COMPLEX STRESS TUTORIAL 5 STRAIN ENERGY

 To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video
Save this PDF as:

Size: px
Start display at page:

Transcription

1 COMPLEX STRESS TUTORIAL 5 STRAIN ENERGY This tutorial covers parts of the Engineering Council Exam D Structural Analysis and further material useful students of structural engineering. You should judge your progress by completing the self assessment exercises. These may be sent for marking or you may request copies of the solutions at a cost (see home page). On completion of this tutorial you should be able to do the following. Define strain energy due to a direct stress. Define strain energy due to pure shear stress. Define strain energy due to pure torsion. Define strain energy due to pure bending. Use strain energy to determine the deflection of simple rectangular and circular structures. Explain the affect of impact loading on the deflection of structures. Explain the affect of a suddenly applied load. Explain the theorem of Castigliano. Calculate the deflection of simple structures by applying Castigliano s theorem. It is assumed that students doing this tutorial are already familiar with the following. Basic stress and strain, The elastic properties of materials Basic bending theory. Basic torsion theory. Calculus including partial differentiation. D.J.Dunn

2 STRAIN ENERGY. INTRODUCTION When an elastic body is deformed, work is done. The energy used up is stored in the body as strain energy and it may be regained by allowing the body to relax. The best example of this is a clockwork device which stores strain energy and then gives it up. We will examine strain energy associated with the most common forms of stress encountered in structures and use it to calculate the deflection of structures. Strain energy is usually given the symbol U.. STRAIN ENERGY DUE TO DIRECT STRESS. Consider a bar of length L and cross sectional area A. If a tensile force is applied it stretches and the graph of force v extension is usually a straight line as shown. When the force reaches a value of F and corresponding extension x, the work done (W) is the area under the graph. Hence W = Fx/. (The same as the average force x extension). Figure Since the work done is the energy used up, this is now stored in the material as strain energy hence Fx/ The stress in the bar is σ = F/A hence F = σa The strain in the bar is ε = x/l hence x = εl For an elastic material up to the limit of proportionality, σ /ε = E (The modulus of elasticity) hence ε = σ /E Substituting we find σaεl/ = σ AL/E The volume of the bar is A L so (σ /E ) x volume of the bar WORKED EXAMPLE No. A steel rod has a square cross section mm x mm and a length of m. Calculate the strain energy when a stress of 4 MPa is produced by stretching it. Take E = GPa SOLUTION A = x = mm or x -6 m. V = AL = x -6 x = x -6 m. σ = 4 x 6 N/m and E = x 9 N/m σ x Volume = E x x 6 ( 4 x ) 6 xx = 8 Joules 9 D.J.Dunn

3 . STRAIN ENERGY DUE TO PURE SHEAR STRESS Consider a rectangular element subjected to pure shear so that it deforms as shown. The height is h and plan area A. It is distorted a distance x by a shear force F. The graph of Force plotted against x is normally a straight line so long as the material remains elastic. The work done is the area under the F - x graph so W = Fx/ Figure The work done is the strain energy stored hence Fx/ The shear stress is τ = F/A hence F = τa The shear strain is γ = x/h hence x = γh Note that since x is very small it is the same length as an arc of radius h and angle γ. It follows that the shear strain is the angle through which the element is distorted. For an elastic material τ/γ = G (The modulus of Rigidity) hence γ= τ/g Substituting we find τaγh/ = τ Ah/G The volume of the element is A h so (τ /G ) x volume Pure shear does not often occur in structures and the numerical values are very small compared to that due to other forms of loading so it is often (but not always) ignored. WORKED EXAMPLE No. Calculate the strain energy due to the shear strain in the structure shown. Take G = 9GPa SOLUTION A = πd /4 = π x. /4 =. x - m τ = F/A = 5/. x - = N/m Volume = A h =. x - x.5 Volume =5.65 x - m (τ /G ) x volume {(56.55) /( x 9 x 9 )} x 5.65 x -.5 x - Joules Note that the structure is also subject to bending. The strain energy due to bending is covered later. Figure D.J.Dunn

4 4. STRAIN ENERGY DUE TO TORSION Consider a round bar being twisted by a torque T. A line along the length rotates through angle γ and the corresponding radial line on the face rotates angle θ. γ is the shear strain on the surface at radius R. Figure 4 The relationship between torque T and angle of twist θ is normally a straight line. The work done is the area under the torque-angle graph. For a given pair of values W = Tθ/ The strain energy stored is equal to the work done hence Tθ/ From the theory of torsion (not covered here) θ = TL/GJ G is the modulus of rigidity and J is the polar second moment of area. J = πr 4 / for a solid circle. Substitute θ = TL/GJ and we get T L/GJ Also from torsion theory T = τj/r where τ is maximum shear stress on the surface. Substituting for T we get the following. (τj/r) /GJ = τ JL/GR Substitute J = πr 4 / τ πr 4 L/4GR = τ πr L/4G The volume of the bar is AL = πr L so it follows that: (τ /4G) x volume of the bar. (τ is the maximum shear stress on the surface) WORKED EXAMPLE No. A solid bar is mm diameter and.8 m long. It is subjected to a torque of Nm. Calculate the maximum shear stress and the strain energy stored. Take G = 9GPa SOLUTION R = mm =. m L =.8 m A = πr = π x. = 4.6 x -6 m Volume of bar = AL = 4.6 x -6 x.8 = 5. x -6 m J = πr 4 / = π (.) 4 / = 5.7 x -9 m 4 τ = TR/J = x./5.7 x -9 = 9. x 6 N/m (τ /4G) x volume of the bar = {(9. x 6 ) /(4 x 9 x 9 )} x 5. x Joules D.J.Dunn 4

