strength of materials - 2 - internal forces in statically determined members
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2. INTERNAL FORCES IN STATICALLY DETERMINATE
MEMBERS
2.1 INTERNAL FORCES COMPUTATION
Consider a body of arbitrary shape acted upon by several external loads (Fig.
2.1). In statics, we would start by determining the resultant of the applied loads to
determine whether or not the body remains at rest. If the resultant is zero, we have
static equilibrium a condition generally
prevailing in structures. If the resultant is
not zero we may apply inertia forces to
bring about dynamic equilibrium. Such
cases will be discussed later under dynamicloading. For the present, we consider only
cases involving static equilibrium.
Generally speaking, when loads are
applied to a certain mechanical structure or
machine, each component of such a
structure or machine is subjected to
external loads of different values (Fig. 2.1). Under the action of the external loads,
internal forces occur inside the involved component (assimilated to the arbitrary
body represented in Fig. 2.1). If these internal forces reach critical values the body(component) will fail.
One of the methods commonly used for the determination of internal forces in
strength of materials is known as the method of sections. In fact the problem remains
the same like that presented in the previous chapter: what does every point of the
body (generically represented in Fig. 2.1) ,,feel when the body is subjected to
external loads in mechanical equilibrium?
The method of sections consists in passing an exploratory plane through an
arbitrary point of the body and with an arbitrary orientation (Fig. 2.2). In this way
two distinct segments of the body will occur (Fig.2.3), the left surface (SL) and the
Fig. 2.1
Fig. 2.2 Fig. 2.3
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right surface (SR) representing the internal plane surfaces of the body, originally in
contact. Since the body represented in Fig. 2.2 is in equilibrium, neither of the two
segments of Fig. 2.3 can be in equilibrium. If we want to bring the segment IIfor
example in the same state it is in Fig. 2.2, the action of segment I on segment II
(which actually exists inside the body represented in Fig. 2.2) has to be considered.
This action may be reduced at the
centroid O of surface SR to a resultant
force R and a resultant moment M(Fig.
2.4).
In other words R and Mrepresent
the action of segmentIon segmentII, as
a global mechanical effect occured at the
level of the entire section SR. In fact this
mechanical effect develops inside the
body represented in Fig. 2.2.
Furthermore it is to be noted that Mand
R represent the effect of the external
loads acting on the segment I(i.e. P1, Pn, M1 Fig. 2.2) which develops inside the
body at the level ofSR.
The resultant force R and the resultant moment Mare called internal forces.
Under the action ofM,R, P2, Pk,Mk the segmentIIof the body is now in mechanical
equilibrium (as it really is in the actual state of Fig. 2.2). Now using the adequate
equilibrium equations for segment II, the values of internal forces R and Mcan be
derived.
A similar reasoning may be also applied to the segmentIof the body, on which
internal forces R and M develop (Fig. 2.5). From action and reaction mechanical
law we may write:
R = - R,
M = - M.
Let us now apply the above
reasoning to a loaded statically
determinate member (Fig. 2.6). It is to
be mentioned that a statically
determinate member is a member for
which all reactions can be completely
computed from statics alone.
After computing the specific
reactions (YA, YB, ZBetc) corresponding to the supporting points A and B, the
member represented in Fig. 2.6 is in fact a body subjected to several external loads in
mechanical equilibrium. Passing an exploratory plane at some arbitrary point of the
member, perpendicular to the axis of the member, and considering only a segment of
Fig. 2.4
Fig. 2.5
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Internal forces in statically determinate members
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the member (just like in the preceding discussion) the internal forces R and Mare
revealed (Fig. 2.7).
We do also attach to the segment
considered in Fig. 2.7 a coordinate
system whose origin is taken at the
centroid O of the exploratory crosssection (SR). Ox is the axis of the
member while Oz and Oy represent the
axes to which the exploratory cross
section of the member is reported.
For convenience, the internal
forces M and R are resolved into
components that are normal and tangent to the cross section considered, within the
chosen coordinate system (Fig. 2.8) -R resolved into componentsN, Ty and Tz, while
Minto componentsMx,Miy,Miz.
