Chapter 8 Structural Design and Analysis. Strength and stiffness 5 types of load: Tension Compression Shear Bending Torsion

Size: px
Start display at page:

Download "Chapter 8 Structural Design and Analysis. Strength and stiffness 5 types of load: Tension Compression Shear Bending Torsion"

Transcription

1 Chapter 8 Structural Design and Analysis 1 Strength and stiffness 5 types of load: Tension Compression Shear Bending Torsion

2 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.

3 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 ()

4 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

5 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

6 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)

7 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

8 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

9 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 Wl EI

10 Cantilever Beam, Concentrated Force 10 w w l 0 1 free-body diagram shear force diagram bending moment diagram y max 3 Wl 3EI

11 Example 1. Calculation of Bending Stress (II) 11 I πd m 6 c m σ max 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 E -1 m. If the load is in axial direction, the deflection is FL δ 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.

12 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.

13 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

14 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 ,010MPa Stress due to tension: A F σ.5mpa Maximum stress (tension) at point A: MPa Minimum stress (compression) at point B: MPa

15 Example 3. Deflection Calculation of Gear Shafts (I) 15 For the three axes, I mm, I 10 mm, I 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

16 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 ( 150mm) N/m mm mm 75mm

17 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 ( 75mm) [( ) mm] 1000N B1 9 B Fc ab 6EIL ( L a b ) N/m ( ) 10 mm 350mm mm [ ( 350mm) ( 75mm) ( 75mm) ] 0.1mm 1500N 75mm mm

18 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 mm + 0.1mm 0.378mm Total separation of axis 1 and axis at the contact point of gears A and B y A + y B mm

19 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 ( 15) 176.7mm N.8 mm.8mpa P P

20 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 mm 1.73MPa 000N D φ10 T 0

21 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.

22 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

23 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) rpm 183 rad/s T πd J 3 τ Tc J.098 π 3 ( N m),098( N mm) ( 10) 98( mm ) ( 098)( 5) max ( MPa)

24 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) m 00mm y max 3 WL 3 EI 1000N N/m m ( 0.m) m A 100mm φ5mm 1000N B

25 Example 6. Combine Torsion and Bending (II) 5 Displacement due to torsion: TL θ (1) GJ J πd 3 π ( 0.05m) m A 00mm 100mm φ5mm 1000N T 1000N 0.1m 100N m θ TL GJ N m 0.m 9 8 N/m m (rad) B Total displacement at point B: m + 0.1m m m m

26 Example 6. Combine Torsion and Bending (III) 6 Stress due to bending: σ MC I 1000N 0.m 0.05 m Pa 8 10 m max Stress due to torsion: 15MPa τ TC J 100N m 0.05 m Pa 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

27 -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.

28 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 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

29 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

30 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

31 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

32 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

33 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.

34 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: τ Consider the Mohr s Circle for x-z plane: 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 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

36 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.

37 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

38 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 ) σ 3 (6) Example 11. Maximum Strain energy Failure in the Pressure Vessel Problem (II) σ ( ) + ( ) ( ) MPa No failure will occur σ S y N MPa 3

39 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

40 0 Stress Concentration Factor Flat Plate with a Central Hole w d

41 Stress Concentration Factor Stepped Round Shaft, Tension F F Kt.5 D r d 1.5 D/d r/d

42 Stress Concentration Factor Stepped Round Shaft, Torsion Kt T T D/d r/d D r d

43 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) mm t max mm 199.6MPa

44 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

45 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

46 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

47 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

48 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

49 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

50 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

51 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.

52 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?

53 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 Johnson 公式 纖細比, K 尤拉公式 Euler formula J.B Johnson Johnson 公式 Formula

Stress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

Stress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy Stress Analysis Lecture 3 ME 276 Spring 2017-2018 Dr./ Ahmed Mohamed Nagib Elmekawy Axial Stress 2 Beam under the action of two tensile forces 3 Beam under the action of two tensile forces 4 Shear Stress

More information

COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4017 COURSE CATEGORY : A PERIODS/WEEK : 6 PERIODS/ SEMESTER : 108 CREDITS : 5

COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4017 COURSE CATEGORY : A PERIODS/WEEK : 6 PERIODS/ SEMESTER : 108 CREDITS : 5 COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4017 COURSE CATEGORY : A PERIODS/WEEK : 6 PERIODS/ SEMESTER : 108 CREDITS : 5 TIME SCHEDULE MODULE TOPICS PERIODS 1 Simple stresses

More information

QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS

QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State Hooke s law. 3. Define modular ratio,