5 WORKED EXAMPLE No.4 A helical spring is constructed by taking a wire of diameter d and length L and coiling it into a helix of mean diameter D with n coils. Show that the stiffness of the helical spring shown below is given by the formula F/y = Gd/8nD SOLUTION Figure 5 When a force F is applied to the end it deflects down by a distance y. Looking at the bottom coil, it can be seen that a torque T = FD/ is twisting the cross section of the wire. This torsion is transmitted throughout the entire length of the wire. Starting with the strain energy due to torsion we have: (τ /4G) x volume of the bar And substituting V = AL and τ = Td/J Td J 4G πd Since J = T d x AL = 6GJ 4 this reduces to U T d x AL = 6GJ T = x L GJ πd x 4 T πd x L = 64GJ 4 x L The work done by a force F is ½ Fy. Equating to U we get: Fy F y 4 4 F Gπd = y 8(nπD)D T F (D/) F D = x L = x L = GJ GJ 8GJ 4 πd 4G 4 4GJ Gd = = = LD LD 8LD and substitute L = nπd x L F Gd = y 8nD This is the well known equation for the stiffness of a helical spring and the same formula may be derived by other methods. D.J.Dunn 5

6 5. STRAIN ENERGY DUE TO BENDING. The strain energy produced by bending is usually large in comparison to the other forms. When a beam bends, layers on one side of the neutral axis are stretched and on the other side they are compressed. In both cases, this represents stored strain energy. Consider a point on a beam where the bending moment is M. Figure 6 Now consider an elementary layer within the material of length x and thickness dy at distance y from the neutral axis. The cross sectional area of the strip is da. The bending stress is zero on the neutral axis and increases with distance y. This is tensile on one side and compressive on the other. If the beam has a uniform section the stress distribution is as shown. Figure 7 Each elementary layer has a direct stress (σ) on it and the strain energy stored has been shown to be (σ/e ) x volume (in section ) The volume of the strip is x da The strain energy in the strip is part of the total so du = (σ/e ) x da From bending theory (not covered here) we have σ = My/I where I is the second moment of area. D.J.Dunn 6

7 Substituting for σ we get (My/I) du = x da and in the limit as x dx E (My/I) M du = dx da = E EI du = {(My/I)/E } x da dx y The strain energy stored in an element of length dx is then M u = dx y da and by definition I = y EI da da so this simplifies to M u = dx EI In order to solve the strain energy stored in a finite length, we must integrate with respect to x. For a length of beam the total strain energy is M dx EI The problem however, is that M varies with x and M as a function of x has to be substituted. WORKED EXAMPLE No.5 Determine the strain energy in the cantilever beam shown. The flexural stiffness EI is knm. Figure 8 SOLUTION This is a bending problem so M dx EI The beam is a simple cantilever so the bending moment at any distance x from the end is simply M = -8 x (The minus sign for hogging makes no difference since it will be squared) EI 4 M dx = 64 x x 64 x x EI 4 (-8x) dx = 64 x dx = x x 4 = 4. Joules 4 5 EI 4 4 x 64 x dx The example was a simple one because the bending moment was easily integrated. D.J.Dunn 7

8 6. DEFLECTION The deflection of simple structures may be found by equating the strain energy to the work done. This is covered in detail later but for the simple cantilever beam it can be demonstrated easily as follows. WORKED EXAMPLE No.6 Calculate the deflection for the cantilever beam in W.E. No.4. SOLUTION The deflection of the beam y is directly proportional to the force F so the work done by the force is W = Fy/ (the aea under the F y graph). Equate the strain energy to the work done and Fy/ = 4. y = 4. x /F y = 4. x /8 =.85 m We can check the answer with the standard formula for the deflection of a cantilever (covered in the beams tutorials). FL 8 x 4 y = = EI x x =.85 m The most severe forms of stress are bending and torsion. If bending or torsion occurs in a structure, they will normally be much larger than that due to direct stress or shear and these are usually neglected D.J.Dunn 8

9 SELF ASSESSMENT EXERCISE No.. A metal rod is 5 mm diameter and.5 m long. It is stretched with a force of N. Calculate the stress level and the strain energy stored in it assuming it has not reached the limit of proportionality. Take E = 8 GPa (Answers 6.98 MPa and. Joules. A bar of metal is mm diameter and.5 m long. It has a shear stress of 4 MPa throughout its entire length. Calculate the strain energy stored. (Answer.4 J). A hollow bar is 6 mm outer diameter, 4 mm inner diameter and.6 m long. It is subjected to a torque of 5 Nm. Calculate the maximum shear stress and the strain energy stored. Take G = 9GPa (Answers 4.7 MPa and.565 J) 4. A cantilever beam is 4 m long and has a point load of KN at the free end. The flexural stiffness is MNm. Determine the strain energy stored and the deflection. (Answers 55.5 J and 7. mm) D.J.Dunn 9

10 7. HARDER BEAM PROBLEMS When the bending moment function is more complex, integrating becomes more difficult and a maths package is advisable for solving them outside of an examination. In an examination you will need to do it the hard way. For example, the bending moment function changes at every load on a simply supported beam so it should be divided up into sections and the strain energy solved for each section. The next example is typical of a solvable problem. WORKED EXAMPLE No.7 Calculate the strain energy in the beam shown and determine the deflection under the load. The flexural stiffness is 5 MNm. SOLUTION Figure 9 First calculate the reactions by taking moments about the ends. RB x 4 = 5 x RB = 7.5 kn RA x 4 = 5 x RA =.5 kn Check that they add up to 5 kn. The bending moment equation is different for section AB and section BC so the solution must be done in parts. The origin for x is the left end. First section AB M = RA x = 5 x EI M dx = (5) x 5 x ( 5) x 5 x 6 6 EI (5x) dx ( 5) x dx = x 5 x = 8.5 Joules 6 x D.J.Dunn

11 Next solve for section BC. To make this easier, let the origin for x be the right hand end. M = RB x = 7 5 x EI M dx = (75) x 5 x ( 75) x 5 x 6 6 EI (75x) dx ( 75) x dx = x 5 x = 9.75 Joules Figure 6 x The total strain energy is 7.5 J The work done by the application of the load is Fy/ = 5 y/ Equating y =.5 m or.5 mm. SELF ASSESSMENT EXERCISE No. Determine the strain energy and for the beam shown and determine the deflection under the load. The flexural stiffness is knm. (Answers. J and 8.4 mm) Figure D.J.Dunn

