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3 Beams3.1 TlCebeam -1
barn A bemn h a member subject to bending. Thisarismfrom foras
transmaeto themember, andapplied moments (Flgum4). Beams arewry
oommonin mginwrinp Someexamples arc door and window lintels,
crankshafts,ships in rough near, bookshelva, floorboards, car bodia
and so on(Figure 5). Note that some of these objcds may not look
like beamsi n t h e ~ t i o d a c n s c ,b u t t h y c a n b e m o
d d l c d a s b e a m s b e a ~ ~ ~ o fthe way8 in which they am
supported and loaded
A beam can be supported in several ways (Figun6).The simple
supportgives reactions along two aner, but no moment If the beam is
fixed, ie.clamped,or solidly cast into say a wall, then there can
also be a momentd o n . This 6 x 4 upport is sometimesd e d
encaatrC. If the end of ab e a m h a s n o ~ o ~ t h e n n e t u r
a l l y i t n c a l k d a ' f n e c n d ' .
simple
SA0 bCan you thinkof any other sort of support reaction? If m,
o.dsacrbc it.
A beam 6x4 t only one end is called a cmuileocl.(Rgure7a). To
locatethe beam against movement in one plane rcquins that it can
neithertranslatenormtate, sotheend mustk rulyflxed(-M). If
thebeamisalso carrying longitudid loads, is . actingalso as a tie
or strut, tbm dueaccount must be taken of these & a8 well
Often, h o w , he tomesalong the beamam notsignilicant, so for this
nectionI am only concanedwith the transverse reactions andmoments.
In thisUnit weam d 4 h g onlywith parallcl forcm acting in one
plam, and moments about axespcpndicular to thisplane. In Figure 7
thisplane is the plane of thepage.How large a fora can a cantilever
aupport at the end,or hall-way slowHow muoh does it bend? Toanswer
thestquestions it isnammry to
lookatthestrep,andstraininsidethebeam.Thiswintakc~~ometime.soIwintcllyou
in ad- that it iswortb the &of4 kcaoscthe raultswinmaMcyou to
deal with all sorts of bending problems, and it a h hrows a
gooddeal of light on the behaviour of memberam w m p d o m .
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3.2 Strengfh of beamsI want to find the forces acting inside the
beam a t a distance X from the freeend. Naturally this requires a
free body exposing the unknowns. Twopossible forces and a moment
may act a t the imaginary cu t a t point X:longitudinalP, hear Pan
d moment M (Figu re 7e). Newton's Third Lawsays that there must be
equal and opposite pairs. My sign convention ispositive shearforce
down on the right, up on the left, clockwise moment onthe right,
anticlockwise on the left (or think of the moment arrows as
upthrough the cut), Figure 8. I shall neglect the weight of the
beam in relationto the externally applied forceF, and choose axes
parallel to the unknownforces. From equilibrium along X, P = 0.
This is always the case fo r a beamthat is not also acting as a
strut, and in future I shall ignore thiscomponent. From equilibrium
along y, F - V= 0, so V= F. Finally,taking moments about the cut (
to avoid involvingP an d V), M - x = 0,so M= Fx. It is this moment
in the beam that causes the bending, andhence it is known as the
bending moment.Note that this internal bending moment is produced
by the external loadsand reactions, be they forces or moments. How
does the beam actuallyexert this bending moment? Th e bottom edge
of the beam is in tension andthe to p edge in compression (Figure
9)- lthough as you have alreadyseen the total sideways force P s
zelo. Because the beam is much longerthan it is deep, the applied
force Fh as a lot of leverage and can cause largeinternal stresses.