More information

Samantha Ramirez, MSE. Stress. The intensity of the internal force acting on a specific plane (area) passing through a point. F 2

Samantha Ramirez, MSE. Stress. The intensity of the internal force acting on a specific plane (area) passing through a point. F 2 Samantha Ramirez, MSE Stress The intensity of the internal force acting on a specific plane (area) passing through a point. Δ ΔA Δ z Δ 1 2 ΔA Δ x Δ y ΔA is an infinitesimal size area with a uniform force

More information

MECH 401 Mechanical Design Applications

MECH 401 Mechanical Design Applications MECH 401 Mechanical Design Applications Dr. M. O Malley Master Notes Spring 008 Dr. D. M. McStravick Rice University Updates HW 1 due Thursday (1-17-08) Last time Introduction Units Reliability engineering

More information

PES Institute of Technology

PES Institute of Technology PES Institute of Technology Bangalore south campus, Bangalore-5460100 Department of Mechanical Engineering Faculty name : Madhu M Date: 29/06/2012 SEM : 3 rd A SEC Subject : MECHANICS OF MATERIALS Subject

More information

: APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4021 COURSE CATEGORY : A PERIODS/ WEEK : 5 PERIODS/ SEMESTER : 75 CREDIT : 5 TIME SCHEDULE

: APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4021 COURSE CATEGORY : A PERIODS/ WEEK : 5 PERIODS/ SEMESTER : 75 CREDIT : 5 TIME SCHEDULE COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4021 COURSE CATEGORY : A PERIODS/ WEEK : 5 PERIODS/ SEMESTER : 75 CREDIT : 5 TIME SCHEDULE MODULE TOPIC PERIODS 1 Simple stresses

More information

Failure from static loading

Failure from static loading Failure from static loading Topics Quiz /1/07 Failures from static loading Reading Chapter 5 Homework HW 3 due /1 HW 4 due /8 What is Failure? Failure any change in a machine part which makes it unable

More information

Downloaded from Downloaded from / 1

Downloaded from   Downloaded from   / 1 PURWANCHAL UNIVERSITY III SEMESTER FINAL EXAMINATION-2002 LEVEL : B. E. (Civil) SUBJECT: BEG256CI, Strength of Material Full Marks: 80 TIME: 03:00 hrs Pass marks: 32 Candidates are required to give their

More information

ME Final Exam. PROBLEM NO. 4 Part A (2 points max.) M (x) y. z (neutral axis) beam cross-sec+on. 20 kip ft. 0.2 ft. 10 ft. 0.1 ft.

ME Final Exam. PROBLEM NO. 4 Part A (2 points max.) M (x) y. z (neutral axis) beam cross-sec+on. 20 kip ft. 0.2 ft. 10 ft. 0.1 ft. ME 323 - Final Exam Name December 15, 2015 Instructor (circle) PROEM NO. 4 Part A (2 points max.) Krousgrill 11:30AM-12:20PM Ghosh 2:30-3:20PM Gonzalez 12:30-1:20PM Zhao 4:30-5:20PM M (x) y 20 kip ft 0.2

More information

D : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each.

D : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each. GTE 2016 Q. 1 Q. 9 carry one mark each. D : SOLID MECHNICS Q.1 single degree of freedom vibrating system has mass of 5 kg, stiffness of 500 N/m and damping coefficient of 100 N-s/m. To make the system

More information

Mechanics of Materials Primer

Mechanics of Materials Primer Mechanics of Materials rimer Notation: A = area (net = with holes, bearing = in contact, etc...) b = total width of material at a horizontal section d = diameter of a hole D = symbol for diameter E = modulus

More information

QUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A

QUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A DEPARTMENT: CIVIL SUBJECT CODE: CE2201 QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State

More information

MAAE 2202 A. Come to the PASS workshop with your mock exam complete. During the workshop you can work with other students to review your work.

MAAE 2202 A. Come to the PASS workshop with your mock exam complete. During the workshop you can work with other students to review your work. It is most beneficial to you to write this mock final exam UNDER EXAM CONDITIONS. This means: Complete the exam in 3 hours. Work on your own. Keep your textbook closed. Attempt every question. After the

More information

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS 2009 The McGraw-Hill Companies, Inc. All rights reserved. Fifth SI Edition CHAPTER 3 MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek Torsion Lecture Notes:

More information

Mechanics of Materials II. Chapter III. A review of the fundamental formulation of stress, strain, and deflection

Mechanics of Materials II. Chapter III. A review of the fundamental formulation of stress, strain, and deflection Mechanics of Materials II Chapter III A review of the fundamental formulation of stress, strain, and deflection Outline Introduction Assumtions and limitations Axial loading Torsion of circular shafts

More information

NAME: Given Formulae: Law of Cosines: Law of Sines:

NAME: Given Formulae: Law of Cosines: Law of Sines: NME: Given Formulae: Law of Cosines: EXM 3 PST PROBLEMS (LESSONS 21 TO 28) 100 points Thursday, November 16, 2017, 7pm to 9:30, Room 200 You are allowed to use a calculator and drawing equipment, only.