12 Now consider a problem involving torsion and bending. The force F will produce a torque on the bar and so F will deflect due to Torsion. In addition, F will deflect because both the bar and lever bends. Remember that distances are usually best measured from the free ends and that this can be changed at will. I order that we may fin the deflection at the force, we must evaluate all strain energies in terms of F. WORKED EXAMPLE No.8 The diagram shows a torsion bar held rigidly at one end and with a lever arm on the other end. Solve the strain energy in the system and determine the deflection at the end of the lever arm. The force is 5 N applied vertically. The following are the relevant stiffnesses. Lever EI = 5 Nm. Bar EI = 6 knm. Bar GJ = 5 knm. SOLUTION The stresses to be considered are LEVER Make the origin for x as shown. The bending moment is M = Fx Figure Bending in the lever. Bending in the bar. Torsion in the bar.. F x 5 x x Figure F F x M dx ( Fx) dx x dx EI = EI = EI = EI.. -9 [. ] = 66.7 x F (numeric value 6.67 J). D.J.Dunn

13 BAR Viewed as shown we can see that the force F acts at the end of the bar as it is transmitted all along the length of the lever to the bar..4 F x 6 x x Figure 4 F F x M dx ( Fx) dx x dx EI = EI = EI = EI TORSION OF BAR [.4 ] = 77.7 x F (numeric value 4.44 J).4 The torque in the bar is T = F x. Figure 5 For torsion T L.4F = GJ GJ.4F x.4 = = 6 x x 5 9 F (numeric value 4 J) The total strain energy is then (66.7F F + 6 F ) x x -9 F The work done is Fy/ so equating y = x 6.5 F x -7 y =. x 5 x -7 =.65 m or 6.5 mm D.J.Dunn

14 WORKED EXAMPLE No.9 Solve the deflection of the curved structure shown. The radius is. m and the force is N. Take EI = 5 Nm. SOLUTION Figure 6 Consider the point shown. The horizontal distance is x. the bending moment is Fx. F M dx (Fx) dx x dx EI = EI = EI In order to solve the problem we must work in terms of angle rather than x. The curved length is Rθ and x Substitute these expressions and integrate between θ = to F EI π/ F R 6EI.67x F R {Rsin θ } x Rcos θ dθ = EI [ sin θ] 6 F π/ -6 [ ] =.67 x F = R sinθ π/ F x. π = sin sin 6 x 5 Differentiate and dx = R cosθos sin θ cos θ dθ π radians (/4 circle). If the final deflection is y, the work done is Fy/. Equate this to the strain energy and we get.67 x -6 F = Fy/ y = x.67 x -6 F = x.67 x -6 x = 6 x -6 m y =.6 mm D.J.Dunn 4

15 SELF ASSESSMENT EXERCISE No.. Solve the deflection at the handle on the cranked lever shown. The crank is 9o. The crank is made from bar mm diameter. Take E = GPa and G = 8 GPa. (Answer.4 mm) Figure 7. Calculate the deflection of the curved member shown. EI = 6 N m. (Answer.864 mm) Figure 8 D.J.Dunn 5

16 8. APPLICATION TO IMPACT LOADS When a load is suddenly applied to a structure (e.g. by dropping a weight on it), the stress and deflection resulting is larger than when a static load is applied. Consider a mass falling onto a collar at the end of a bar as shown. The bar has a length L and a cross sectional area A. The mass falls a distance z. At the moment the bar is stretched to its maximum the force in the bar is F and the extension is x. The corresponding stress is σ = F/A The strain is ε = x/l. The relationship between stress and strain is E = σ /ε hence x = σl/e The strain energy in the bar is σal/e The potential energy given up by the falling mass is P.E. = mg(z + x) Figure 9 8. SIMPLIFIED SOLUTION If the extension x is small compared to the distance z then we may say P.E. = mgz Equating the energy lost to the strain energy gained we have mgze Hence σ = AL mgz = σal/e 8. EXACT SOLUTION Equating P.E and Strain energy we have mg{z + x) = σ(al/e) Substitute x = σl/e mg{z + (σl/e)} = σ(al/e) mgz + mgσl/e= σ(al/e) Rearrange into a quadratic equation σ(al/e) - (mgl/e)σ - mgz = Solving with the quadratic equation we find σ = mgl E m σ = mg ± A mgl E ( AL ) mg a + 4ALmgz E E + mgze AL D.J.Dunn 6

17 8. SOLVING DEFLECTION The static deflection (xs) is the extension due to the weight mg when resting on the collar. ε = and σ = εe = Equate for σ L L mg xse mgl = Hence xs = A L AE Equating strain and potential energy again we have mg{z + x) = σ(al/e) Substituting σ = Ex/L σ = x F A x s = mg A Ex AL AEx mg(z + x) = = L E L mgl x = ( z + x) substitute x AE = x s E { x }(z + x) = x z + x x Rearrange into a quadratic equation x -xsx - xsz = Applying the quadratic formula 8.4 SUDDENLY APPLIED LOADS s x = xs + {xs + xs z}½ x = xs{ + ( + z/xs)½} s s s = mgl AE A suddenly applied load occur when z =. This is not the same as a static load. Putting z = yields the result: x = x s It also follows that the instantaneous stress is double the static stress. This theory also applies to loads dropped on beams where the appropriate solution for the static deflection must be used. WORKED EXAMPLE No. A mass of 5 kg is dropped from a height of. m onto a collar at the end of a bar mm diameter and.5 m long. Determine the extension and the maximum stress induced. E = 5 GPa. SOLUTION A = π x./4 = 4.59 x -6 m. xs = MgL/AE = 5 x 9.8 x.5 /(5 x 9 x 4.59 x -6) =.4x -6 x = xs + xs{ + z/xs}½ x =.4 x x -6{ + x./.4 x -6}½ x = 88.9 x - 6m σ = x E/L = 88.9 x - 6 x 5 x 9/.5 =.8 MPa D.J.Dunn 7