You can see already that the depth of the-beam will be ofgreat
importance. The bending moment will be bigger the further we gofrom
the point of application of F, so if the beam is of uniform
cross-section we would expect failure to occur a t the section with
largest b en d igmoment. To examine the load-carrying capabilities
of beams you need tobe able to find the bending moment at all
sections along the beam, andcompare the maximum bending moment with
that which will cause thebending stresses to reach a,.Returning to
the cantilever in Figure 7, the bending moment is associatedwith a
bend in the beam. Figure 10 shows in more detail a short piece
ofthe beam around the section at X. This piece of the beam is bent
to a radiusof curvature R. Somewhere between the extreme tension a
t the bottoman d compressionat the to p must be a plane of zero
stress.This is called theneutral plane of the beam. Th e zero
stress means that it must also have zerostrain. Its length, in
Figure 10, is R9 because it is an arc of a circle. Nowlook at the
'fibre' a t distance y from the neutral plane. (By 'fibre' I am no
timplying that the material is actually fibrous, it is just a
convenient way ofrefemng to a long thin element.) The beam
cross-section,Figure 11,showsthe 'fibre' with a cross-sectional
area 6A . The beam cross-section is shownas rectangular for
convenience, but it need no t be so. Th e intersection ofthe
neutral plane and cross-section is called the neutral axis. It can
beshown that the neutral axis passes through the centre of area
(centroid) ofthe cross-section,if the loading is as described in
Section 3.1 above.
(a) What is the radius of curvature of the fibre because of the
beambending?(b) What is the length of the fibre?
p;bsitiGsignconvention for momenrs
effectofpos,tfvs tnternalmoment
effectof negative nternalmomentFigure 8
Figure 9
Figure I0
Figure J J
When the beam was completely unstressed, the length of every
fibre in thispiece was Re. Th e stressed length of the fibre is (R
+ y)9.
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SA0 7(a) What is the extension of the fibre?(b) What is the
strain of the fibre?
Figure I2
I 0- - --
i /. , ., ...., , . . ,ltwl rein orccornme
cast iron woodFlgure I4 Bcmn sectiom (designedwith tenvile
&aces on top)
The fibre has an extension of y0 and a corresponding strain
The strain is therefore proportional to y, the distana from the
neutralplane. For fibres with negative y (above the neutral plane
in this case) thestrain is negative, implying compression, so the
mathematics has handledthat for us automatically. Figure 12 shows
how the strain varies linearlyam088 the depth of the beam, and is
zero at the neutral plane.Provided the material of the beam is
within its linear elastic ngion, wecan deduce the stress
distribution by using the stms-strain relationshipa =Ee. S i = y/R,
then a= Ey/R (Figurc 13).The distribution is linear.The material
near to the neutral axis has relatively low stress. The
bendiingmoment is mostly carried by the material near to the top
and bottom ofthe beam. Hence for the most efficient use of mats it
should beconantrated away from the neutral plane. Thecommon
mUed-steel joist(RSJ) beam section (Figurc 14)h a good example.
This takes maximumadvantage of the pmpmties of steel. You can also
see the reason whyconcrete beams ~IC d o & with steel rods.
Concrete has very poortensile propertits, but part of the beam is
in high tension. Concrete (andmasonry generally) is therefore a
poor beam matcrial. However, if anappropriate quantity of steel rod
is inserted on the tensile side, an efficientbeam will result. If
it is used the wrong way up, the performana is terrible!Cast iron
is weakex in tension than compression, although the differenceisnot
as drasticas it is with concrete. Cast iron beams for beat
performancewould therefore have extra material on the tensile side.
Wood s strongerin tension than compression, so for best
stmgth/weight perlormana thetension side will be the one with least
material Note that the beams inFigure 14am bending opposite to the
way shown in Figurc 10,and tensionoccurs at the top, compression at
the bottom.Already you can explain a good deal about beam design,
but we nccd toquantify it. Returning to the cantilever with its
radius R at the section X(Figure 10). what is the actual bending
moment? Look again at the fibre ofelemental area &A,distance y
from the neutral plane. The stress on thatfibre is
a= Ee = Ey/RThe fora is given by stms times area, so
The tension in this fibre contributes to the bending moment. The
momentabout the neutral axis is
The total bendiing moment is found by integrating over the
cross-sectionalarea:M= dM= -dAI IrThe radius of curvature R of the.