Each component reflects a certain effect of the applied loads on the member
and is given a special name as follows:
N: axial force (R component along Ox axis, or, more briefly, x axis). This
component measures the pulling (or pushing) action perpendicular to the section
considered. A pull represents a tensile force that tends to elongate the member,
whereas a push is a compressive force that tends to shorten it.
Ty, Tz: shearing forces (R components alongy axis andz axis respectively). These are
components of the total resistance to sliding the portion to one side of the
exploratory section past the other. The resultant shearing force (acting on zOy plane)
is usually denoted by T, and its components by Ty and Tz to identify their directions.
Mx: twisting couple (twisting moment or torque) (Mcomponent along x axis). This
component measures the resistance to twisting the member and is commonly given
the symbolMt.
Miy, Miz: bending moments. These components measure the resistance to bending themember about they orz axes and are often denoted merely byMiy andMiz.
Fig. 2.6
Fig. 2.7 Fig. 2.8
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The quantities N, Ty, Tz, Mx, Miy, Miz are also called internal forces. Each of
them produces a certain type of mechanical effect on the involved member:
.bending:,
;torsion:;loadingshearing:,
;loadingaxial:
iziy
t
zy
MM
MTT
N
The simultaneous presence on the current member cross section of two or
more types of internal forces determines a combined loading.
Although the type of coordinate system used within such analysis is, in a way,
controversial, we shall use the following sign convention:
N, Ty and Tz should be considered positive if orientated to the opposite senseof the axes;
Mt, Miy and Miz should be considered positive if orientated to the sense ofthe axes.
From the preceding discussion, it is obvious that the internal effect of a given
loading depends upon the selection and orientation of the exploratory section. In
particular, if the loads act in a single plane, say the xy plane as is frequently the case,
the six components of Fig. 2.8 reduce to only three namely, the axial force (N), theshearing force (T) and the bending moment Miz. This in why, in case of plane
problems (when plane members are subjected to loads contained in the same plane)
the internal forces refer to only three components whose positive sign convention
should be taken as follows:
The positive sign convention represented in Fig. 2.9 should be used for
plotting the axial forces, shearing forces and bending moments diagrams. As it will
be explained later, the positive sign convention corresponding to the face SR is used
when the member is covered from the left to the right while the positive sign
convention corresponding to the face SL is used when the member is covered fromthe right to the left.
Fig. 2.9
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2.2 DIFFERENTIAL RELATIONS AMONG LOAD, SHEAR AND
BENDING MOMENT
Let us now consider a simply supported beam AB of spanl
, carrying adistributed loadp per unit length (Fig. 2.10a), and let Cand C be two points of the
beam at an infinitely small distance dx from each other.
We shall detach the portion CC of the beam and draw its free body diagram
(Fig. 2.10b). The forces exerted on the free body include a load of magnitude pdx
and the internal forces at Cand C as shown. The shear and bending moment at C
will be denoted by T and M respectively, and will be assumed positive while the
shear and bending moment at C will be denoted by T+ dTandM+ dMrespectively.
Since the shear and bending moment are assumed to be positive, the internal forceswill be directed as shown in Fig. 2.10 b. It is also to be mentioned that since the
distance dx between C and C is considered infinitely small, the load p may be
assumed uniformly distributed per length dx and may be replaced by a resultantpdx,
(Fig. 2.11).
From the mechanical equilibrium of the detached segment CC we write:
the summation of forces about the vertical direction is zero:0dd0 =++= TxpTTFy .
a. b.
Fig. 2.10
Fig.2.11.
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We write
Txp dd = .Dividing by dx the two members of the equation, we have:
px
T=
d
d.
(2.1)
writing now that the summation of moments about C is zero, we have:0d
2
ddd0
'=++= xTM
xxpMMM
C.
The third member of this equation( )
2
d2
xpbeing an infinitely small quantity
compared with the others, it may be neglected and we may write:
xTM dd = .
Dividing now the two members of the equation by dx we obtain:
Tx
M=
d
d. (2.2)
Relations (2.1) and (2.2) may be written in a single one as follows:
px
T
x
M==
d
d
d
d2
2
. (2.3)
The above presented relations may be successfully used for plotting the shear
and the bending moment diagrams. Generally speaking, internal forces diagrams (i.e.
diagrams of axial forces, shearing forces, torsion and bending moments) are a
graphical representation of the successive values of axial force N, shearing force T,
torque Mt and bending moment Mi in the various sections against the distance
measured from one end of the involved member.