18 SELF ASSESSMENT EXERCISE No.4. Calculate the stress and extension produced in a wire mm diameter and 5 m long if a load of 5 g is suddenly applied. E = GPa. Show that this is twice the static values. (Answers x s =.9 σ s = 56. kpa, x =.78 mm and σ =. kpa). A rod is 5 mm diameter and.6 m long and is suspended vertically from a rigid support. A mass of kg falls vertically mm onto a collar at the end. Calculate the following. i. The static deflection (.5 mm) ii. The static stress (99.8 kpa) iii. The maximum deflection (. mm) iv. The maximum stress. (9.95 MPa) Take E = GPa. D.J.Dunn 8

19 9. CASTIGLIANO'S THEOREM Castigliano takes the work so far covered and extends it to more complex structures. This enables us to solve the deflection of structures which are subjected to several loads. Consider the structure shown. The structure has three loads applied to it. Consider the first point load. If the force was gradually increased from zero to F, the deflection would increase from zero to y and the relationship would be linear as shown. The same would be true for the other two points as well. Figure Figure The work done by each load is the area under the graph. The total work is the sum of the three and this is equal to the strain energy hence: W = ½ Fy + ½ Fy + ½ Fy... (A) Next consider that F is further increased by δf but F and F remain unchanged. The deflection at all three points will change and for simplicity let us suppose that they increase as shown by δy, δy and δy respectively. Figure D.J.Dunn 9

20 The increase in the work done and hence the strain energy δu is represented by the shaded areas (the increase in the areas) under the graphs. Note the first one is a tall rectangle with a small triangle on top and the other two are just tall rectangles. δ Fδy + δfδy/+ Fδy + Fδy The second term (the area of the small triangle) is very small and is ignored. δ Fδy + Fδy + Fδy...(B) Now suppose that the same final points were arrived at by the gradual application of all three loads as shown. Figure The work done and hence the strain energy is the area under the graphs. ½ (F + δf)(y + δy) + ½ (F)(y + δy) + ½ (F)(y + δy)...(c) The change in strain energy is found this time by subtracting (A) from (C). This may be equated to (B). This is a major piece of algebra that you might attempt yourself. Neglecting small terms and simplifying we get the simple result y =δu/δf Since this was found by keeping the other forces constant, we may express the equation in the form of partial differentiation since this is the definition of partial differentiation. y = U/ F If we repeated the process making F change and keeping F and F constant we get: y = U/ F If we repeated the process making F change and keeping F and F constant we get: y = U/ F This is Castigiano s theorem the deflection at a point load is the partial differentiation of the rain energy with respect to that load. Applying this is not so easy as you must determine the complete equation for the strain energy in the structure with all the forces left as unknowns until the end. If the deflection is required at a point where there is no load, an imaginary force is placed there and then made zero at the last stage. D.J.Dunn

21 WORKED EXAMPLE No. The diagram shows a simple frame with two loads. Determine the deflection at both. The flexural stiffness of both sections is MNm. SOLUTION Figure 4 It is important to note from the start that section AB bends and the bending moment at B turns the corner and section BC bends along its length due to both forces. Also, section BC is stretched but we will ignore this as the strain energy will be tiny compared to that produced by bending. Consider each section separately. SECTION AB Measure the moment arm x from the free end. Figure 5 M = F x (x measured from the free end). F M dx ( F x) dx x EI = = EI EI F x EI.5 x. 9 F F = x x Joules. 6.. dx D.J.Dunn

22 SECTION BC The bending moment at point B is. F. This is carried along the section BC as a constant value. The moment am x is measured from point B. The second force produces additional bending moment of F x. Both bending moments are in the same direction so they add. It is important to decide in these cases whether they add or subtract as deciding whether they are hogging (minus) or sagging (plus) is no longer relevant. M =. F + F x EI EI.5.5 x x M dx = EI {.9F x + F x +.6F F } 6.9F.5 Figure 6 (.F + F x) dx = (.F ).5 F + dx = EI EI.6F F x { + (F x ) + (.6F F x) }.9F F x x +.5F x F x FFx -9 The total strain energy is.6f F x +.5F x F x FFx x -9 F.5F x F x FF x -9.5 dx To find y carry out partial differentiation with respect to F. y =δu/δf = 7F x F x -9 Insert the values of F and F and y = 7.8 x -6 m To find y carry out partial differentiation with respect to F. y =δu/δf = +.84F x F x -9 Insert the values of F and F and y = 7 x -6 m D.J.Dunn

23 SELF ASSESSMENT EXERCISE No.5 Find the vertical deflection at the corner of the frame shown (point B). EI =.8 x 6 Nm for both sections. (Answer 5 x -9 m) Figure 7 D.J.Dunn

OUTCOME 3 - TUTORIAL 1 STRAIN ENERGY

UNIT 6: Unit code: QCF level: 5 Credit value: 5 STRENGTHS OF MATERIALS K/6/49 OUTCOME - TUTORIAL STRAIN ENERGY Be able to determine the behavioural characteristics of loaded structural members by the consideration

MECHANICS OF SOLIDS - BEAMS TUTORIAL 1 STRESSES IN BEAMS DUE TO BENDING. On completion of this tutorial you should be able to do the following.

MECHANICS OF SOLIDS - BEAMS TUTOIAL 1 STESSES IN BEAMS DUE TO BENDING This is the first tutorial on bending of beams designed for anyone wishing to study it at a fairly advanced level. You should judge

MECHANICS OF SOLIDS COMPRESSION MEMBERS TUTORIAL 1 STRUTS. On completion of this tutorial you should be able to do the following.

MECHANICS OF SOLIDS COMPRESSION MEMBERS TUTORIAL 1 STRUTS You should judge your progress by completing the self assessment exercises. On completion of this tutorial you should be able to do the following.

CH 4: Deflection and Stiffness

CH 4: Deflection and Stiffness Stress analyses are done to ensure that machine elements will not fail due to stress levels exceeding the allowable values. However, since we are dealing with deformable

MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS

MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS This is the second tutorial on bending of beams. You should judge your progress by completing the self assessment exercises.

