CITY AND GUILDS 9210 UNIT 135 MECHANICS OF SOLIDS Level 6 TUTORIAL 5A - MOMENT DISTRIBUTION METHOD

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1 Outcome 1 The learner can: CITY AND GUIDS 910 UNIT 15 ECHANICS OF SOIDS evel 6 TUTORIA 5A - OENT DISTRIBUTION ETHOD Calculate stresses, strain and deflections in a range of components under various load conditions 1. Use ohr s Circle to determine: a. stresses on inclined planes b. combined bending torsion and axial loading.. Use and position on components strain gauge rosettes.. Use calculations and or graphic means to determine: a. shear force and bending moments in laterally loaded beams b. bending stress and shear stress distribution in beams c. deflection of beams d. solution of statically indeterminate beams e. centre of shear in beams. 4. Extend shear force, bending moment, bending stress, shear stress and deflection analysis to: a. beams of asymmetric cross section b. composite beams c. beams of elastic-perfectly plastic material. 5. Determine shear stress and twist of: a. circular solid sections b. thin walled cylinders c. simple open sections. 6. Apply Euler critical loads to determine buckling for a combination of: a. free conditions b. pinned conditions c. built in end conditions. 7. Determine limiting stress condition. 9. Apply ame equations to problem solving. 10. Employ Finite Element Analysis: a. discretisation b. types of elements c. relationship between i nodal forces ii nodal displacements iii stiffness matrix. 11. Represent examples of linear elements using springs. 1. Obtain stiffness matrix using: a. one-dimensional quadratic elements b. displacement functions c. shape functions d. principle of virtual work. 1. Determine stresses from primary unknown nodal displacements. 14. Explain the underlying assumptions and approximate nature of the results of Finite Element ethod. 15. Analyse engineering materials behaviour when loadings and service conditions: a. involve b. fatigue c. yield criteria d. fracture mechanics e. creep f. viscoelasticity. 8. Use analytical methods to determine stresses and displacements in rings, cylinders and discs under 16. Assess and select materials for axi-symmetric loading: a. internal/external pressure b. shrink fits c. rotation. applications: a. plastics b. composites c. ceramics d. modern materials. This is a stand alone tutorial on an alternative method for finding the bending moment in a beam. 1

OENT DISTRIBUTION ETHOD This is a stand-alone tutorial for students studying structures. It is about a method of finding the bending moment in beams that cannot easily be solved by other methods.

The work can be used to solve the bending moment in frames but this is not covered here.

2 OENT DISTRIBUTION ETHOD This is a stand-alone tutorial for students studying structures. It is about a method of finding the bending moment in beams that cannot easily be solved by other methods. The theory is due to the work of Professor Hardy Cross - the very same man who evolved the theory for solving pipe networks. The work can be used to solve the bending moment in frames but this is not covered here. At the heart of this method is a way to determine how a moment applied at a rigid joint (such as that shown) is distributed to the members. When the moment is applied the joint will rotate a tiny angle θ and this is the same for all the members at the joint and leads to the solution. The fraction of that is distributed to any given member is D where D is called the distribution factor. The fraction depends on the rotational stiffness of each member and this depends on:- 1. The length, material and cross section (E, I and ). The way the other end of the member is fixed. A fraction of is also carried over to the opposite end of the member and this depends on the way the end is supported. The moment carried over to the other end is C D where C is the carry over factor. Usually we use x = C y where y is the moment at the joint end and x the moment at the other end. ROTATIONA STIFFNESS Click here for the full derivation ROTATIONA STIFFNESS Relative Rotational Stiffness for each member is the ratio where is the moment in the member θ at the joint. 4EI When the other end of the member is rigidly fixed. (e.g. at end B of AB). θ EI When the other end is simply supported. (e.g. at end B of CB). θ At a free end (e.g. C on the diagram) the stiffness is zero as there is no moment = 0. At a rigid end (wall) (e.g. A on the diagram) the stiffness is infinity since θ = 0.

3 OENT DISTRIBUTION FACTOR Click here for the derivation DISTRIBUTION FACTOR The moment applied at the joint is distributed to the members in a proportion that depends on the stiffness of the member as seen from the joint. The moment distributed to each member at joint B is BA, BC and BD. At the joint the sum of all the moments is zero so it follows BA + BC + BD = The proportion of distributed to a member is called the DISTRIBUTION FACTOR which we will designate D. BA = ( D ) BA BC = ( D ) BC BD = ( D ) BD At a free end D = 1 and at a rigidly fixed end D = 0 CARRY OVER FACTOR When a portion of is distributed to a member, a moment is produced at the opposite end of the member. The proportion of at the opposite end is called the CARRY OVER FACTOR and we will define it as C. oment carried over = C x e.g. at A the moment is A = ( C ) A x B The derivation shows that when the other end is rigidly fixed C = ½ of the moment at the joint end. When the other end is simply supported C = 0 (no moment possible at a simple support or pin joint).