neutral plane is a constant for the section.E, the material s t i l
k ~ould vary if different materials wen used (a'compound' beam) but
I shall only consider the case of constant E. Nowthe constants E
and R can be removed from the integd, so
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What does the term Jy'dA represent? Figure 10 shows that y is
the'moment arm' of the area dA from the neutral axis, so J dA is
called thesecond moment of area ('second' because y is squared). In
Block 4, second moment o f a mDynamics, you met something similar-
he second moment of mass, I.The same symbol I is also used for
second moment of area. Should youever be in danger of confusing the
two you should use subscripts, I , andl,, but normally it should be
clear from the context which is meant.Finding 1 or a given beam
section is usually fairly easy: there are simpleformulae for simple
shapes, and more complicated ones can often belooked up in
handbooks. Table 2 shows the ones that are commonlyencountered in
beam design. Note that the sections in Table 2 are subjectto a
vertical loading plane at right angles to the neutral axis XX
passingthrough the centre of area.Table 2 Properties of c
ross-sections
All the sections shown can be analysed using the same data if
they aresubject to a ho rizontal bending plane at right angles to
the neutral axis YY(shown in the I-section diagram). Fo r the
circular sections the same dataapply; for the rectangular sections
the same formulae apply provided thewidth and depth terms are
correctly substituted, and for the I-sectionseparate formulae are
appropriate, a s shown.
Tx-i-x-k-T-C
U&/ notes:11) If or the whole section is the sum of h forthe
parts l ~ a e, or I-bsam smionl(21 1 for a 'hollow' section is the
I for the outer shape minus the Ifor the
missing pana [wehollow nmnple, hollow circle, h or I-beam
i.*ionl(31 Caution: for the rmlangls l=ab1/l2, the dimension cubed
ia the one
perpendicular to the axis
ab' ccrl. -77 -2 -
m' fl1" =2,7+ j-
b2
4
I/ V
{/V
ab - cd
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SA0 8(1) Draw the stress and strain distributions across a beam
in bending andmark the neutral axis. Which part of the section
carria most of thebending moment? Which part carries least?(2)
Explain the cross-sectional shapes of the beams in Figure 15.
0 ) (b) (0 ) (d )Figure I5
(3) Indicate for each of the beams in Figure 15 the surface
which isintended t o be in tension, if the materials are: (a)
steel, (b)wood, (c)rein for ad concrete, (d) cast iron.
Writing I for r y 2 dA in our equation M = ( E / R ) J y a A
gives us M = EI/Ror the more common form M11 = EIR. For a given
beam design andmaterial we know I and E, so this equation gives us
a re lationsh ip betweenthe bending moment M and the radius of
curvature R. I want to find,however, the maximum bending moment
that a beam can exert withoutsuffering permanent distortion .
Having found this I can work back to 6ndthe maximum safe loads the
beam can carry.You remember tha t for a fibre a t a distance y from
the neutral axis E = y/R,but a = Ee, so a= Ey/R an d a / y = E/R.
Now I have just shown you tha tM /I = E/R, so
a M EY l R
The greatest stress for the section, a,,, will occur on the
fibre withgreatest strain. This is the one furthest from the
neutral axis. This fibrecarrying stress a,. is at the distance y,,,
soa,, a M-=-a-y,, Y I
. . Y-This is the equation that tells us the maximum beam moment
because a,.must not exceed the material capabilities. This could be
the yield stress,a,,for a ductile material or the fracture stress,
a ,, for a brittle one, or someother chosen design limit. The term
I / y , frequently c r o p up and is giventhe name section modulus,
symbol Z , so
Iz = -Y, .The maximum moment is therefore given by
M = o Zwhere a is the approp riate stress limit and Z s the
section modulus. Notethat the maximum moment is proportional to the
permissible stress andthe section modulus. If the stress limit is
the yield stress, the beam couldprovide a greater moment, but only
by suffering permanent plasticdeformation, so our maximum moment is
known as the maximum elasticmoment.
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Note tnat tne neutra axu u m tne mlaale 01 a symmetnm secuon, so
it ishalf-way up a square, rectangle, circle, ellipse, round tube,
square tube, etc.Finally, take care with the units.SA 0 SWhat are
the S1units of (a) second moment of area, (b) section modulus?
Figure 16 shows a solid rod. (a) What is its second moment of
area?(b) What is its section modulus?SA0 11Figure 17 shows a hollow
tube. (a) What is its second moment of area?(b) What is its
sectionmodulus?What diameter solid rod would have(c) thesame I, (d)
the same Z?