MECHANICS OF SOLIDS - BEAMS TUTORIAL 3 THE DEFLECTION OF BEAMS

MECHANICS OF SOLIDS - BEAMS TUTORIAL THE DEECTION OF BEAMS This is the third tutorial on the bending of beams. You should judge your progress by completing the self assessment exercises. On completion

8.2 Elastic Strain Energy

Section 8. 8. Elastic Strain Energy The strain energy stored in an elastic material upon deformation is calculated below for a number of different geometries and loading conditions. These expressions for

MECHANICAL PRINCIPLES HNC/D PRELIMINARY LEVEL TUTORIAL 1 BASIC STUDIES OF STRESS AND STRAIN

MECHANICAL PRINCIPLES HNC/D PRELIMINARY LEVEL TUTORIAL 1 BASIC STUDIES O STRESS AND STRAIN This tutorial is essential for anyone studying the group of tutorials on beams. Essential pre-requisite knowledge

Stress and Deformation Analysis. Representing Stresses on a Stress Element. Representing Stresses on a Stress Element con t

Stress and Deformation Analysis Material in this lecture was taken from chapter 3 of Representing Stresses on a Stress Element One main goals of stress analysis is to determine the point within a load-carrying

ENGINEERING COUNCIL CERTIFICATE LEVEL

ENGINEERING COUNCIL CERTIICATE LEVEL ENGINEERING SCIENCE C103 TUTORIAL - BASIC STUDIES O STRESS AND STRAIN You should judge your progress by completing the self assessment exercises. These may be sent

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. UNIT I STRESS STRAIN DEFORMATION OF SOLIDS PART- A (2 Marks)

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK SUB CODE/NAME: CE1259 STRENGTH OF MATERIALS YEAR/SEM: II / IV 1. What is Hooke s Law? 2. What are the Elastic Constants?

ENGINEERING COUNCIL DIPLOMA LEVEL MECHANICS OF SOLIDS D209 TUTORIAL 10 - TORSION

ENGINEERING COUNCI IPOMA EVE MECHANICS OF SOIS 20 TUTORIA 10 - TORSION You should judge your progress by completing the self assessment exercises. On completion of this tutorial you should be able to do

ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P

ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P This material is duplicated in the Mechanical Principles module H2 and those

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

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA FURTHER MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 11 - NQF LEVEL 3 OUTCOME 2 - STRESS AND STRAIN

EDEXCEL NATIONAL CERTIICATE/DIPLOMA URTHER MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 11 - NQ LEVEL 3 OUTCOME - STRESS AND STRAIN TUTORIAL 1 - SHEAR CONTENT Be able to determine the stress in structural

MATHEMATICS FOR ENGINEERING INTEGRATION TUTORIAL 1 BASIC INTEGRATION

MATHEMATICS FOR ENGINEERING INTEGRATION TUTORIAL 1 ASIC INTEGRATION This tutorial is essential pre-requisite material for anyone studying mechanical engineering. This tutorial uses the principle of learning

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA ADVANCED MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 18 NQF LEVEL 3

EEXCE NATIONA CETIFICATE/IPOMA AVANCE MECHANICA PINCIPES AN APPICATIONS UNIT 18 NQF EVE OUTCOME 2 BE ABE TO ETEMINE THE STESS UE TO BENING IN BEAMS AN TOSION IN POWE TANSMISSION SHAFTS TUTOIA - SHEA STESS

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME 2 ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS

ENGINEERING COMPONENTS EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS Structural members: struts and ties; direct stress and strain,

Problem P5.2: A 1 Mg container hangs from a 15 mm diameter steel cable. What is the stress in the cable?

Problem P5.: A 1 Mg container hangs from a 15 mm diameter steel cable. What is the stress in the cable? Find the cross sectional area in terms of diameter using Equation (5.1). Calculate the tensile stress

Mechanics of Materials Stress-Strain Curve for Mild Steel

Stress-Strain Curve for Mild Steel 13-1 Definitions 13-2a Hooke s Law Shear Modulus: Stress: Strain: Poisson s Ratio: Normal stress or strain = " to the surface Shear stress = to the surface Definitions

MECHANICS OF SOLIDS - BEAMS TUTORIAL TUTORIAL 4 - COMPLEMENTARY SHEAR STRESS

MECHANICS OF SOLIDS - BEAMS TUTORIAL TUTORIAL 4 - COMPLEMENTARY SHEAR STRESS This the fourth and final tutorial on bending of beams. You should judge our progress b completing the self assessment exercises.

ENGINEERING COUNCIL CERTIFICATE LEVEL ENGINEERING SCIENCE C103 TUTORIAL 3 - TORSION

ENGINEEING COUNCI CETIFICATE EVE ENGINEEING SCIENCE C10 TUTOIA - TOSION You should judge your progress by completing the self assessment exercises. These may be sent for marking or you may request copies

Chapter 12 Elasticity

If I have seen further than other men, it is because I stood on the shoulders of giants. Isaac Newton 12.1 The Atomic Nature of Elasticity Elasticity is that property of a body by which it experiences

OUTCOME 1 - TUTORIAL 1 COMPLEX STRESS AND STRAIN

UNIT 6: STRNGTHS OF MATRIALS Unit code: K/60/409 QCF level: 5 Credit value: 5 OUTCOM - TUTORIAL COMPLX STRSS AND STRAIN. Be able to determine the behavioural characteristics of engineering components subjected

COMPLEX STRESS TUTORIAL 3 COMPLEX STRESS AND STRAIN

COMPLX STRSS TUTORIAL COMPLX STRSS AND STRAIN This tutorial is not part of the decel unit mechanical Principles but covers elements of the following sllabi. o Parts of the ngineering Council eam subject

Structures and Stiffness

Structures and Stiffness ENGR 10 Introduction to Engineering Ken Youssefi/Thalia Anagnos Engineering 10, SJSU 1 Wind Turbine Structure The Goal The support structure should be optimized for weight and

Week 8 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

Double integrals. Notice: this material must not be used as a substitute for attending the lectures

ouble integrals Notice: this material must not be used as a substitute for attending the lectures . What is a double integral? Recall that a single integral is something of the form b a f(x) A double integral

CHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS

CHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS Concepts of Stress and Strain 6.1 Using mechanics of materials principles (i.e., equations of mechanical equilibrium applied to a free-body diagram),