More information

[5] Stress and Strain

[5] Stress and Strain [5] Stress and Strain Page 1 of 34 [5] Stress and Strain [5.1] Internal Stress of Solids [5.2] Design of Simple Connections (will not be covered in class) [5.3] Deformation and Strain [5.4] Hooke s Law

More information

Mechanical Design. Design of Shaft

Mechanical Design. Design of Shaft Mechanical Design Design of Shaft Outline Practical information Shaft design Definition of shaft? It is a rotating member, in general, has a circular cross-section and is used to transmit power. The shaft

More information

NORMAL STRESS. The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts.

NORMAL STRESS. The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts. NORMAL STRESS The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts. σ = force/area = P/A where σ = the normal stress P = the centric

More information

Name :. Roll No. :... Invigilator s Signature :.. CS/B.TECH (CE-NEW)/SEM-3/CE-301/ SOLID MECHANICS

Name :. Roll No. :... Invigilator s Signature :.. CS/B.TECH (CE-NEW)/SEM-3/CE-301/ SOLID MECHANICS Name :. Roll No. :..... Invigilator s Signature :.. 2011 SOLID MECHANICS Time Allotted : 3 Hours Full Marks : 70 The figures in the margin indicate full marks. Candidates are required to give their answers

More information

R13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PART-A

R13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PART-A SET - 1 II B. Tech I Semester Regular Examinations, Jan - 2015 MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) Time: 3 hours Max. Marks: 70 Note: 1. Question Paper consists of two parts (Part-A and Part-B)

More information

Structural Analysis I Chapter 4 - Torsion TORSION

Structural Analysis I Chapter 4 - Torsion TORSION ORSION orsional stress results from the action of torsional or twisting moments acting about the longitudinal axis of a shaft. he effect of the application of a torsional moment, combined with appropriate

More information

Tuesday, February 11, Chapter 3. Load and Stress Analysis. Dr. Mohammad Suliman Abuhaiba, PE

Tuesday, February 11, Chapter 3. Load and Stress Analysis. Dr. Mohammad Suliman Abuhaiba, PE 1 Chapter 3 Load and Stress Analysis 2 Chapter Outline Equilibrium & Free-Body Diagrams Shear Force and Bending Moments in Beams Singularity Functions Stress Cartesian Stress Components Mohr s Circle for

More information

MEMS Project 2 Assignment. Design of a Shaft to Transmit Torque Between Two Pulleys

MEMS Project 2 Assignment. Design of a Shaft to Transmit Torque Between Two Pulleys MEMS 029 Project 2 Assignment Design of a Shaft to Transmit Torque Between Two Pulleys Date: February 5, 206 Instructor: Dr. Stephen Ludwick Product Definition Shafts are incredibly important in order

More information

PERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR - VALLAM THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK

PERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR - VALLAM THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK PERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR - VALLAM - 613 403 - THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Sub : Strength of Materials Year / Sem: II / III Sub Code : MEB 310

More information

Stress Analysis Lecture 4 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

Stress Analysis Lecture 4 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy Stress Analysis Lecture 4 ME 76 Spring 017-018 Dr./ Ahmed Mohamed Nagib Elmekawy Shear and Moment Diagrams Beam Sign Convention The positive directions are as follows: The internal shear force causes a

More information

3. BEAMS: STRAIN, STRESS, DEFLECTIONS

3. BEAMS: STRAIN, STRESS, DEFLECTIONS 3. BEAMS: STRAIN, STRESS, DEFLECTIONS The beam, or flexural member, is frequently encountered in structures and machines, and its elementary stress analysis constitutes one of the more interesting facets

More information

SOLUTION (17.3) Known: A simply supported steel shaft is connected to an electric motor with a flexible coupling.

SOLUTION (17.3) Known: A simply supported steel shaft is connected to an electric motor with a flexible coupling. SOLUTION (17.3) Known: A simply supported steel shaft is connected to an electric motor with a flexible coupling. Find: Determine the value of the critical speed of rotation for the shaft. Schematic and

More information

Sample Questions for the ME328 Machine Design Final Examination Closed notes, closed book, no calculator.