A particular size of rectangular section beam of depth a and
width b carriesits maximum elastic moment. Sketch the stress
distribution showing a,.How is the maximum elastic moment changed
by the following separatemodifications?(Trynot to refer to
formulae. Draw the stress distributionon the new bigger beam and
think about forces and moment arms.)(a) Doubling the width. (b)
Trebling the width. (c) Doubling the depth.
(a) For the beams of the previous question, how much is the
weight in-creased ineach case? (b) Does doubling the depth change
the strengthfweightratio? If so, how?
A uniform rectangular section cantilever carries a vertical
force at the freeend which just causes the root ( 6 x 4 end) to be
at its elastic moment limit.You need to save weight hut cannot
increase the depth. Can you suggestways of removing material
safely?SA0 l5A rectangular section steel beam (E= 210GNm-=,a, = 300
M N m-2) isintended to carry a bending moment of 250 N m. The
section is 20mmdeep and 10mm wide. (a) Is this section adequate?
(h) What is the loadsafety factor? Find the safety factor if the
beam is: (c) doubled in width,(d) doubled in depth (original
width).
diametu30mmFigure 16
externaldiameter30mm
Figure 17
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Figure I9
A beam, as you saw earlier, genmlly has a shear fora in it too.
Figurc 18shows a slia out of our cantilever from Figurc 7. The
ekment at the topedge (exaggerated in size) cannot have a shear
fora acting on the frec f aaA, so bccausc of the equality of
complementary sheam, that on f aa B mustbe zero too. The element at
the bottom must also have zmo shear.However, an element away from
the e d m cm , be in shear. Detailedanalysis shows that the shear
is distributed pambolically in a rsctaagular-section k m . Hence it
is a uselul rule of thumb to member that thebending moment in a
beam is camed by forces at the top and bottom, butthe shear is
&ed by the material near the neutral plane.Generally the shear
stnsscs in a beam m an order of magnitude smallerthan the bending
stnsscs, and together with the fact that they areconcentrated away
from the bending stresses for a given cross-&on, wedo not often
consider them, a t a i n l y at the early stage ofdesigning
abeam.Bending stresses are by far the most signhicant. At the
preh&&y tages,thcrdon, we nced only to design the beam to
carry the maximum bcndingsvcsscs at thed o n whm the bending moment
is a maximum. Hence wenccd to fiod where the maximum bending moment
occurs, and what itsvalue is.
3.3 Bendlng-moment dlagramsTo decide whet& a beam is strong
enough or has a sufficiently k g csafety lactor you need to know
both the capabiities of the beam, whichyou have just covered in
step 1 to 6 above, and also the bending momentsthat would be
produad by the proposed loads, stcp 7. It is narssary toensure that
the beam is strong enough at every point. The bendingmoments in the
beam a n most conveniently represented by the bending-moment ( B M
) diagram. This is simply a graph showing the bendingmoment plotted
along the beam, for various significant points.You already know how
to h d he bending moment at any single point ofthe beam.Block 2,
Statics,gaveyou the method. The beam must be 'cut' toexpose the
unknowns, and then equilibrium is used. figure 19(a) showsour
cantilever again. To 6nd the bending moment at the point X the
beamiscut,giving the frec body. Takiing moments about an axis
through the cut,M - Fx =0 , so M =Fx. For this cantilever the
bending moment isproportional to the distana x from the load. At
the root the moment is200 N X 1m =200N m. F i 9(c) is the complete
BM diagram.Figure 2 q a ) shows a slightly more complicated
example. This time I ncedto know the support reaction at one end
(or both) which I can 6nd byequilibrium of the wholebeam, giving
Figure Wh). Choosing to measureX from the right-hand end again, for
X < 2 m the free body is Figurc 2qc).This time I have not
bothered to mark the shear force because I know thatI am going to
be taking moments about the cut:
M - R , x = O so M = R , xThe BM digram tberefon stam at M =0
when X =0 and rtaches400 N X 2 m = 800N m when X = 2 m (Figure
20d). Beyond that distana,I need a new fne body- nccd to include
the fora F of magnitude60 0 N (Figurea).ow, by dockwisc
equilibrium,
M - R , x + F ( x - 2 ) = OM = R , x - F ( x - 2 ) - R B x - F x
+ Z F
= 4 0 0 x - 6 0 0 x + l200= 1 2 0 0 - m
When x - m, M = 800 N m, agreeing with the existing graph. When
x =6 m, M =0so the rest of theBM diagram is a svaight
linedescending fromM = 800 N m at X = 2 to zmo at the support at B
Figurc 200.64
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Both the BM diagrams that I have drawn comprise straight lines.