4 APPICATION TO BEAS The moment distribution method may be used to solve difficult problems that cannot be solved by other means because there are too many unknowns (Indeterminate beam or structure). In this tutorial we will only study beams with joints such one shown. The beam is rigidly held at the wall C and simply supported at A and B. There will be three vertical reactions due to the point load F and the uniform load w. are too many unknowns to solve by normal means. as the These We start be treating the beam as two parts AB and BC. It might be that the lengths AB and BC have a different flexural stiffness EI and this can be accommodated in this method. Next we clamp the end of each section rigidly so there is no deflection or rotation. This enables us to tackle each section separately to begin. When a joint is fixed there must be a fixing moment at each end A, B and C. Fb a Fa b It can be shown that for a single point load F A and B w And for a single uniformly distributed load w A 1 and w B 1 In this analysis, clockwise is positive and anticlockwise is negative which is consistent with the other tutorials in the series where x is measured from the left and an upwards force causes sagging which is always positive. For any other combination of loading, the principle of superposition may be used. This principle allows us to calculate the moments for each load separately and then add them together for the combined affect. Click here for the derivation FIXING OENT Since the joints are not in reality rigidly fixed, we now consider what happens if we release the fixing clamps and restore the beam to its correct condition. The beam would take up its natural position. In this example, the moment at A must be zero because it is a free end so a moment equal and opposite of A must be added and distributed at A. There will be a carry over to end B and we should take this into consideration before balancing the joint B. Joint C is rigid so we do not need to add a balancing moment and there is nothing to distribute. We calculate the imbalance at B and add an equal and opposite moment to balance it and then distribute it. This would equalise the moment at B but a fraction of the moment is carried over to the ends and this upsets the balance again so we have to redistribute the moment again and again until the moment at a joint is balanced. A worked example will explain it better. 4

5 WORED EXAPE No. 1 Find the bending moment at A, B and C for the beam shown. The EI values are the same for both sections. First treat the beam as two separate sections with rigid ends. CACUATE FIXING OENTS and STIFFNESS FACTORS. Fb a 150 x 10 x 1 x 1 EBER AB A 7.5 x 10 Fa b 150 x 10 x 1 x 1 B 7.5 x 10 Note that B and B are equal and opposite only because F is in the middle. EI EI BA 1.5EI w 15x10 x EBER CB B 11.5 x w 4EI 4EI C 11.5 x 10 BC 1.EI 1 CACUATE THE DISTRIBUTION FACTORS At end B of BA D = BA 1.5EI EI 1.EI At end B of BC D = BC 1.EI EI 1.EI Note is another way to get it. Calculate the CARRY OVER FACTOR When a balancing moment is applied at a point, ½ of the moment is carried over to the opposite end unless the opposite end is free in which case it is zero. A rigid end does not carry over any moment to the other end. If we released the joints that were fixed we have to add moments in the opposite direction in order to balance them. At the free end there should be no moment so we must add a moment of -7.5 knm. At end C there is no moment to be added and no carry over to B. We start by balancing point A. 5

6 Joint A Joint B Joint C D 0/ /1 Fixing oment (X) Balance A&C (Y) x 0 Carry over to B -7.5 x ½ Totals (X+Y) Imbalance at B -45 Balance B +45 Distribute using D Carry Over x x ½ Total At B we have achieved equal and opposite moments. The moment at A is zero, the moment at B is.445 knm and at the wall (C) 0.65 knm. This problem is in fact solvable with acaulay s method and this gives similar answers. We did not need extra correction cycles in this problem. 6