In particular, relations (2.1), (2.2) and (2.3) bring us several important rules
concerning the shear and bending moment diagrams:
The distributed force p measures the tangent slope of the shear curve (sheardiagram). Ifp = 0, the shearing force will be constant;
It should be observed that Eq. (2.1) is not valid at a point where a concentratedforce is applied. At such a point the shear curve is discontinues and a sudden
change occurs in the diagram. The value of the sudden change in the shear
diagram, when a concentrated force is applied, equals the value of that
concentrated force;
Equation (2.2) indicates that the slope xMidd of the bending moment diagram isequal to the value of the shearing force. This is true at any point where the
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shearing force has a well defined value, i.e. at any point where no concentrated
load is applied;
Equation (2.2) does also show that T= 0 at points where M is maximum. Thisproperty facilitates the determination of the points where the beam (a member in
bending is often referred to a beam) is likely to fail under bending;
If a concentrated couple is applied at an arbitrary point of the beam, a suddenchange in the bending moment diagram occurs, the change value being equal to
the applied concentrated moment (couple);
Equation (2.3) shows that the shear and the bending moment curves will alwaysbe, respectively, one or two degrees higher than the load curve. For example if the
load curve is a horizontal straight line (the case of an uniformly distributed load
p), the shear curve is an oblique straight line and the bending moment curve is aparabola. If the load curve is an oblique straight line (first degree), the shear curve
is a parabola (second degree) and the bending moment curve is a cubic (third
degree).
With the above rules in mind, we should be able to sketch the shear and the
bending moment diagrams without actually determining the function T(x) and M(x)
along the member, once a few values of the shear and the bending moment have been
computed. The sketches obtained will be more accurate if we make use of the fact
that, at any points where the diagrams are continuous, the slope of the shear curve isequal to (-p) and the slope of the bending moment curve is equal to T.
For plotting the internal forces diagrams, the following steps have to be
covered:
a) Denoting of the important points. An important point of a member is a pointwhere a certain change (geometrical, loading, etc) occurs. The supporting points
are usually denoted by capital letters A,B, C, etc. and the other important points
by figures 1, 2, 3 etc.;b) Two successive important points define a portion of the member;c) Determination (when necessary) the magnitude of the reactions at the supports;d) A covering sense of the member has to be chosen (from the left to the right, from
the right to the left or both);
e) For each distinct portion of the member a current cross section at distancex fromone end of the involved portion has to be considered;
f) For the current cross section considered, each distinct internal force (N, T,Mt,Mi)has to be mathematically expressed as a function ofx:N(x), T(x),Mt(x),Mi(x);
g) Plotting the functions N(x), T(x), Mt(x), M
i(x) along the entire member, the
internal forces diagrams are finally obtained.
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2.3 EXAMPLES CONCERNING THE MAIN TYPES OFDIAGRAMS
2.3.1 AXIAL FORCES DIAGRAMS
Example 1
Draw the axial force diagram for the horizontal member with one fixed end
and uniform cross section, shown in Fig. 2.12.
a. b.
Fig. 2.12
Step 1 important (main) points: 1, 2 andA;Step 2
main portions of the member: 1 - 2; 2 - A;Step 3 the magnitude of reactions may be determinated using the condition ofmechanical equilibrium:
Fx = 0 P + 2P -XA= 0 XA = 3P ;
Step 4 the covering sense of the member: let us say it is from the left to theright;
Step 5 we first consider the first portion of the member (1 -2) and anexploratory current cross section located at distance x from end 1 of the
portion. Looking to the left one can conclude that the single axial forcecomponent acting upon the current cross section considered is equal to
P. For any value ofx this component remains constant. This is why, for
portion 1 - 2, the axial force will be constant (NP). The correspondingaxial force diagram of portion 1 - 2 has to be hachured perpendicularly
to a reference horizontal line. Since the covering sense of the member
was chosen from the left to the right, the positive sign convention Ihas
been used (Fig. 2.12a).
In the same manner, the axial force for the second portion 2 - A of the member
is:N2-A = P + 2P = 3P = ct.
