EDEXCEL NATIONAL CERTIFICATE/DIPLOMA FURTHER MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 11 - NQF LEVEL 3 OUTCOME 1 - FRAMES AND BEAMS

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1 EDEXCEL NATIONAL CERTIFICATE/DIPLOMA FURTHER MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 11 - NQF LEVEL 3 OUTCOME 1 - FRAMES AND BEAMS TUTORIAL 2 - BEAMS CONTENT Be able to determine the forces acting in pin-jointed framed structures and simply supported beams Pin-jointed framed structures: solution e.g. graphical (such as use of Bow s notation, space and force diagram), analytical (such as resolution of joints, method of sections, resolution of forces in perpendicular directions (F x = F Cosθ, F y = F Sinθ), vector addition of forces, application of conditions for static equilibrium (ΣF x = 0, ΣF y = 0, ΣM = 0 )) Forces: active forces e.g. concentrated loads; uniformly distributed loads; reactive forces e.g. support reactions, primary tensile and compressive force in structural members Simply supported beams: distribution of shear force and bending moment for a loaded beam e.g. concentrated loads, uniformly distributed load (UDL); types of beam arrangement e.g. beam without overhang, beam with overhang and point of contraflexure. It is assumed that the student has studied Mechanical Principles and Applications Unit 6 and is already familiar with equilibrium, forces and reactions, point and uniform loads but we start with a brief revision of these. D.J.Dunn 1

1. REVISION OF BEAMS AND LOADS You might be asking the question, what is a beam? Well a beam is a structure that is loaded laterally (sideways) to its length.

2 1. REVISION OF BEAMS AND LOADS You might be asking the question, what is a beam? Well a beam is a structure that is loaded laterally (sideways) to its length. These loads produce bending and bending is the most severe way of stressing a component. Suppose you were given a simple rod or a ruler and asked to break it. You would struggle to break it by stretching it or twisting it but it would be easily to break it by bending it. SIMPLY SUPPORTED BEAM A beam may be obvious when it is a girder or a length of timber like a plank but a beam can be any structure with side loads. You are only required to study simply supported beams in this unit so what is a simply supported beam? Quite simply it is a beam that rests on supports without being attached to them. We usually idealise the supports as sharp edges or rollers so that the force acting at these points can only be normal (90 o ) to the length. The forces that act at the resting points are called the reaction forces and you need to be able to calculate these first. When a beam bends, one surface is compressed and the other is stretched as shown. In addition the transverse forces produce shear forces on a given section as indicated. The purpose of this tutorial is to enable you to calculate the bending moments and shear forces in a simply supported beam. POINT LOADS A point load is a load or force that acts at a single point on a structure and it is depicted by a single arrow on diagrams. WORKED EXAMPLE No. 1 The free body diagram shown is for a simply supported beam with point loads. Calculate the reaction forces R a and R b. MOMENTS ΣM = 0 about any point so it is convenient here to use the left end at the point where R a acts. Since the reaction at that point cannot produce a moment, it is eliminated from the balance. (Remember clockwise is negative) (R b x 5) - (60 x 4) - (20 x 2) = 0 (R b x 5) = (60 x 4) + (20 x 2) R b = 280/5 = 56 kn VERTICAL FORCES ΣF y = 0 (remember up is positive) R a + R b = 0 R a + R b = 80 kn R a + 56 = 80 kn R a = = 24 kn D.J.Dunn 2

UNIFORMLY DISTRIBUTED LOADS (u.d.l.) A uniform load is one which is evenly distributed along a length such as the weight of the beam or a wall built on top of a beam.

