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					                   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 (Fx = F Cosθ, Fy = F Sinθ), vector addition of forces, application of
conditions for static equilibrium (ΣFx= 0, ΣFy = 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 www.freestudy.co.uk                  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. 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 (90o) 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 Ra and Rb.

     MOMENTS ΣM = 0 about any point so it is convenient here to use the left end at the point
     where Ra acts. Since the reaction at that point cannot produce a moment, it is eliminated from
     the balance. (Remember clockwise is negative)
     (Rb x 5) - (60 x 4) - (20 x 2) = 0
     (Rb x 5) = (60 x 4) + (20 x 2) Rb = 280/5 = 56 kN
                              VERTICAL FORCES ΣFy = 0 (remember up is positive)
     Ra + Rb - 20 - 60 = 0     Ra + Rb = 80 kN
     Ra + 56 = 80 kN          Ra = 80 - 56 = 24 kN


© D.J.Dunn www.freestudy.co.uk                       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. 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 Rb we have
     (10 x 2) - (Ra x 4 ) = 0          4Ra = 20                Ra = 5 kN
     Balancing vertical forces Ra + Rb - 10 = 0        5 + Rb - 10 = 0         Rb = 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 Ra we have
     (Rb x 4) - (3.75 x 0.75) = 0      Rb = 2.8125/4 = 0.703 kN

     Balancing vertical forces Rb + Ra = 3.75         Ra = 3.75 - 0.703 = 3.05 Kn




© D.J.Dunn www.freestudy.co.uk                   3
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 Ra 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 F1 acting down.
   o
The total load to the left is F = Ra – wx – F1

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 www.freestudy.co.uk                      4
     WORKED EXAMPLE No. 4

     Draw the shear force diagram for the simply
     supported beam shown.

     SOLUTION

     First work out the reaction forces R1 and R2.
     Take moments about the right end.
     R1 x 2.5 = 400 x 2 + 800 x 0.5
     R1 = 1200/2.5 = 480 kN
     Take moments about the left end.
     R2 x 2.5 = 400 x 0.5 + 800 x 2
     R2 = 1800/2.5 = 720 kN
     Check vertical force balance
     480 + 720 = 1200 kN up
     400 + 800 = 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 480 - 400 = 80 kN
     At x = 2 m the SF changes to 80 - 800 = -720 kN
     At x = 2.5 m the SF changes to -720 + 720 = 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 R1 and R2.
     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 www.freestudy.co.uk                    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
     2.5 from the end.
     Total load down = 250 + 100 = 350 N.

     Balance moments about left end.
     (R2)(5) – (50)(5)(5/2) – (100)(1) = 0

     R2= 145 N                   R1= 350 – 145 = 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 = 205 – 50 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 = 205 – 100 – wx = 105 – 50 x 2 = 105 – 100 = 5 N
     At x = 3, F = 205 – 100 – wx = 105 – 50 x 3 = 105 – 150 = -45 N
     At x = 4, F = 205 – 100 – wx = 105 – 50 x 4 = 105 – 200 = -95 N
     At x = 5, F = 205 – 100 – wx = 105 – 50 x 5 = 105 – 250 = -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 www.freestudy.co.uk                  6
     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 R1 and R2.
     Take moments about R2.
     R1 x 4 + 200 x 1 = (50 x 4) x 2 R1= 200/4 = 50 kN
     Take moments about R1.
     R2 x 4 = 200 x 5 + (50 x 4) x 2 R2= 1400/4 = 350 kN
     Check vertical force balance
     Up we have 50 + 350 = 400 kN
     Down we have (50 x 4) + 200 = 400 kN
     The SF diagram is like this.




© D.J.Dunn www.freestudy.co.uk                 7
      SELF ASSESSMENT EXERCISE No. 1

     1.   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 www.freestudy.co.uk                   8
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 Ra produces a moment of Ra x and this is sagging so the bending moment at the
section is plus.

     o The force F1 produces a moment of F1 (x-A) and this is hogging so it is minus.
     o Sum the moments and we get M = Ra x - F1 (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 www.freestudy.co.uk                    9
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 R1 and R2.
     Total downwards load = 100 N.
     Moments about left end. (R2)(5) = (100)(1)
     R2 = 20 N           R1=100 - 20 = 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 www.freestudy.co.uk                    10
     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) + 100 = 350 N.
     Moments about left end. (R2)(5) = (50)(5)(5/2) + (100)(1)
     R2 = 145 N               R1= 350 - 145 = 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 =          96.25 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 www.freestudy.co.uk                   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.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.              (R2)(4) = (500)(3) + (100)(4)2/2 = 575N

     VERTICAL BALANCE               R1+R2= 500 + (100)(4) = 900
                                    R1= 900 - 575 = 325 N
     Next evaluate the bending moments at 1 m intervals.
     x (m)       0        1       2        3        3.5    4
     M (Nm)       0       275     450      525      275    0

     The bending moment diagram is like this. The blue line is the smoothed curve.




© D.J.Dunn www.freestudy.co.uk                 12
     SELF ASSESSMENT EXERCISE No. 2

1.   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 www.freestudy.co.uk             13
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 www.freestudy.co.uk                 14
     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.
     R1 = R2 = 800/2 = 400 N
     SHEAR FORCE
     x = 0 F goes from 0 to - 200 instantly.
     x =1 F changes from -200 to 200
     x = 2.5 F changes from 200 to -200.
     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 www.freestudy.co.uk                 15
     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 962.5 N and 2687.5 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 1537.5 N 10 m from left end, BM 2 937.5 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
     R1 and R2 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 -38.5 kN and 18 kNm)




© D.J.Dunn www.freestudy.co.uk                  16

				
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