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It is to be mentioned that portion 2 - A for example, could have been covered
from the right to the left as well. In such a case the current cross section is taken at
distancex fromA and, looking to the right, we have:
NA-2 =XA = 3P (the same value as above).
When covering the member from the right to the left, the positive signconventionIIshould be taken (Fig.2.12b).
The above presented algorithm for plotting the axial forces diagrams remains
unchanged even if the loading or the geometry are much more complicated or the
internal forces are not axial but shearing forces or bending moments.
The following examples will be accompanied by no supplementary
explanations.
Example 2
Draw the axial force diagram
for the member supported and
axially loaded as shown in Fig. 2.13.
Portion 1 - 2 or, more simple, 1 - 2:
N(x) = 0;
Portion 2 - A:
N(x) = 3P = ct.
Example 3
Draw the axial force diagram
for the member shown in Fig. 2.14.
Fx = 0;XA - 20 - 10 - 52= 0;
XA = 40 kN.
Portion 1 - 2:
N(x) = 20 kN;
Portion 2-A:
N(x) = 20 + 10 + 5x = 30 +5x ;
==
==
.40;2
;30;0 2
kNNmx
kNNx
A
2.3.2 SHEAR AND BENDING-MOMENT DIAGRAMS
Fig. 2.15 shows a simply supported beam that carries a concentrated load P,
being held in equilibrium by the reactions YA and YB. For the time being we neglect
the mass of the beam and consider only the effect of load P. Applying the method ofsections, let us assume that a cutting plane d - d, located at a distancex from pointA,
divides the beam into two segments.
Fig. 2.13
Fig. 2.14
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Fig.2.15
The free-body diagram of the left segment (Fig. 2.16) shows that the externally
applied load is YA . To maintain equilibrium in this segment of the beam the internal
forces occurring at the level of the exploratory section d - dmust supply the resisting
forces necessary to satisfy the conditions of static equilibrium. In this case, the
external load is vertical, so the condition Fx= 0 (the x axis is horizontal) is
automatically satisfied.
Fig. 2.16
Since the left segment of the beam is in equilibrium, the resisting shearing
force Tacting on the left segment has to be numerically equal to YA. In other words,the shearing force in the beam may be determined from the summation of all vertical
components of the external loads acting on either side of the section. However, it is
simpler to restrict this summation to the loads that act on the segment to the left of
the section. This definitions of the shearing force (also called vertical shear or just
shear) may be expressed mathematically as:
LyFT = , (2.4)
the subscript L emphasizing that the vertical summation includes only the external
loads acting on the beam segment to the left of the section being considered.In computing T, when the beam is covered from the left to the right, upward
acting forces and loads are considered as positive (see also the sign convention
presented in the preceding section). This rule of sign produces the effect shown in
Fig. 2.17, in which a positive shearing force tends to move the left segment upward
with respect to the right segment, and vice versa.
Fig. 2.17
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For a complete equilibrium of the free-body diagram in Fig. 2.15 and Fig. 2.16
the summation of moments must also balance. In this discussion YA and Tare equal,
thereby producing a coupleMi that is equal to YAx and is called the bending moment
because it tends to bend the beam.
Analogous to the computation of Tat the current cross section, the bending
moment is defined as the summation of moments about the centroidal axis of anyselected cross section of all loads acting either to the left or to the right side of the
section, being expressed mathematically as:
( ) ( ) ,RLi MMM == (2.5)
where the subscriptL indicates that the bending moment is computed in terms of the
loads acting to the left of the section, while the subscript R referring to loads acting
to the right of the section.
Why the centroidal axis of the exploratory section must be chosen as the axis
of bending moment may not be clear at this moment; this will be explained later.
To many engineers, bending moment is positive if it produces bending of the
beam concave upward, as in Fig. 2.18.
Fig. 2.18
We prefer to use an equivalent convention, which states that the upward acting
external forces cause positive bending moments with respect to any section while
downward forces cause negative bending moments. Therefore, if the left segments of
the beam is concerned (Fig. 2.16), this is equivalent to taking clockwise moments
about the bending axis as positive, as indicated by the moment sense of YA. With
respect to the right segment of the beam (Fig. 2.16) this convention means that the
moment sense of the upward reaction YB
is positive in counterclockwise direction.