Sample Questions for the ME328 Machine Design Final Examination Closed notes, closed book, no calculator. Sample Questions for the ME328 Machine Design Final Examination Closed notes, closed book, no calculator. The following is from the first page of the examination. I recommend you read it before the exam.

More information

2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at A and supported at B by rod (1). What is the axial force in rod (1)?

2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at A and supported at B by rod (1). What is the axial force in rod (1)? IDE 110 S08 Test 1 Name: 1. Determine the internal axial forces in segments (1), (2) and (3). (a) N 1 = kn (b) N 2 = kn (c) N 3 = kn 2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at

More information

D : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown.

D : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown. D : SOLID MECHANICS Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown. Q.2 Consider the forces of magnitude F acting on the sides of the regular hexagon having

More information

ME 354, MECHANICS OF MATERIALS LABORATORY COMPRESSION AND BUCKLING

ME 354, MECHANICS OF MATERIALS LABORATORY COMPRESSION AND BUCKLING ME 354, MECHANICS OF MATERIALS LABATY COMPRESSION AND BUCKLING PURPOSE 01 January 2000 / mgj The purpose of this exercise is to study the effects of end conditions, column length, and material properties

More information

Use Hooke s Law (as it applies in the uniaxial direction),

Use Hooke s Law (as it applies in the uniaxial direction), 0.6 STRSS-STRAIN RLATIONSHIP Use the principle of superposition Use Poisson s ratio, v lateral longitudinal Use Hooke s Law (as it applies in the uniaxial direction), x x v y z, y y vx z, z z vx y Copyright

More information

Chapter 3. Load and Stress Analysis

Chapter 3. Load and Stress Analysis Chapter 3 Load and Stress Analysis 2 Shear Force and Bending Moments in Beams Internal shear force V & bending moment M must ensure equilibrium Fig. 3 2 Sign Conventions for Bending and Shear Fig. 3 3

More information

CE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR

CE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR CE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR 2014-2015 UNIT - 1 STRESS, STRAIN AND DEFORMATION OF SOLIDS PART- A 1. Define tensile stress and tensile strain. The stress induced

More information

Table of Contents. Preface...xvii. Part 1. Level

Table of Contents. Preface...xvii. Part 1. Level Preface...xvii Part 1. Level 1... 1 Chapter 1. The Basics of Linear Elastic Behavior... 3 1.1. Cohesion forces... 4 1.2. The notion of stress... 6 1.2.1. Definition... 6 1.2.2. Graphical representation...

More information

Chapter 5: Torsion. 1. Torsional Deformation of a Circular Shaft 2. The Torsion Formula 3. Power Transmission 4. Angle of Twist CHAPTER OBJECTIVES

Chapter 5: Torsion. 1. Torsional Deformation of a Circular Shaft 2. The Torsion Formula 3. Power Transmission 4. Angle of Twist CHAPTER OBJECTIVES CHAPTER OBJECTIVES Chapter 5: Torsion Discuss effects of applying torsional loading to a long straight member (shaft or tube) Determine stress distribution within the member under torsional load Determine

More information

DEPARTMENT OF CIVIL ENGINEERING

DEPARTMENT OF CIVIL ENGINEERING KINGS COLLEGE OF ENGINEERING DEPARTMENT OF CIVIL ENGINEERING SUBJECT: CE 2252 STRENGTH OF MATERIALS UNIT: I ENERGY METHODS 1. Define: Strain Energy When an elastic body is under the action of external

More information

[8] Bending and Shear Loading of Beams

[8] Bending and Shear Loading of Beams [8] Bending and Shear Loading of Beams Page 1 of 28 [8] Bending and Shear Loading of Beams [8.1] Bending of Beams (will not be covered in class) [8.2] Bending Strain and Stress [8.3] Shear in Straight

More information

Russell C. Hibbeler. Chapter 1: Stress

Russell C. Hibbeler. Chapter 1: Stress Russell C. Hibbeler Chapter 1: Stress Introduction Mechanics of materials is a study of the relationship between the external loads on a body and the intensity of the internal loads within the body. This

More information

Sub. Code:

Sub. Code: Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may

More information

FINAL EXAMINATION. (CE130-2 Mechanics of Materials)

FINAL EXAMINATION. (CE130-2 Mechanics of Materials) UNIVERSITY OF CLIFORNI, ERKELEY FLL SEMESTER 001 FINL EXMINTION (CE130- Mechanics of Materials) Problem 1: (15 points) pinned -bar structure is shown in Figure 1. There is an external force, W = 5000N,