7 WORED EXAPE No. Find the moments at A, B and C for the beam below. The section AB has an EI value twice that of section BC. CACUATE FIXING OENTS and STIFFNESS FACTORS. Fb a 00 x 10 x x 1 EBER AB A x 10 Fa b 00 x 10 x 1 x B x 10 (EI) 6EI BA EI w 40x10 x 4 EBER CB B 5. x w 4EI 4EI C 5. x 10 BC EI 1 4 CACUATE THE DISTRIBUTION FACTORS At end B of BA D = BA EI EI 1EI At end B of BC D = BC 1EI 0. EI 1EI Joint A Joint B Joint C D Fixing oment (X) Balance A (Y) Carry over A to B x ½ Totalise (X+Y) Inbalance at B Balance B Distribute using D Carry Over x x ½ 5.9 Total At B we have achieved equal and opposite moments. The moment at A is zero, the moment at B is knm and at the wall (C) knm. This problem is in fact solvable with acaulay s method and this gives similar answers. 7

8 SEF ASSESSENT EXERCISE No Solve the moments at A B and C for the beam shown. Section AB has an EI value three times that of BC. Answers 0, 4.4 knm and 4.8 knm. Repeat problem 1 if section BC has an EI value three times that of AB. (Answers 0, 9.4 knm and 45. knm). Solve the bending moment at the supports for the beam shown. Answers 1.81 knm, 8.61 knm and 0 Now we need to try harder problems with more than one joint. This requires more correction cycles. 8

9 WORED EXAPE No. Calculate the bending moment in the beam at A, B, C and D. The point loads at mid span. The middle span (BC) has an EI value twice that of the other two spans. CACUATE FIXING OENTS and STIFFNESS FACTORS FOR B. Fb a 0 x 10 x.5 x.5 EBER AB A 1.5 x 10 5 Fa b 0 x 10 x.5 x x 10 5 Note that B and B are equal and opposite only because F is in the middle. EI EI BA 0.6EI 5 w 5 x10 x 5 EBER CB B x w C x (EI) 8EI BC 1.6EI 5 Note it is still rigidly fixed at C at this stage. CACUATE THE DISTRIBUTION FACTORS for B At end B of BA D = BA 0.6EI EI 1.6EI At end B of BC D = BC 1.6EI EI 0.6EI CACUATE FIXING OENTS and STIFFNESS FACTORS FOR C. EBER BC w 5 x10 x 5 B x w C x (EI) 8EI CB 1.6EI 5 4EI 4EI CD 0.8EI 5 Note both ends are rigid when C is released. 9

10 CACUATE THE DISTRIBUTION FACTORS for C At end C of BC D = CB 1.6EI EI 0.8EI At end C of CD D = CB 0.8EI EI 0.8EI EBER CD Fb a 0 x 10 x.5 x.5 C 1.5 x 10 5 Fa b 0 x 10 x.5 x.5 D 1.5 x 10 5 Note that B and B are equal and opposite only because F is in the middle. The following table shows the process of balancing and correcting until we get very close to zero difference at B and C. This shows that the moments at A, BC and D are 0, knm, knm and 1.66 knm. The table was produced using a spread sheet. A B C D D Fix om Balance A Carry over ½ x B Total B Difference Balance B Distribute B Carry Over Totals Difference Balance Distribute Carry Over ½ x ½ x Total Difference Balance Distribute Carry Over ½ x Total Difference Balance Distribute Carry Over ½ x ½ x Total Difference Balance Distribute Carry Over 0 ½ x Total Difference Balance Distribute Carry Over Total Difference Balance Distribute Carry Over Total Difference Balance Distribute Carry Over Total

11 SEF ASSESSENT EXERCISE No. Find the bending moments in the beams shown at A, B, C and D. 1. Answers 0, 45.8 kn, 59 kn and 8 kn Answers 0, 19.1 kn, 18 kn and.8 kn 11

12 FIXING OENTS Fixing moments have been covered in the tutorial on deflection of beams under the heading ENCASTRÉ BEAS. A fixing moment is the moment at the end of a beam where it is rigidly fixed to prevent any deflection or rotation. Finding them can be quite complicated depending on the loading of the beam. For those wishing to get to grips with the derivation the following is given. et the beam span be designated A to B and the moments at the ends A and B. There will also be reactions R A and R B. We shall use acaulay s method to solve the slope and deflection and start with the derivation for a single point load. POINT OAD The bending moment at distance x from the left end is d y EI RAx Fx a A dx dy x F x a Integrate EI RA dx dy The slope at both ends is zero so 0 dx Put x = 0 and A is zero. a x Fx a dy x F x EI R A Ax...(1) Integrate again EIy R A dx 6 6 Since thedeflection is zero at both ends then puttingy 0 and x 0 yields that B 0 A x A x B (The x F x a x EIy R A A...() 6 6 constants of integration A and B are always zero for an encastré beam). Equations 1 and give the slope and deflection. Before they can be solved, the fixing moment must be found by using another boundary condition. Remember the slope and deflection is zero at both ends of the beam so we have two more boundary conditions to use. A suitable condition is that y = 0 at x =. From equation this yields:- F a A EI(0) RA and note that a = b for simplicity 6 6 Fb A RA Fb 0 RA A 6 6 We still have an unknown R A so using the other boundary condition that dy/dx = 0 when x = then from equation () dy F b RA Fb EI 0 RA A and substitute A dx Fb R Fb A RA Fb RA Fb R A 0 R A 6 Fb Fb Fb Fb R A A 1