This convention has the advantage of permitting a bending moment to be computed,
without any confusion in sign, in terms of the forces to either the left or the right of a
section, depending on which requires the least mathematical work. We never need
think about whether a moment is clockwise or counterclockwise; upward acting
forces always cause positive bending moments regardless of whether they act to the
left or to the right of the exploratory section.
The definition of shearing force and bending moment may be summarized
mathematically as follows:
RL yy FFT == ;( ) ( ) ,RL MMMi ==
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in which positive effects are produced by upward forces and negative effects by
downward forces.
This rule of sign will be used exclusively hereafter. To avoid conflict with this
rule, we must compute vertical shear in terms of the forces lying to the left of the
exploratory section. If the forces acting to the rightof the section were used, it would
be necessary to take downward forces as positive so as to agree with the sign
convention shown in Fig. 2.17.
Example 1
Draw the shear and bending-
moment diagrams for the cantilever beam
shown in Fig. 2.19. (A cantilever beam is
a beam with a fixed end, subjected at its
free end to a single concentrated force P).
We observe that the internal forces
exerted on a current cross section at
distance x from the free end 1 are
represented by:
a shearing force Tof magnitude T= -P (see the positive sign convention);
a bending moment
Mi= - P
x:
==
==
.
;001
ll PMx
Mx
Ai
i
We note (Fig. 2.19) that the
negative values corresponding to the
bending-moment diagram are represented
above the reference line. In this way the
bending-moment diagram shows us how
the involved beam deforms under the
action of the external loads.
Example 2
Draw the shear and bending-
moment diagram for a simply supported
beam AB, of span l subjected to a single
concentrated load P (Fig. 2.20) the case
of Fig. 2.15.
We first determine the reactions at the supports from the free-body diagram of
the entire beam (Fig. 2.20); we find that:
l
bPYA = ;l
aPYB = .
Fig. 2.19
Fig. 2.20
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For the portionA - 1, cutting the beam at distance x from endA, we have:
T= YA = constant;
Mi= YA x:
=
===
==
.
;00
1 ll
Paba
bPaYMax
Mx
A
A
i
i
While the bending moment increases linearly fromM= 0 atx = 0 tol
PabM= at x =
a,we note that the shear has a constant value.Even if the problem is quite simple, it
is more convenient to cover the second portion of the beam from the right to the left.
Therefore, cutting the beam at distance x from end B and using the adequate
sign convention we have:
B - 1:
l
aPYT B == ;
Mi= YB x:
==
==
.
;00
1 l
PabMbx
Mx
i
iB
We can now complete the shear and bending-moment diagrams (Fig. 2.20).
For portionB - 1 the shear has a negative constant value while the bending moment
increases linearly fromM= 0 atB to
Pab
M l=
at 1 (forx = b).
Remarks
If a concentrated traverse force acts at a section of the beam, a sudden change inthe shear diagram at that section occurs, the sudden change value being equal to
that concentrated force. In our case of Fig. 2.20, at point 1, the sudden change is:
( ).P
PbaPPa
Pb =
=
+=+
l
l
lll
If, for a certain portion of the beam, the shear is constant, the bending-momentdiagram is linear;
Covering the beam from the left to the right within theportionA - 1 and then fromthe right to the left within the portion B - 1, and since at point 1 there is no
concentrated moment, there will be no sudden change in the bending-moment
diagram at point 1. This is why we have obtained the same value of the maximum
bending moment at 1;
The covering sense of the beam, when plotting such diagrams, has no importance.It may be chosen from the left to the right, or from the right to the left orcombined, as it is convenient to us;
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When designing a beam like that presented in Fig. 2.20, we must note that thestrength of the beam is usually controlled by the maximum absolute value Mimax
of the bending moment in the beam (in our casePab
Mmaxi
l= ).
We note from the foregoing example that, when a beam is subjected only toconcentrated forces, the shear is constant between the applied forces while the
bending-moment varies linearly between the forces. In such situations, therefore, the
shear and bending-moment diagrams may easily be drawn, once the values of Tand
Mi have been obtained at sections selected just to the left and just to the right of the
points where the loads and reactions are applied.