3 UNIFORMLY DISTRIBUTED LOADS (u.d.l.) A uniform load is one which is evenly distributed along a length such as the weight of the beam or a wall built on top of a beam. It is depicted by a series of arrows as shown. We usually denote the loading as w Newtons per metre. We deal with uniform loads by replacing them with an equivalent point load. If the load extends over a length x metres then the total load is F = wx Newton. We replace the u.d.l. with a single point load at the middle a distance of x/2 metres from the end. WORKED EXAMPLE No. 2 A uniform load on a beam is shown below. It has a value of 2.5 kn per metre. Calculate the reaction forces. SOLUTION If the load is 2.5 kn for each metre of length then the total load is 4 x 2.5 = 10 kn This load is equivalent to a single point load at the middle of 10 kn. It should be apparent that in this case, the reactions must be equal to half the total load so are both 5 kn acting up. Let s do the calculation anyway. Taking moments about R b we have (10 x 2) - (R a x 4 ) = 0 4R a = 20 R a = 5 kn Balancing vertical forces R a + R b - 10 = R b - 10 = 0 R b = 5 kn WORKED EXAMPLE No. 3 A beam 4 m long rests on simple supports and carries a uniform load of 2.5 kn/m over the first 1.5 m as shown. Calculate the reaction forces. The total load is now 1.5 x 2.5 = 3.75 kn This acts at the middle of the length 0.75 m from the end. Balancing moments about R a we have (R b x 4) - (3.75 x 0.75) = 0 R b = /4 = kn Balancing vertical forces R b + R a = 3.75 R a = = 3.05 Kn D.J.Dunn 3

4 2. SHEAR FORCE DIAGRAMS The forces on a beam produce shearing at all sections along the length. The sign convention for shear force in beams is as shown. The best way is to remember this is that up on the left is positive. Consider the shear force in a section x metres from the end as shown. Only consider the forces to the left of the section. DEFINITION The shear force is the sum of all the force acting to the left of the section. Since the beam is in equilibrium, it must also be the sum of all the forces acting to the right If the beam is cut at this section as shown, a force F must be placed on the end to replace the shear force that was exerted by the material when joined. List all the forces to the left. Remember up is plus. o There is a reaction force R a up. o There is a uniform load over the length x metres and this is equivalent to a downwards load of wx Newton. o There is a point load F 1 acting down. o The total load to the left is F = R a wx F 1 If the result for F is positive (up) then it produces positive shear. We normally measure the position along the beam as the distance x form the left end or the left support. A shear force diagram is simply a graph of shear force plotted against x. This is best demonstrated with several worked examples. D.J.Dunn 4

5 WORKED EXAMPLE No. 4 Draw the shear force diagram for the simply supported beam shown. SOLUTION First work out the reaction forces R 1 and R 2. Take moments about the right end. R 1 x 2.5 = 400 x x 0.5 R 1 = 1200/2.5 = 480 kn Take moments about the left end. R 2 x 2.5 = 400 x x 2 R 2 = 1800/2.5 = 720 kn Check vertical force balance = 1200 kn up = 1200 kn down The shear force is simply the total vertical force acting to the left of a given point. There will be a sudden change from one side to the other of a point load. The SF is constant between point loads. Up is positive. At x = 0 the SF changes from 0 to 480 kn At x = 0.5 m the SF suddenly changes to = 80 kn At x = 2 m the SF changes to = -720 kn At x = 2.5 m the SF changes to = 0 The graph is shown. WORKED EXAMPLE No. 5 Draw the shear force diagram for the simply supported beam shown. SOLUTION First work out the reaction forces R 1 and R 2. The total downwards load is 40 x 5 = 200 kn so the reactions are 100 kn each since the layout is symmetrical. At x = 0, the SF changes from 0 to 100 kn it then diminishes by 40 kn/m and at x = 5 the SF is -100 kn and changes back to zero. The graph is as shown. D.J.Dunn 5

WORKED EXAMPLE No. 6 Draw the shear force diagram for the simply supported beam shown. SOLUTION It is necessary to first calculate the beam reactions. Total downwards load due the u.d.l.= w x length = (50 x 5) = 250 N This will act at the middle 2.