SAMPLE OF THE STUDY MATERIAL PART OF CHAPTER 1 STRESS AND STRAIN

SAMPLE OF THE STUDY MATERIAL PART OF CHAPTER 1 STRESS AND STRAIN 1.1 Stress & Strain Stress is the internal resistance offered by the body per unit area. Stress is represented as force per unit area. Typical

STRENGTH OF MATERIALS - II

STRENGTH OF MATERIALS - II DEPARTMENT OF CIVIL ENGINEERING INSTITUTE OF AERONAUTICAL ENGINEERING UNIT 1 TORSION TORSION OF HOLLOW SHAFTS: From the torsion of solid shafts of circular x section,

MECHANICS OF MATERIALS Plastic Deformations of Members With a Single Plane of Symmetry

Plastic Deformations of Members With a Single Plane of Smmetr Full plastic deformation of a beam with onl a vertical plane of smmetr. The neutral axis cannot be assumed to pass through the section centroid.

MECHANICS OF MATERIALS

2009 The McGraw-Hill Companies, Inc. All rights reserved. Fifth SI Edition CHAPTER 4 MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek Pure Bending Lecture

There are three principle ways a load can be applied:

MATERIALS SCIENCE Concepts of Stress and Strains Stress-strain test is used to determine the mechanical behavior by applying a static load uniformly over a cross section or a surface of a member. The test

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

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 1 NON-CONCURRENT COPLANAR FORCE SYSTEMS 1. Be able to determine the effects

Chapter 3 THE STATIC ASPECT OF SOLICITATION

Chapter 3 THE STATIC ASPECT OF SOLICITATION 3.1. ACTIONS Construction elements interact between them and with the environment. The consequence of this interaction defines the system of actions that subject

Laboratory Weeks 9 10 Theory of Pure Elastic Bending

Laboratory Weeks 9 10 Theory of Pure Elastic Bending Objective To show the use of the Sagital method for finding the Radius of Curvature of a beam, to prove the theory of bending, and find the elastic

SAMPLE OF THE STUDY MATERIAL PART OF CHAPTER 1 STRESS AND STRAIN

1.1 Stress & Strain SAMPLE OF THE STUDY MATERIAL PART OF CHAPTER 1 STRESS AND STRAIN Stress is the internal resistance offered by the body per unit area. Stress is represented as force per unit area. Typical

P4 Stress and Strain Dr. A.B. Zavatsky MT07 Lecture 4 Stresses on Inclined Sections

4 Stress and Strain Dr. A.B. Zavatsky MT07 Lecture 4 Stresses on Inclined Sections Shear stress and shear strain. Equality of shear stresses on perpendicular planes. Hooke s law in shear. Normal and shear

Stresses in Beam (Basic Topics)

Chapter 5 Stresses in Beam (Basic Topics) 5.1 Introduction Beam : loads acting transversely to the longitudinal axis the loads create shear forces and bending moments, stresses and strains due to V and

R A = R B = = 3.6 kn. ΣF y = 3.6 V = 0 V = 3.6 kn. A similar calculation for any section through the beam at 3.7 < x < 7.

ENDNG STRESSES & SHER STRESSES N EMS (SSGNMENT SOLUTONS) Question 1: 89 mm 3 mm Parallam beam has a length of 7.4 m and supports a concentrated load of 7.2 kn, as illustrated below. Draw shear force and

Worked Examples of mathematics used in Civil Engineering

Worked Examples of mathematics used in Civil Engineering Worked Example 1: Stage 1 Engineering Surveying (CIV_1010) Tutorial - Transition curves and vertical curves. Worked Example 1 draws from CCEA Advanced

STRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION

Chapter 11 STRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION Figure 11.1: In Chapter10, the equilibrium, kinematic and constitutive equations for a general three-dimensional solid deformable

Mechanics of Materials Qualifying Exam Study Material

Mechanics of Materials Qualifying Exam Study Material The candidate is expected to have a thorough understanding of mechanics of materials topics. These topics are listed below for clarification. Not all

Chapter 4 Strain and Material Relations

CIVL 222 STRENGTH OF MATERIALS Chapter 4 Strain and Material Relations Terminology 1. Displacement 2. Deformation 3. Strain 4. Average Axial Strain 5. Shearing Strain 6. Poisson s Ratio 7. Mechanical Properties

Section 16: Neutral Axis and Parallel Axis Theorem 16-1

Section 16: Neutral Axis and Parallel Axis Theorem 16-1 Geometry of deformation We will consider the deformation of an ideal, isotropic prismatic beam the cross section is symmetric about y-axis All parts

Strength of Materials Prof: S.K.Bhattacharya Dept of Civil Engineering, IIT, Kharagpur Lecture no 29 Stresses in Beams- IV

Strength of Materials Prof: S.K.Bhattacharya Dept of Civil Engineering, IIT, Kharagpur Lecture no 29 Stresses in Beams- IV Welcome to the fourth lesson of the sixth module on Stresses in Beams part 4.

Fy = P sin 50 + F cos = 0 Solving the two simultaneous equations for P and F,

ENGR0135 - Statics and Mechanics of Materials 1 (161) Homework # Solution Set 1. Summing forces in the y-direction allows one to determine the magnitude of F : Fy 1000 sin 60 800 sin 37 F sin 45 0 F 543.8689

SOLID MECHANICS DYNAMICS TUTORIAL NATURAL VIBRATIONS ONE DEGREE OF FREEDOM

SOLID MECHANICS DYNAMICS TUTORIAL NATURAL VIBRATIONS ONE DEGREE OF FREEDOM This work covers elements of the syllabus for the Engineering Council Exam D5 Dynamics of Mechanical Systems, C05 Mechanical and

MCE380: Measurements and Instrumentation Lab. Chapter 9: Force, Torque and Strain Measurements

MCE380: Measurements and Instrumentation Lab Chapter 9: Force, Torque and Strain Measurements Topics: Elastic Elements for Force Measurement Dynamometers and Brakes Resistance Strain Gages Holman, Ch.

