σ min
Fatigue Test, S-N Plot, Endurance Limit 6 S ( stress ( 應力 ) ) Endurance limit is obtained under reversed, repeated bending load. Endurance 忍受限 limit S n 0.50Sut 700MPa S S ut ut 100MPa > 100MPa 10 6 N Number ( ( 旋轉數 of ) ) rotation
Fatigue Failure Prediction: Soderberg Criterion 力振Stress 幅S amplitude n 破壞線應S n N Horizontal axis: static load Vertical axis: reversed, repeated load. Design 安全應力線 stress line Failure 破壞區 zone Failure line 7 Safe 安全區 zone S y N S S y 平均應力 Mean stress
Buckling 8 x Force y A column is a structural member that carries an axial compressive load and that tends to fail by elastic instability, or buckling, rather than by crushing the y x material due to high stress. Columns are relatively long and slender. Slenderness ratio: Slenderness ratio L e r KL min r min (33) r min : minimum radius of gyration in the whole column L e : effective length
Radius of Gyration 9 r I A (31) D I πd 6 A πd y I πd 6 r A πd D x x h x-x axis: I 3 ht 1 3 r I th 1 h 0. A th 1 89 h y t y-y axis: I 3 th 1 3 r I ht 1 t 0. t A th 1 89
Effective Length 50 L e KL (3) P P P P P Theoretical value: K1.0 K0.5 K.0 K0.7 Practical value: K1.0 K0.65 K.1 K0.8
Long Column Analysis: The Euler Formula 51 For long columns, Euler formula gives the critical load P cr, at which the column would begin to buckle. P cr π EA ( KL r) (3) Notice that the buckling critical load is dependent only on the geometry (length and cross section) of the column and the stiffness of the material represented by the modulus of elasticity. The strength of the material is not involved at all. Therefore, it is of no benefit to use a high-strength material in a long column application.
Short Column Analysis: The J.B. Johnson Formula 5 For short columns, J.B. Johnson formula gives the critical load P cr, at which the column would begin to buckle. P cr ( KL r) S 1 y AS y π E (35) The critical load for a short column is affected by the strength of the material in addition to its stiffness. At a very low value for the slenderness ratio, the second term of the equation approaches zero and the critical load approaches the yield load. P cr S y A How to differentiate long columns and short columns?
Column Constant 53 The slenderness ratio at the intersection between Euler formula and J.B. Johnson formula is called column constant. C c π E S y (36) Long column: slenderness ratio larger than C c. Short column: slenderness ratio lower than C c. Pcr/A 臨界負荷 / 面積 1000 900 800 700 600 500 00 300 00 100 Johnson 公式 0 0 50 100 纖細比, K 尤拉公式 Euler formula J.B Johnson Johnson 公式 Formula