6 WORKED EXAMPLE No. 6 Draw the shear force diagram for the simply supported beam shown. SOLUTION It is necessary to first calculate the beam reactions. Total downwards load due the u.d.l.= w x length = (50 x 5) = 250 N This will act at the middle 2.5 from the end. Total load down = = 350 N. Balance moments about left end. (R 2 )(5) (50)(5)(5/2) (100)(1) = 0 R 2 = 145 N R 1 = = 205 N Now calculate the shear force at 1 m intervals. At x = 0, the shear force suddenly changes from zero to 205 N up F = 205 N At x = 1, F = 205 wx = x 1 = 155 N At this point the shear force suddenly changes as the 100 N acts down so there is a sudden change from 155 to 55 N. At x = 2, F = wx = x 2 = = 5 N At x = 3, F = wx = x 3 = = -45 N At x = 4, F = wx = x 4 = = -95 N At x = 5, F = wx = x 5 = = -145 N At the left end, the reaction force is 145 N up to balance the shear force of 1455 N down. The diagram looks like this. D.J.Dunn 6

7 WORKED EXAMPLE No. 7 Draw the shear force diagram for the simply supported beam with an overhang shown. SOLUTION First work out the reaction forces R 1 and R 2. Take moments about R 2. R 1 x x 1 = (50 x 4) x 2 R 1 = 200/4 = 50 kn Take moments about R 1. R 2 x 4 = 200 x 5 + (50 x 4) x 2 R 2 = 1400/4 = 350 kn Check vertical force balance Up we have = 400 kn Down we have (50 x 4) = 400 kn The SF diagram is like this. D.J.Dunn 7

8 SELF ASSESSMENT EXERCISE No A beam is loaded as shown below. Calculate the reactions and draw the shear force diagram. (Answers 310 N and 210 N) 2. A beam is loaded as shown below. Calculate the reactions and draw the shear force diagram. (Answers 600 N and 600 N) 3. A simply supported beam is loaded as shown in the diagram. Draw the bending moment diagram. (Answers 1150 N, 950 N) D.J.Dunn 8

9 3. BENDING MOMENTS FOR POINT LOADED BEAMS DEFINITION The bending moment acting at a point on a beam is the resultant turning moment due to all the forces acting to one side of the point. Normally we draw the beam horizontally with the usual conventions of up being positive (y) and left to right being positive (x). On the diagram, the point considered is x metres from the left end. It is usual to only consider the forces to the left of the point. If the forces act up then they will produce a clockwise turning moment and bend the beam up on the left. This is called SAGGING and the bending moment is positive. If the force on the left is down, the moment produced about the point is anti-clockwise and the beam is bent down. This is called HOGGING and the bending moment is negative. The reason for using the names sagging and hogging is that you might decide to consider the forces to the right of the point instead of the left. The numerical value of the bending moment will be the same but in this case the anti-clockwise moment produces sagging and the clockwise moment produces hogging. In any analysis, you must decide if the moment is sagging or hogging. Bending stress is not part of this unit but you should know that a sagging moment produces tensile stress on the bottom and compressive stress on the top. A hogging moment produces tension on the top compression on the bottom. Consider the simply supported beamshown. If the beam was cut as shown, in order to keep the left section in place, not only would a vertical force have to be applied (the shear force) to stop it moving up, but an anti-clockwise moment M would have to be applied to stop it rotating. This moment must have been previously exerted by the material and accounts for the bending stress in the material. For equilibrium the total moment must be zero at the point considered (and for any other point on the beam). In this tutorial we will solve the moment by only considering forces to the left of the point. List all the turning moments due to the forces to the left of the point. The reaction R a produces a moment of R a x and this is sagging so the bending moment at the section is plus. o The force F 1 produces a moment of F 1 (x-a) and this is hogging so it is minus. o Sum the moments and we get M = R a x - F 1 (x-a) We would get the same result if we summed the moments due to all the forces on the right. It is worth noting that the bending moment must be zero at a free end. In the case discussed, the bending moment diagram is like this. D.J.Dunn 9