The BMd i g r a m resulting from min t loads is always a series of
straight lineswhichcha& angle a t the f o r i application
points. The diagram can therefore beplotted by calculating the
bending moment at the points of application ofthe loads only,
plotting the values and joining them up with straight lines.This is
a marvellous simplification and saves a lo t of work. In the
lastexample a l l that is needed is to calculate the bending moment
a t the threeforces. It must be zero a t the outside ones because
there are no furtherloads to be supported. By equilibrium of the
whole beam, R, = 400N, andthen the bending moment a t C is + 800 N
m. Simply plot the three pointsand join them up. Try this method on
these SAQs.SA0 l8(a) Plot the BM diagram for the beam in Figure
21.(b) What is the maximum bending moment?(c) What should the beam
maximum elastic moment be for a bendingmoment safety factor of
five?
It is intended to use a steel beam with an elastic bending
moment capacityof 2 kN m to support an 80 kg person at F in Figure
22. (a) Would youexpect the beam to yield? (b) What bending moment
capability is neededfor a load safety factor of four? (c) How far
out can the person stand for aload safety factor of two? Figure
22
Applied point forces cause sharp turns in the BM diagram.
Appliedmoments cause abrupt changes of value equal to the applied
moment. Forexample, Figure 23 shows a cantilever with a
singleapplied moment. Fromthe free-body diagrams the bending moment
is zero to the right, an d500 N m to the left of the application
point. It remains at 500 N m up tothe mounting, which applies 500 N
m clockwise to give overallequilibrium.To obtain the correct BM
diagram in such cases we need to calculate twobending moment values
at the point of application of the moment. It isperhaps better to
think of calculating the bending moment a very smalldistance each
side, so that it is clear which one includes the appliedmoment.
LFigure 23
EramplePlot the BM diagram for the beam of Figure
24.SolullonApplying equilibrium to the free-body
diagrams:Immediately to the right of C: MC,*, = 0Immediately to the
left of C: MC,, - 00=0, so MC,&,= +200N mAtB: M,-200=0, s o M B
= + 2 1 M N mAt A: M,,-200-300x 1=0 , so M,= + 5 0 0 N mWhen
drawing the diagram, be careful to join up the M,,,, value to
theright and M , , , to the left; M,,,, joins up to M, not to M,.
Figure 24With practice it is no t strictly necessary to draw each
free-body diagram,but it does help to avoid errors, so unless you
are really confident please dothem. Note that the load force acting
a t the point being cut has zeromoment at that point.
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You now have simple rules for draw ing the BM diagram for a beam
withapplied point load f o r m an d moments: calculate the bcnding
moment ateach point load force and on each side of each moment, and
then plot thepoints and join up with straigh t lines. One more
note: you can use eitherside of the beam as thefm body- usually
choose the simpler side withfewer forces. Just bc carefulto mark
the M as up through the cut regardlessof side.BA0 l8Sketch the BM
diagrams for the beams of Figure 25. For a vertical beamuse the
sign convention positive M through the cut towards th e left.
Figure 25
SA0 1s(a) Plot the BM diagram for the beam of Figure 26.(b) What
is the maximum bending moment?(c) What is the load safety factor
against yield if the beam has a maximumelastic moment of 250 N
m?
Figure 28
SA0 l0Figure 27 shows a beam with a bending-moment capability of
32kN m.(a) Plot the BM diagram. b) What is the maximum bending
moment?(c) Will the beam collapse?