Impact Load Factors for Static Analysis

Impact Load Factors for Static Analysis Often a designer has a mass, with a known velocity, hitting an object and thereby causing a suddenly applied impact load. Rather than conduct a dynamic analysis

Bending Stress in Beams

936-73-600 Bending Stress in Beams Derive a relationship for bending stress in a beam: Basic Assumptions:. Deflections are very small with respect to the depth of the beam. Plane sections before bending

THREE DIMENSIONAL ACES MODELS FOR BRIDGES

THREE DIMENSIONAL ACES MODELS FOR BRIDGES Noel Wenham, Design Engineer, Wyche Consulting Joe Wyche, Director, Wyche Consulting SYNOPSIS Plane grillage models are widely used for the design of bridges,

Strength of Materials

FE Review Strength of Materials Problem Statements Copyright 2008 C. F. Zorowski NC State E490 Mechanics of Solids 110 KN 90 KN 13.5 KN A 3 = 4.5x10-3 m 2 A 2 = 2x10-3 m 2 A 1 = 5x10-4 m 2 1. A circular

Pin jointed structures are often used because they are simple to design, relatively inexpensive to make, easy to construct, and easy to modify.

4. FORCES in PIN JOINTED STRUCTURES Pin jointed structures are often used because they are simple to design, relatively inexpensive to make, easy to construct, and easy to modify. They can be fixed structures

Torsion Testing. Objectives

Laboratory 4 Torsion Testing Objectives Students are required to understand the principles of torsion testing, practice their testing skills and interpreting the experimental results of the provided materials

CHAPTER 4 TENSILE TESTING

CHAPTER 4 TENSILE TESTING EXERCISE 28, Page 7 1. What is a tensile test? Make a sketch of a typical load/extension graph for a mild steel specimen to the point of fracture and mark on the sketch the following:

Chapter 9 Deflections of Beams

Chapter 9 Deflections of Beams 9.1 Introduction in this chapter, we describe methods for determining the equation of the deflection curve of beams and finding deflection and slope at specific points along

Welcome to the first lesson of third module which is on thin-walled pressure vessels part one which is on the application of stress and strain.

Strength of Materials Prof S. K. Bhattacharya Department of Civil Engineering Indian Institute of Technology, Kharagpur Lecture -15 Application of Stress by Strain Thin-walled Pressure Vessels - I Welcome

Remember that the information below is always provided on the formula sheet at the start of your exam paper

Maths GCSE Linear HIGHER Things to Remember Remember that the information below is always provided on the formula sheet at the start of your exam paper In addition to these formulae, you also need to learn

Figure 12 1 Short columns fail due to material failure

12 Buckling Analysis 12.1 Introduction There are two major categories leading to the sudden failure of a mechanical component: material failure and structural instability, which is often called buckling.

BENDING STRESS. Shear force and bending moment

Shear force and bending moment BENDING STESS In general, a beam can be loaded in a way that varies arbitrarily along its length. Consider the case in which we have a varying distributed load w. This will

ANALYTICAL METHODS FOR ENGINEERS

UNIT 1: Unit code: QCF Level: 4 Credit value: 15 ANALYTICAL METHODS FOR ENGINEERS A/601/1401 OUTCOME - TRIGONOMETRIC METHODS TUTORIAL 1 SINUSOIDAL FUNCTION Be able to analyse and model engineering situations

Rotational Dynamics. Luis Anchordoqui

Rotational Dynamics Angular Quantities In purely rotational motion, all points on the object move in circles around the axis of rotation ( O ). The radius of the circle is r. All points on a straight line

CHAPTER MECHANICS OF MATERIALS

CHPTER 4 Pure MECHNCS OF MTERLS Bending Pure Bending Pure Bending Other Loading Tpes Smmetric Member in Pure Bending Bending Deformations Strain Due to Bending Stress Due to Bending Beam Section Properties

Chapter 2: Load, Stress and Strain

Chapter 2: Load, Stress and Strain The careful text- books measure (Let all who build beware!) The load, the shock, the pressure Material can bear. So when the buckled girder Lets down the grinding span,

CE 1252 STRENGTH OF MATERIALS

CE 1252 STRENGTH OF MATERIALS TWO MARK QUESTION & ANSWERS Prepared by K.J.Jegidha, M.E. Lecturer, Civil UNIT : I ENERGY METHODS 1.Define: Strain Energy When an elastic body is under the action of external

Course 1 Laboratory. Second Semester. Experiment: Young s Modulus

Course 1 Laboratory Second Semester Experiment: Young s Modulus 1 Elasticity Measurements: Young Modulus Of Brass 1 Aims of the Experiment The aim of this experiment is to measure the elastic modulus with

A=b h= 83 in. 45ft. ft. = ft.2

The Easiest Way to Convert Units in an Algebraic Equation Physics professors teach you to convert everything into standard SI units, solve the problem, and hope the units come out right. In Chemistry and

Deflections. Question: What are Structural Deflections?

Question: What are Structural Deflections? Answer: The deformations or movements of a structure and its components, such as beams and trusses, from their original positions. It is as important for the

B.TECH. (AEROSPACE ENGINEERING) PROGRAMME (BTAE) Term-End Examination December, 2011 BAS-010 : MACHINE DESIGN

No. of Printed Pages : 7 BAS-01.0 B.TECH. (AEROSPACE ENGINEERING) PROGRAMME (BTAE) CV CA CV C:) O Term-End Examination December, 2011 BAS-010 : MACHINE DESIGN Time : 3 hours Maximum Marks : 70 Note : (1)

Indeterminate Analysis Force Method 1

Indeterminate Analysis Force Method 1 The force (flexibility) method expresses the relationships between displacements and forces that exist in a structure. Primary objective of the force method is to

Announcements. Moment of a Force

Announcements Test observations Units Significant figures Position vectors Moment of a Force Today s Objectives Understand and define Moment Determine moments of a force in 2-D and 3-D cases Moment of

EML 5526 FEA Project 1 Alexander, Dylan. Project 1 Finite Element Analysis and Design of a Plane Truss

Problem Statement: Project 1 Finite Element Analysis and Design of a Plane Truss The plane truss in Figure 1 is analyzed using finite element analysis (FEA) for three load cases: A) Axial load: 10,000