Numerical examples
1.Draw the shear and bending-moment diagrams for a simply supported beam subjected to twoconcentrated loads (forces) as shown in Fig. 2.21.
Determination of the reactions at the supports
Fy= 0 ; YA -5 - 10 + YB= 0 YA + YB = 15 kN ;
MA = 0 ; YB 4 - 10 3 - 5 1 = 0 YB = 8,75 kN ;
MB = 0 ; YA 4 - 5 3 - 10 1 = 0 YA = 6,25 kN .
Portion A - 1
T= YA = 6,25 kN;
Mi= YA x;
==
==
.25,61
;00
1mkNMmx
Mx
i
iA
Portion 1 - 2
T= YA - 5 = 6,25 - 5 = 1,25 kN;
Mi= YA(1 +x) - 5x .
This means that
( )
=+==
==
.75,8252125,62
;25.60
2
1
mkNMmx
mkNMx
i
i
For the last portion it is much more convenient to
cover the beam from the right to the left.
Portion B - 2
T= -YB = - 8,75 kN;
Mi= YB x
===
==
.75,8175,81
;00
2mkNMmx
iMx
iB
We obtain therefore the shear and bending-moment diagrams shown in Fig. 2.21.
Fig. 2.21
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2. Draw the axial force, shear and bending - moment diagrams for the beam shown in Fig. 2.22.The beam represented in Fig. 2.22 can
be drawn in a simplified manner as shown
in Fig. 2.23.
As in preceding example, the reactions
are determined by considering the entirebeam as a free body, they are:
XB = 10 kN; YA = 22,5 kN; YB = 2,5 kN.
Portion 1 - A
;1045cos210 kNN == o
;1045sin210 kNT == o
xxMA
i 102
2210
1==
:
==
==
.101
;001
mkNMmx
Mx
Ai
i
Portion A - 2
;1045cos210 kNN == o
=+= AYTo45sin210
;5,125,2210 kN=+=
( ) =++=
xYxM AiA1
2
2210
2
( ) xx ++= 5,22110 :
==
==
.5,21
;100
2mkNMmx
mkNMx
i
iA
It is more convenient to us to cover the last portion of the beam from the right to the left.Portion B 2
;10 kNXN B ==
;5,2 kNYT B ==
xxYM BBi==
5,2
2:
==
==
.5,21
;00
2mkNMmx
Mx
i
iB
We can now complete the axial force, shear and bending-moment diagrams of Fig. 2.23. We
note that the axial force has a constant value along the beam; the shear has also constant values
between the important points of the beam while the bending moment varies linearly. At points
(sections) where concentrated forces act, sudden changes in the shear diagram occur (whose valuesequals the applied concentrated forces). Since there are no concentrated moments on the beam there
will be no sudden changes in the bending-moment diagram.
Fig. 2.22
Fig. 2.23
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Example 3
Draw the shear and bending-moment diagrams for a beam, simply supported
at its ends and subjected to a uniformly distributed load p (Fig. 2.24).
Due to the symmetry of loading and geometry, the reactions are:
2l== pYY BA .
As usually, we cut the beam at distance
x fromA and note that:
pxp
pxYT A == 2l
:
==
==
==
.2
;
;0;2
;2
;0
ll
l
l
pTx
Tx
pTx
B
A
222
2
pxxpxxpxYM Ai == l :
== == .0;;0;0
B
A
MxMx
l
Within the calculus, the distributed load over the current portion of the beam
has been replaced by its resultantpxapplied at the midpoint of the involved portion.
Since at the midpoint of the beam the shear equals zero, the bending moment
reaches a maximum value at that point:
.822222
22
max
lllll pppMM i =
=
=
We do also note that the shear diagram is represented by an oblique straight line (Fig.
2.24), while the bending-moment diagram by a parabola. In the section where T= 0,
the bending-moment has a maximum value.
Example 4
Draw the shear and bending-moment diagrams for a beam, simply supported
at its ends and subjected to a linearly distributed load (Fig. 2.25).