10 BENDING MOMENT DIAGRAM A Bending Moment diagram is simply a graph of bending moment plotted vertically against distance x from the left end. It involves working out the bending moment at strategic points along the beam. This is best demonstrated with a series of worked examples. WORKED EXAMPLE No. 8 Construct the bending moment diagram for the beam shown. SOLUTION First it is necessary to calculate the reactions R 1 and R 2. Total downwards load = 100 N. Moments about left end. (R 2 )(5) = (100)(1) R 2 = 20 N R 1 = = 80 N Let's now work out the bending moments at key positions for the same example. x=0 M =0 and this stays constant upto x = 1 m x=1 M =(80)(1) = 80 Nm x=2 M =(80)(2) - (100)(1) = 60 Nm x=3 M =(80)(3) - (100)(2) = 40 Nm x=4 M =(80)(4) - (100)(3) = 20 Nm x=5 M =(80)(5) - (100)(4) = 0 Nm The bending moment diagram is shown. Note for this simple case we only needed to calculate the value at x = 1 m (the point load) in order to draw it since the diagrams is made from straight lines. Also note that M = 0 at a free unrestrained end. D.J.Dunn 10

11 WORKED EXAMPLE No. 9 Repeat the last example but this time a uniform load of 50 N/m is added along the entire length of the beam. Calculate the reactions. Total downwards load = (50)(5) = 350 N. Moments about left end. (R 2 )(5) = (50)(5)(5/2) + (100)(1) R 2 = 145 N R 1 = = 205 N Let's now work out the bending moments at key positions for the same example. x=0 M =0 x= 0.5 M =(205)(0.5) - (50)(0.5) 2 /2 = Nm x=1 M =(205)(1) - (50)(1) 2 /2 = 180 Nm x=2 M =(205)(2) - (50)(2) 2 /2 - (100)(1) = 210 Nm x=3 M =(205)(3) - (50)(3) 2 /2 - (100)(2) = 190 Nm x=4 M =(205)(4) - (50)(4) 2 /2 - (100)(3) = 120 Nm x=5 M =(205)(5) - (50)(5)2 /2 - (100)(4) = 0 Nm Note that the bending moments at the free ends of a simply supported beam must be zero since there are no loads to the one side of that point. If the last figure had not come out to zero, then the calculations would have to be checked thoroughly. In this example the bending moment is always positive since the beam sags along its entire length. The graph is like this. The blue line shows the smoothed curve which is judged as there is not enough points for an accurate plot. Note that the maximum bending moment occurs at about x = 2 m. To determine this precisely would require more points to be plotted or the use of max and min theory in conjunction with the equation. This is not expected of you but later on we will see how this precise point may be located from the shear force diagrams. D.J.Dunn 11

Beam problems usually involve finding the stress due to the bending moment. Bending stress is covered in tutorial 1.The next problem brings in the stress equation. WORKED EXAMPLE No.

12 Beam problems usually involve finding the stress due to the bending moment. Bending stress is covered in tutorial 1.The next problem brings in the stress equation. WORKED EXAMPLE No.10 Draw the bending moment diagram for the beam shown. SOLUTION First we must find the maximum bending moment by drawing a bending moment diagram. MOMENTS ABOUT LEFT END. (R 2 )(4) = (500)(3) + (100)(4)2/2 = 575N VERTICAL BALANCE R 1 +R 2 = (100)(4) = 900 R 1 = = 325 N Next evaluate the bending moments at 1 m intervals. x (m) M (Nm) The bending moment diagram is like this. The blue line is the smoothed curve. D.J.Dunn 12

13 SELF ASSESSMENT EXERCISE No A beam is loaded as shown below. Calculate the reactions and draw the Bending Moment diagram. Determine the maximum bending moment. (Answers 310 N. 210 N and 275 Mm) 2. A beam is loaded as shown below. Calculate the reactions and draw the bending moment diagram. Determine the greatest ending moment. (Answers 600 N, 600 N and 1188 Nm) D.J.Dunn 13