l6kN 8kN
Figure 27
So far I have only dealt with loads applied at a point -
oncentratedloads. Of course the loads that we normally draw without
hesitation as'concen trated' are d y istributed over a finite
length. For example, thefloor joists of my house are supported by
the wall with a contac t areaabout 50 mm by 100mm. When analysing
the joist a s a beam this wouldbe represented by a wncentrated
reaction force, acting at an estimatedpoint. At a n even more fund
amental level, the force is the sum of a lo t ofconcentrated in tm
to m ic forces. Th e distributed force is itself really
onlysomething of a convenient fiction.So how are we to analyse the
effect of a force which is distributed over alarge portion of the
beam? Figure 28 shows an example, a cantilever withdistributed load
of 240newtons per metre (240N m-') o n the outer half, atotal load
of 240 N m-' X 0.5 m = 120N. T h e n are several possibk waysto
analyse this. The way I shall show you h e n is to model the
distributedforce by concentrated forcca Then you can use the
methods that you havealready learned to get the BM diagram. How
valid is this? We are going tomodel the distributed force
thwretically as on e or more concentratedfo rm . Like any model of
reality it has limitations, bu t it is capable ofadequate accuracy
if used with care. Ao interesting equivalent is th ephysical
modelling of a distributed force by concentrated foras . Figure
29shows an aircraft wing.Th e forest of rods attached to the wing
are uscd toapply forces to simulate the distributed aerodynamic
forces.
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Figure 29 Wing @ a Concorde aircr@ undergoing testsReturning to
our cantilever of Figure 28, to find the bending moment inthe
region outside the distributed load, it is sufiicient to replace
the oad bya single force (at the centre ofgravity, the middle of a
uniform force ditri-bution) -Figure 28(b). However, this is not
good enough within the regionof the load. A much better estimate in
this region is obtained by dividingthe load up into several
sections, and replacing each section by a pointload. The solid line
in Figure 28(c) shows the result inside the load if manysections
are used.ExampleFigure 30 shows a simply-supported beam with a
uniformly-distributedload. Plot the BM diagram.solullonI shall
divide the distributed load into four equal parts, each of1M)ON m-'
X 0.5 m= 500N. placing each concentrated force at thecentre of the
part (Figure 30c). I can now calculate the bending momentat each of
the loads by the usual method, and plot the BM diagram(Figure 3W).
(Actually, because the beam is symmetrical I could havecalculated
just one half.) Thismodel shows amaximum bendingmomentof 500 N m.
If I had used a different number of sections, then the resultingBM
diagram would be slightly diercnt (say by 10%) but as long as
youuse several sections, not just one or two, the error is not very
critical.SA0 21Draw the BM diagram for the beam of the previous
example, modellingthe distributed load by (a) a single point load
(b) two equal point loads.
For approximate manual calculations, and certainly for this
course, usingfour sections will be accurate etlough unless you are
told otherwise.What happens if the load isnot uniform? You can
tilldivide it into equallength sections, and replaa each with an
appropriate point load of thesame value as the total of the
distributed load in that section. Now youknow how to cope with
anysort of applied or point load force.
Figure 30
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lWN 80N BON+R , - 130N R , - 1 lONFigure 33
If you end up with a large number of point loads then the
calculations canbe rather tedious by hand, but you have the method.
The more com-plicated loadiigsan ot really any more di5cult, they
just take mon careand patience. Computers a n well suited to using
this method, and allow alarge number of sections to be used.
Draw the tnabody d h p m of the beam and dstsrmine allapplied
l& and @M (forcer and momcatr).2 Replaa distributsd load8 by
rcvcral point ( u s d y lonrmtions). (The rq$oo outside
tbedistributed loadcdabs plottedusing a mingle qlacmomt paint
load.) ICalcuIutcthebmbgmoment at each point load, and uoh ideof
any a momentn4 Plot the heknown valuar, oin up with straight
linsr
810 IPFigwe 31 shows a non-uniformly distributed load on a
bridge. Calculateand sketch the four-section point load
equivalent.
Appmximatiq the 2 kN m-' distributed load with two point
loads:(a) draw the BM diagram for the beam of figure 32, part of a
bridge.(b) What is themaximum bending moment? (c) If thebeam
material is tobe mild steel, what is the minimum section modulus
for a load safety factorof four7BA0WFigure 33 shows the loads on aM
olding hi-fi equipment. (a) Plot theBM diagram. (b) If tbe &elf
iwood with a safe working stress limit of15MN m-2, what is the
minimum d o n modulus? (c) For 0.4 m width,what thickme
isneeded?810 PbPlot the BM diagram for the beam of Figure 34.