More information

4. SHAFTS. A shaft is an element used to transmit power and torque, and it can support

4. SHAFTS. A shaft is an element used to transmit power and torque, and it can support 4. SHAFTS A shaft is an element used to transmit power and torque, and it can support reverse bending (fatigue). Most shafts have circular cross sections, either solid or tubular. The difference between

More information

STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS

STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1 UNIT I STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define: Stress When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The

More information

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS GE SI CHAPTER 3 MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek Torsion Lecture Notes: J. Walt Oler Texas Tech University Torsional Loads on Circular Shafts

More information

SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA

SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA (Declared as Deemed-to-be University under Section 3 of the UGC Act, 1956, Vide notification No.F.9.9/92-U-3 dated 26 th May 1993 of the Govt. of

More information

Comb resonator design (2)

Comb resonator design (2) Lecture 6: Comb resonator design () -Intro Intro. to Mechanics of Materials School of Electrical l Engineering i and Computer Science, Seoul National University Nano/Micro Systems & Controls Laboratory

More information

CHAPTER 2 Failure/Fracture Criterion

CHAPTER 2 Failure/Fracture Criterion (11) CHAPTER 2 Failure/Fracture Criterion (12) Failure (Yield) Criteria for Ductile Materials under Plane Stress Designer engineer: 1- Analysis of loading (for simple geometry using what you learn here

More information

FME461 Engineering Design II

FME461 Engineering Design II FME461 Engineering Design II Dr.Hussein Jama Hussein.jama@uobi.ac.ke Office 414 Lecture: Mon 8am -10am Tutorial Tue 3pm - 5pm 10/1/2013 1 Semester outline Date Week Topics Reference Reading 9 th Sept 1

More information

(48) CHAPTER 3: TORSION

(48) CHAPTER 3: TORSION (48) CHAPTER 3: TORSION Introduction: In this chapter structural members and machine parts that are in torsion will be considered. More specifically, you will analyze the stresses and strains in members

More information

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS 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

More information

Chapter 5 Elastic Strain, Deflection, and Stability 1. Elastic Stress-Strain Relationship

Chapter 5 Elastic Strain, Deflection, and Stability 1. Elastic Stress-Strain Relationship Chapter 5 Elastic Strain, Deflection, and Stability Elastic Stress-Strain Relationship A stress in the x-direction causes a strain in the x-direction by σ x also causes a strain in the y-direction & z-direction

More information

Static Failure (pg 206)

Static Failure (pg 206) Static Failure (pg 06) All material followed Hookeʹs law which states that strain is proportional to stress applied, until it exceed the proportional limits. It will reach and exceed the elastic limit

More information

Chapter 4 Deflection and Stiffness

Chapter 4 Deflection and Stiffness Chapter 4 Deflection and Stiffness Asst. Prof. Dr. Supakit Rooppakhun Chapter Outline Deflection and Stiffness 4-1 Spring Rates 4-2 Tension, Compression, and Torsion 4-3 Deflection Due to Bending 4-4 Beam

More information

CIVL222 STRENGTH OF MATERIALS. Chapter 6. Torsion

CIVL222 STRENGTH OF MATERIALS. Chapter 6. Torsion CIVL222 STRENGTH OF MATERIALS Chapter 6 Torsion Definition Torque is a moment that tends to twist a member about its longitudinal axis. Slender members subjected to a twisting load are said to be in torsion.

More information

[7] Torsion. [7.1] Torsion. [7.2] Statically Indeterminate Torsion. [7] Torsion Page 1 of 21

[7] Torsion. [7.1] Torsion. [7.2] Statically Indeterminate Torsion. [7] Torsion Page 1 of 21 [7] Torsion Page 1 of 21 [7] Torsion [7.1] Torsion [7.2] Statically Indeterminate Torsion [7] Torsion Page 2 of 21 [7.1] Torsion SHEAR STRAIN DUE TO TORSION 1) A shaft with a circular cross section is

More information

STRENGTH OF MATERIALS-I. Unit-1. Simple stresses and strains

STRENGTH OF MATERIALS-I. Unit-1. Simple stresses and strains STRENGTH OF MATERIALS-I Unit-1 Simple stresses and strains 1. What is the Principle of surveying 2. Define Magnetic, True & Arbitrary Meridians. 3. Mention different types of chains 4. Differentiate between

More information

MECE 3321 MECHANICS OF SOLIDS CHAPTER 1

MECE 3321 MECHANICS OF SOLIDS CHAPTER 1 MECE 3321 MECHANICS O SOLIDS CHAPTER 1 Samantha Ramirez, MSE WHAT IS MECHANICS O MATERIALS? Rigid Bodies Statics Dynamics Mechanics Deformable Bodies Solids/Mech. Of Materials luids 1 WHAT IS MECHANICS