Mechanical Principles

Unit 4: Mechanical Principles Unit code: F/60/450 QCF level: 5 Credit value: 5 OUTCOME 3 POWER TRANSMISSION TUTORIAL BELT DRIVES 3 Power Transmission Belt drives: flat and v-section belts; limiting coefficient

DESIGN & ANALYSIS OF AUTOMOBILE CHASSIS A.HARI KUMAR, V.DEEPANJALI

DESIGN & ANALYSIS OF AUTOMOBILE CHASSIS A.HARI KUMAR, V.DEEPANJALI Abstract The objective of paper is to find out best material and most suitable cross-section for an Eicher E2 TATA Truck ladder chassis

Week #15 - Word Problems & Differential Equations Section 8.1

Week #15 - Word Problems & Differential Equations Section 8.1 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 25 by John Wiley & Sons, Inc. This material is used by

Bending Beam. Louisiana State University. Joshua Board

Bending Beam Louisiana State University Joshua Board Table of Contents: Table of Figures:... 4 Purpose... 5 Introduction... 5 Apparatus and Test Procedures... 11 Summary of Data... 14 Discussion of Results...

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

Module Analysis of Statically Indeterminate Structures by the Matrix Force Method esson 11 The Force Method of Analysis: Frames Instructional Objectives After reading this chapter the student will be able

www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x

Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity

Torsion Sign convention Right Hand Rule Turbine: Shaft: Generator:

orsion orsion is the twisting of a straight bar when it is loaded by twisting moments or torques that tend to produce rotation about the longitudinal axes of the bar. For instance, when we turn a screw

cos 2u - t xy sin 2u (Q.E.D.)

09 Solutions 46060 6/8/10 3:13 PM Page 619 010 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 9 1. Prove that

MATH SOLUTIONS TO PRACTICE FINAL EXAM. (x 2)(x + 2) (x 2)(x 3) = x + 2. x 2 x 2 5x + 6 = = 4.

MATH 55 SOLUTIONS TO PRACTICE FINAL EXAM x 2 4.Compute x 2 x 2 5x + 6. When x 2, So x 2 4 x 2 5x + 6 = (x 2)(x + 2) (x 2)(x 3) = x + 2 x 3. x 2 4 x 2 x 2 5x + 6 = 2 + 2 2 3 = 4. x 2 9 2. Compute x + sin

9.1 Rotational Kinematics: Angular Velocity and Angular Acceleration

Ch 9 Rotation 9.1 Rotational Kinematics: Angular Velocity and Angular Acceleration Q: What is angular velocity? Angular speed? What symbols are used to denote each? What units are used? Q: What is linear

EQUILIBRIUM AND ELASTICITY

Chapter 12: EQUILIBRIUM AND ELASTICITY 1 A net torque applied to a rigid object always tends to produce: A linear acceleration B rotational equilibrium C angular acceleration D rotational inertia E none

MECHANICAL PRINCIPLES HNC/D MOMENTS OF AREA. Define and calculate 1st. moments of areas. Define and calculate 2nd moments of areas.

MECHANICAL PRINCIPLES HNC/D MOMENTS OF AREA The concepts of first and second moments of area fundamental to several areas of engineering including solid mechanics and fluid mechanics. Students who are

10 Space Truss and Space Frame Analysis

10 Space Truss and Space Frame Analysis 10.1 Introduction One dimensional models can be very accurate and very cost effective in the proper applications. For example, a hollow tube may require many thousands

Chapter (1) Fluids and their Properties

Chapter (1) Fluids and their Properties Fluids (Liquids or gases) which a substance deforms continuously, or flows, when subjected to shearing forces. If a fluid is at rest, there are no shearing forces

POWER SCREWS (ACME THREAD) DESIGN There are at least three types of power screw threads: the square thread, the Acme thread, and the buttress thread. Of these, the square and buttress threads are the most

11 Vibration Analysis

11 Vibration Analysis 11.1 Introduction A spring and a mass interact with one another to form a system that resonates at their characteristic natural frequency. If energy is applied to a spring mass system,

Applications of Integration Day 1

Applications of Integration Day 1 Area Under Curves and Between Curves Example 1 Find the area under the curve y = x2 from x = 1 to x = 5. (What does it mean to take a slice?) Example 2 Find the area under

People s Physics book 3e Ch 25-1

The Big Idea: In most realistic situations forces and accelerations are not fixed quantities but vary with time or displacement. In these situations algebraic formulas cannot do better than approximate

STRENGTH OF MATERIALS LABORATORY

STRENGTH OF MATERIALS LABORATORY 1. TESTING OF MATERIALS IN TORSION OBJECTIVE To apply an increasing torque to straight cylindrical specimens of material, to observe the deformation of the specimens, and

6 1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft.

06 Solutions 46060_Part1 5/27/10 3:51 PM Page 329 6 1. Draw the shear and moment diagrams for the shaft. The bearings at and exert only vertical reactions on the shaft. 250 mm 800 mm 24 kn 6 2. Draw the

Strength of materials Lab. Manual. Production Engineering

1 Strength of materials Lab. Manual Production Engineering 2 Strength of materials lab. manual Contents S.No. Title Pg.no 1. Rockwell Hardness test 3 2. Brinell hardness test. 5 3. Impact test 8 4. Tension

Chapter 7 Bend and crush: strength-limited design

Lecture notes version 11 Oct 2011 Chapter 7 Bend and crush: strength-limited design Elastic design, avoiding plasticity, ensures that the cabin of the car does not deform in a crash. Plasicity absorbs

Lecture Slides. Chapter 10. Mechanical Springs

Lecture Slides Chapter 10 Mechanical Springs The McGraw-Hill Companies 2012 Chapter Outline Mechanical Springs Exert Force Provide flexibility Store or absorb energy Helical Spring Helical coil spring

CHAPTER 1 INTRODUCTION

CHAPTER 1 INTRODUCTION 1.1 Background of the research Beam is a main element in structural system. It is horizontal member that carries load through bending (flexure) action. Therefore, beam will deflect