13 R A Fb Substitute into A Fb Fb Fb Fb Fb Fb A Fb Fb A Fb b Fb Fb a Fa b A 1 b and it can be shown that B UNIFORY DISTRIBUTED OADS A uniform load across the entire span will produce equal and opposite moments at either end of the beam. In this case the reactions are R A = R B = w/ The bending moment at distance x from the left end is : d y wx d y wx wx EI R Ax A EI A dx dx Integrate dy Since the slopeis zero at both ends it then putting dx dy EI dx wx 4 wx 6 0 and x 0 yields that A 0 A x A dy wx wx wx wx Ax EI Ax...(1) Integrate again EIy dx Since the deflection is zero at both ends then puttingy 0 and x 0 yields that B 0 4 wx wx Ax EIy...() 1 4 As in the other case, A and B are zero but we must find the fixing moment by using the other boundary condition of y = 0 when x = w EI(0) 1 w 0 4 A 4 w 1 w 4 A 4 4 A If we worked out B we would get B w 1 4 B 1

14 ROTATIONA STIFFNESS and CARRY OVER When a moment is applied to the end of a beam without any deflection, the end will rotate a small angle θ. The ratio /θ is the stiffness and this depends on the values of E, I and. It also depends how the other end of the beam is fixed. FIXED END Consider a section of beam with a moment at the left end and fixed rigidly at the right end. The beam will deflect as shown. The left end must not deflect so an upwards force R must be applied also. The bending moment any distance x from the end is:- d y x EI Rx dx Integrate dy Rx EI x A but at x = the slope is zero. dx R R 0 A A dy Rx R EI x dx Integrate again. x Rx R x EIy x B 6 The deflection at x = 0 must also be zero hence B = 0 The deflection at x = is zero hence:- R R 0 6 R 0 hence R The gradient at the end is the rotation θ and at x = 0 we have:- R EI θ substitute R EI θ 4 4 4EI The rotational stiffness is hence θ CARRY OVER FACTOR At end B the moment is B = - A + R B A A B The fraction ½ is called the CARRY OVER FACTOR 14

15 SIPY SUPPORTED END Consider a section of beam with a moment at the left end and fixed rigidly at the right end. The beam will deflect as shown. The reaction at the support must be equal and opposite to R. It follows that = R d y x EI Rx - x dx Integrate dy x EI xdx A Integrate again. x x EIy - Ax B 6 The deflection at x = 0 must be zero hence B = 0 x x EIy - Ax 6 The deflection at x = is zero hence:- 0 - A A A 6 A - dy x x x EI - dx 6 The gradient at the end is the rotation θ and at x = 0 we have:- EIθ - EI The rotational stiffness is hence θ CARRY OVER FACTOR The moment at the simply supported end is zero so the carry over factor is zero. 15

DISTRIBUTION FACTOR The diagram shows three members AB, DB and CB rigidly joined at B. When a moment is applied at B the joint will twist an angler θ.

16 DISTRIBUTION FACTOR The diagram shows three members AB, DB and CB rigidly joined at B. When a moment is applied at B the joint will twist an angler θ. The stiffness of each member as seen from B is θ The moment distributed to each member is BA, BC and BD. At the joint the sum of all the moments is zero so it follows BA + BC + BD = BA BC BD θ BA θ BC θ BD BAθ BCθ BDθ BA BC θ BD Hence θ BA θ and if we substitute BA BC Similarly BC and BD BA The moment is distributed to each member by the distribution factor BA = ( D ) BA BC = ( D ) BC BD = ( D ) BD BD BA D At a free end, (e.g. C) CB (D) CB 1 0 CB At a rigidly fixed end (e.g. A) AB AB (D) AB 0 16

Module 4 : Deflection of Structures Lecture 4 : Strain Energy Method

Module 4 : Deflection of Structures Lecture 4 : Strain Energy Method Objectives In this course you will learn the following Deflection by strain energy method. Evaluation of strain energy in member under