More information

MECHANICS OF MATERIALS REVIEW

MECHANICS OF MATERIALS REVIEW MCHANICS OF MATRIALS RVIW Notation: - normal stress (psi or Pa) - shear stress (psi or Pa) - normal strain (in/in or m/m) - shearing strain (in/in or m/m) I - area moment of inertia (in 4 or m 4 ) J -

More information

Critical Load columns buckling critical load

Critical Load columns buckling critical load Buckling of Columns Buckling of Columns Critical Load Some member may be subjected to compressive loadings, and if these members are long enough to cause the member to deflect laterally or sideway. To

More information

Mechanical Design in Optical Engineering

Mechanical Design in Optical Engineering OPTI Buckling Buckling and Stability: As we learned in the previous lectures, structures may fail in a variety of ways, depending on the materials, load and support conditions. We had two primary concerns:

More information

MECE 3321: Mechanics of Solids Chapter 6

MECE 3321: Mechanics of Solids Chapter 6 MECE 3321: Mechanics of Solids Chapter 6 Samantha Ramirez Beams Beams are long straight members that carry loads perpendicular to their longitudinal axis Beams are classified by the way they are supported

More information

March 24, Chapter 4. Deflection and Stiffness. Dr. Mohammad Suliman Abuhaiba, PE

March 24, Chapter 4. Deflection and Stiffness. Dr. Mohammad Suliman Abuhaiba, PE Chapter 4 Deflection and Stiffness 1 2 Chapter Outline Spring Rates Tension, Compression, and Torsion Deflection Due to Bending Beam Deflection Methods Beam Deflections by Superposition Strain Energy Castigliano

More information

EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 3 Torsion

EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 3 Torsion EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 3 Torsion Introduction Stress and strain in components subjected to torque T Circular Cross-section shape Material Shaft design Non-circular

More information

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV UNIT I STRESS, STRAIN DEFORMATION OF SOLIDS PART A (2 MARKS)

More information

Principal Stresses, Yielding Criteria, wall structures

Principal Stresses, Yielding Criteria, wall structures Principal Stresses, Yielding Criteria, St i thi Stresses in thin wall structures Introduction The most general state of stress at a point may be represented by 6 components, x, y, z τ xy, τ yz, τ zx normal

More information

Note: Read section (12-1) objective of this chapter (Page 532)

Note: Read section (12-1) objective of this chapter (Page 532) References: Machine Elements in Mechanical Design by Robert L. Mott, P.E. (Chapter 12 Note: Read section (12-1 objective of this chapter (Page 532 Page 1 of 29 Shaft Design Procedure (Sec. 12-2, Page 532

More information

Mechanical Properties of Materials

Mechanical Properties of Materials Mechanical Properties of Materials Strains Material Model Stresses Learning objectives Understand the qualitative and quantitative description of mechanical properties of materials. Learn the logic of

More information

MECHANICS OF SOLIDS. (For B.E. Mechanical Engineering Students) As per New Revised Syllabus of APJ Abdul Kalam Technological University

MECHANICS OF SOLIDS. (For B.E. Mechanical Engineering Students) As per New Revised Syllabus of APJ Abdul Kalam Technological University MECHANICS OF SOLIDS (For B.E. Mechanical Engineering Students) As per New Revised Syllabus of APJ Abdul Kalam Technological University Dr. S.Ramachandran, M.E., Ph.D., Mr. V.J. George, M.E., Mr. S. Kumaran,

More information

CO~RSEOUTL..INE. revisedjune 1981 by G. Frech. of..a.pqij~t(..~ttsa.fidteconol.q.gy. Sault ",Ste'...:M~ri,e.: SAUl. ir.ft\,nl~t';~l' G ". E b:.

CO~RSEOUTL..INE. revisedjune 1981 by G. Frech. of..a.pqij~t(..~ttsa.fidteconol.q.gy. Sault ,Ste'...:M~ri,e.: SAUl. ir.ft\,nl~t';~l' G . E b:. -/ 1/ /.. SAUl. ir.ft\,nl~t';~l' G ". E b:.~~~~~, of..a.pqij~t(..~ttsa.fidteconol.q.gy. Sault ",Ste'...:M~ri,e.: ',' -.\'~. ~ ;:T.., CO~RSEOUTL..INE ARCHITECTURAL ENGINEERING II ARC 200-4 revisedjune 1981

More information

Stress Transformation Equations: u = +135 (Fig. a) s x = 80 MPa s y = 0 t xy = 45 MPa. we obtain, cos u + t xy sin 2u. s x = s x + s y.