14 DETERMINING THE POSITION OF GREATEST BENDING MOMENT. It can be shown that where ever the shear force changes from positive to negative, the bending moment is a maximum and where it changes from negative to positive the bending moment is a minimum. This means that anywhere on the SF diagram where the value goes through zero, the bending moment reaches a peak. This is illustrated by combining worked example 6 and worked example 9 which contains the solutions for the same beam. The combined diagrams should be drawn as shown. Note that the maximum bending moment occurs just over 2 metres from the end at the point where the SF diagram passes through zero. On more complex beams, the SF diagram might pass through the zero value at several points resulting in several maximum and/or minimum points on the BM diagram. The bending moment that produces maximum stress could be the greatest positive value or the greatest negative value. POINT OF CONTRAFLEXURE A point of contraflexure occurs where the bending moment in a beam is zero and hence there is no bending stress. Advanced studies will also reveal that it is a point significant in the shape of the bent beam. D.J.Dunn 14

15 WORKED EXAMPLE No.11 A simply supported beam is loaded as shown in the diagram. Draw the B.M. and S.F. diagrams. Determine the point of maximum bending moment and the points of contraflexure. SOLUTION The beam is symmetrical so the reactions are equal and half the total load. R 1 = R 2 = 800/2 = 400 N SHEAR FORCE x = 0 F goes from 0 to instantly. x =1 F changes from -200 to 200 x = 2.5 F changes from 200 to x = 4 F changes from -200 to 200 x = 5 F = changes from 200 to 0 instantly. BENDING MOMENTS. Remember that we only calculate moments of the forces to the left of the point considered and x is measured from the extreme left end. x = 0 M = 0 x = 1 M = -(200)(1) = -200 Nm x = 2.5 M = -(200)(2.5) + (400)(1.5) = 100 Nm x = 4 M = -(200)(4) + (400)(3) -(400)(1.5) = -200 Nm x = 5 M = -(200)(5) + (400)(4) - (400)(2.5) +(400)(1) = 0 Nm as expected for a free end. The points of maximum bending occur at 'M' and these are at 1 m, 2.5 m and 4 m. The points of contraflexure occur at C and these are at 2 m and 3 m. D.J.Dunn 15

16 SELF ASSESSMENT EXERCISE No.3 1 Draw the SF diagram for the beam below. Determine the greatest bending moment and the position at which it occurs. 2. A light alloy tube rests horizontally on simple supports 4 m apart. A concentrated load of 500N is applied to the tube midway between the supports. Sketch diagrams of shear force and bending moments due to the applied load. Determine the maximum bending moment. (Answers 500 Nm) 3. Part of a test rig consists of a 15 m long elastic beam which is simply supported at one end and rests on a frictionless roller located 5 m from the other end, as shown. The beam has a uniformly distributed load of 150 N/m due to its own weight and is subjected to concentrated loads of 1000 N and 400 N as shown. (i) Calculate the reactions. (Answers N and N) (ii) Sketch the shear force and bending moment diagrams and determine the magnitude and location of the maximum values of the shear force and bending moment. (Answers SF N 10 m from left end, BM Nm 5 m from left end). 4. A horizontal beam 4.2 m long, resting on two simple supports 3.0 m apart, carries a uniformly distributed load of 25 kn/m between the supports with concentrated loads of 15 kn and 20 kn at the ends, as shown. Assuming the weight of the beam is negligible determine the reactions R 1 and R 2 at the supports. (Answers 51.5 kn and 58.5 N) Sketch diagrams of shear force and bending moments and indicate the point of maximum bending moment. State the greatest bending moment and shear force. (Answers kn and 18 knm) D.J.Dunn 16

ENGINEERING COUNCIL DIPLOMA LEVEL MECHANICS OF SOLIDS D209 TUTORIAL 3 - SHEAR FORCE AND BENDING MOMENTS IN BEAMS

ENGINEERING COUNCIL DIPLOMA LEVEL MECHANICS OF SOLIDS D209 TUTORIAL 3 - SHEAR FORCE AND BENDING MOMENTS IN BEAMS You should judge your progress by completing the self assessment exercises. On completion