Stress Transformation Equations: u = +135 (Fig. a) s x = 80 MPa s y = 0 t xy = 45 MPa. we obtain, cos u + t xy sin 2u. s x = s x + s y. 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 9 7. Determine the normal stress and shear stress acting

More information

Review Lecture. AE1108-II: Aerospace Mechanics of Materials. Dr. Calvin Rans Dr. Sofia Teixeira De Freitas

Review Lecture. AE1108-II: Aerospace Mechanics of Materials. Dr. Calvin Rans Dr. Sofia Teixeira De Freitas Review Lecture AE1108-II: Aerospace Mechanics of Materials Dr. Calvin Rans Dr. Sofia Teixeira De Freitas Aerospace Structures & Materials Faculty of Aerospace Engineering Analysis of an Engineering System

More information

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS STATICS AND MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr, John T. DeWolf David E Mazurek \Cawect Mc / iur/» Craw SugomcT Hilt Introduction 1 1.1 What is Mechanics? 2 1.2 Fundamental

More information

ENT345 Mechanical Components Design

ENT345 Mechanical Components Design 1) LOAD AND STRESS ANALYSIS i. Principle stress ii. The maximum shear stress iii. The endurance strength of shaft. 1) Problem 3-71 A countershaft carrying two-v belt pulleys is shown in the figure. Pulley

More information

PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [ ] Introduction, Fundamentals of Statics

PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [ ] Introduction, Fundamentals of Statics Page1 PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [2910601] Introduction, Fundamentals of Statics 1. Differentiate between Scalar and Vector quantity. Write S.I.

More information

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS EDEXCEL NATIONAL CERTIICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQ LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS 1. Be able to determine the effects of loading in static engineering

More information

High Tech High Top Hat Technicians. An Introduction to Solid Mechanics. Is that supposed to bend there?

High Tech High Top Hat Technicians. An Introduction to Solid Mechanics. Is that supposed to bend there? High Tech High Top Hat Technicians An Introduction to Solid Mechanics Or Is that supposed to bend there? Why don't we fall through the floor? The power of any Spring is in the same proportion with the

More information

18.Define the term modulus of resilience. May/June Define Principal Stress. 20. Define Hydrostatic Pressure.

18.Define the term modulus of resilience. May/June Define Principal Stress. 20. Define Hydrostatic Pressure. CE6306 STREGNTH OF MATERIALS Question Bank Unit-I STRESS, STRAIN, DEFORMATION OF SOLIDS PART-A 1. Define Poison s Ratio May/June 2009 2. What is thermal stress? May/June 2009 3. Estimate the load carried

More information

Torsion of shafts with circular symmetry

Torsion of shafts with circular symmetry orsion of shafts with circular symmetry Introduction Consider a uniform bar which is subject to a torque, eg through the action of two forces F separated by distance d, hence Fd orsion is the resultant

More information

Unit III Theory of columns. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE, Sriperumbudir

Unit III Theory of columns. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE, Sriperumbudir Unit III Theory of columns 1 Unit III Theory of Columns References: Punmia B.C.,"Theory of Structures" (SMTS) Vol II, Laxmi Publishing Pvt Ltd, New Delhi 2004. Rattan.S.S., "Strength of Materials", Tata

More information

Chapter 3. Load and Stress Analysis. Lecture Slides

Chapter 3. Load and Stress Analysis. Lecture Slides Lecture Slides Chapter 3 Load and Stress Analysis 2015 by McGraw Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner.

More information

MAE 322 Machine Design. Dr. Hodge Jenkins Mercer University

MAE 322 Machine Design. Dr. Hodge Jenkins Mercer University MAE 322 Machine Design Dr. Hodge Jenkins Mercer University What is this Machine Design course really about? What you will learn: How to design machine elements 1) Design so they won t break under varying

More information

3 Hours/100 Marks Seat No.

3 Hours/100 Marks Seat No. *17304* 17304 14115 3 Hours/100 Marks Seat No. Instructions : (1) All questions are compulsory. (2) Illustrate your answers with neat sketches wherever necessary. (3) Figures to the right indicate full

More information

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS CHAPTER 2 MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek Lecture Notes: J. Walt Oler Texas Tech University Stress and Strain Axial Loading 2.1 An Introduction

More information

UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation.

UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation. UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The magnitude

More information

7. Design of pressure vessels and Transformation of plane stress Contents

7. Design of pressure vessels and Transformation of plane stress Contents 7. Design of pressure vessels and Transformation of plane stress Contents 7. Design of pressure vessels and Transformation of plane stress... 1 7.1 Introduction... 7. Design of pressure vessels... 7..1

More information

Chapter 5 Torsion STRUCTURAL MECHANICS: CE203. Notes are based on Mechanics of Materials: by R. C. Hibbeler, 7th Edition, Pearson

Chapter 5 Torsion STRUCTURAL MECHANICS: CE203. Notes are based on Mechanics of Materials: by R. C. Hibbeler, 7th Edition, Pearson STRUCTURAL MECHANICS: CE203 Chapter 5 Torsion Notes are based on Mechanics of Materials: by R. C. Hibbeler, 7th Edition, Pearson Dr B. Achour & Dr Eng. K. El-kashif Civil Engineering Department, University

More information

Advanced Structural Analysis EGF Section Properties and Bending

Advanced Structural Analysis EGF Section Properties and Bending Advanced Structural Analysis EGF316 3. Section Properties and Bending 3.1 Loads in beams When we analyse beams, we need to consider various types of loads acting on them, for example, axial forces, shear

More information

MECE 3321: MECHANICS OF SOLIDS CHAPTER 5

MECE 3321: MECHANICS OF SOLIDS CHAPTER 5 MECE 3321: MECHANICS OF SOLIDS CHAPTER 5 SAMANTHA RAMIREZ TORSION Torque A moment that tends to twist a member about its longitudinal axis 1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT Assumption If the

More information

ME 243. Mechanics of Solids

ME 243. Mechanics of Solids ME 243 Mechanics of Solids Lecture 2: Stress and Strain Ahmad Shahedi Shakil Lecturer, Dept. of Mechanical Engg, BUET E-mail: sshakil@me.buet.ac.bd, shakil6791@gmail.com Website: teacher.buet.ac.bd/sshakil

More information

Spherical Pressure Vessels

Spherical Pressure Vessels Spherical Pressure Vessels Pressure vessels are closed structures containing liquids or gases under essure. Examples include tanks, pipes, essurized cabins, etc. Shell structures : When essure vessels

More information

Comb Resonator Design (2)

Comb Resonator Design (2) Lecture 6: Comb Resonator Design () -Intro. to Mechanics of Materials Sh School of felectrical ti lengineering i and dcomputer Science, Si Seoul National University Nano/Micro Systems & Controls Laboratory

More information

PURE BENDING. If a simply supported beam carries two point loads of 10 kn as shown in the following figure, pure bending occurs at segment BC.

PURE BENDING. If a simply supported beam carries two point loads of 10 kn as shown in the following figure, pure bending occurs at segment BC. BENDING STRESS The effect of a bending moment applied to a cross-section of a beam is to induce a state of stress across that section. These stresses are known as bending stresses and they act normally

More information

Solid Mechanics Chapter 1: Tension, Compression and Shear

Solid Mechanics Chapter 1: Tension, Compression and Shear Solid Mechanics Chapter 1: Tension, Compression and Shear Dr. Imran Latif Department of Civil and Environmental Engineering College of Engineering University of Nizwa (UoN) 1 Why do we study Mechanics

More information

2012 MECHANICS OF SOLIDS

2012 MECHANICS OF SOLIDS R10 SET - 1 II B.Tech II Semester, Regular Examinations, April 2012 MECHANICS OF SOLIDS (Com. to ME, AME, MM) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~

More information

Module 5: Theories of Failure

Module 5: Theories of Failure Module 5: Theories of Failure Objectives: The objectives/outcomes of this lecture on Theories of Failure is to enable students for 1. Recognize loading on Structural Members/Machine elements and allowable

More information

Aircraft Stress Analysis and Structural Design Summary

Aircraft Stress Analysis and Structural Design Summary Aircraft Stress Analysis and Structural Design Summary 1. Trusses 1.1 Determinacy in Truss Structures 1.1.1 Introduction to determinacy A truss structure is a structure consisting of members, connected

More information

Pressure Vessels Stresses Under Combined Loads Yield Criteria for Ductile Materials and Fracture Criteria for Brittle Materials

Pressure Vessels Stresses Under Combined Loads Yield Criteria for Ductile Materials and Fracture Criteria for Brittle Materials Pressure Vessels Stresses Under Combined Loads Yield Criteria for Ductile Materials and Fracture Criteria for Brittle Materials Pressure Vessels: In the previous lectures we have discussed elements subjected

More information

December 10, PROBLEM NO points max.

December 10, PROBLEM NO points max. PROBLEM NO. 1 25 points max. PROBLEM NO. 2 25 points max. B 3A A C D A H k P L 2L Given: Consider the structure above that is made up of rod segments BC and DH, a spring of stiffness k and rigid connectors

More information