The entire beam is taken as a free body, and, from the conditions of
equilibrium, we write:
Fig. 2.24
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Internal forces in statically determinate members
37
Fy = 0 ; ;20 l=+ pYY BA
MB = 0 ; ;032
0 =
ll
lp
YA
;60 l=
pYA
MA= 0 ; ;03
2
2
0 =
ll
lp
YB
.3
0 l=p
YB
Using the first equation of equilibrium (Fy = 0) we check that the valuesalready obtained for YA and YB are correct.
Now writing the mathematical expressions of the shear and bending-moment
at an arbitrary section at distancex from endA, we have:
=
=
.32
;
2 xxpxYM
xpYT
xAi
xA
From similar triangles we may write:
,00 ll
xpp
x
p
px
x ==
which substituted in the preceding expressions of TandMi, leads to :
l
l
l
l
26262
2
0000 xppxxppxpYT xA === ;
=
;3
-=26
=;=
;6
=;0
02
00
0
llll
l
pppTx
pTx
B
A
l
=
==
=
;0=;=
;0=;0:
666632
300
2
00
B
A
i
ixAi
Mx
Mxxpxpxxpx
pxxpxYM
ll
l
l
l
Fig. 2.25
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Strength of Materials
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The shear curve is thus a parabola while the bending-moment curve is a third degree
function. The shear curve intersects thex axis at a distance given by equation:
30
260
200 l
l
l=== x
xppT .
Therefore, the maximum value of the bending moment occurs at3
l=x , since T
(and thusx
Mi
d
d) is zero for this value ofx:
.3936363
20
300 ll
l
lll pppMi =
=
Example 5Draw the shear and bending-moment diagrams for a simply supported beam,
subjected to a concentrated moment M0 applied at point 1 (Fig. 2.26).
The entire beam is taken as a free body and
we have:
;0
l
MYA = .
0
l
MYB =
The negative sign of YB indicates that the
real sense of this reaction is opposite to
that represented in Fig. 2.26.
The shear at any section is constant
and equal to M0/ l. Since a concentrated
moment (couple) is applied at 1, the
bending-moment diagram is discontinuousat 1; the bending-moment decreases
suddenly by an amount equal toM0.
Remark
The concentrated moment in Fig.
2.26 symbolizes for example the action of
two equal and opposite concentrated forces
as shown in Fig. 2.27, whereM0 = P d.
Fig. 2.26
Fig. 2.27
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Internal forces in statically determinate members
39
A complex sample problem
Sketch the shear and bending-moment diagrams for the simply supported beam shown in
Fig. 2.28.
Considering the entire beam as a
free body, we determine the reactions as
follows:
Fy = 0 ;YA + YB + 5 - 10 1 = 0;
YA + YB = 5;
MB= 0 ; 53-10 12,5+ YA 2 +15 = 0;
YA = - 2,5 kN;
MA = 0 ; 51 -1010,5 + 15- YB 2 = 0;
YB = 7,5 kN.
Using the first equation of
equilibrium (Fy = 0) we check that thetwo values obtained for YA and YB are
valid.
Next we draw the shear and bending-moment diagrams. The sketches obtained will be more
accurate if we make use of the fact that, at any point where the curves are continuous, the slope ofthe shear curve is equal to -p while the slope of the bending-moment curve is equal to T.
Portion 1 - A
==
===
.5;1
;5;0:105
1
kNTmx
kNTxxT
A
This means that at the midpoint between 1 andA (forx = 0,5 m) the shear is zero, and, therefore, the
bending-moment reaches a maximum value. It is to be mentioned that this point of maximum for
the bending moment is valid only for the involved portion (i.e. 1 - A). Within the other portions of
the beam the bending-moment could reach grater values as well. This is why, the maximum value of
the bending reached within a certain portion of the beam is called a local maximum. There are casesin which a local maximum does also represent a global maximum too.
==
==
==
==
.25,1;5,0
;0;1
;0;0
:552
1051
2
mkNMmx
Mmx
Mx
xxx
xxM
MAXi
ii A
i
Portion A - 2
( ) kNYT A 5,75,21051105 =+=+= ;
( ) ( ) ==
==++= .5,7;1
;0;0:5,0110152
mkNMmxMxxxM
i
ii
A
Fig. 2.28
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Strength of Materials
40
2iM = - 7,5 kN m tells us that close to the end 2 of the portionA - 2, the bending moment reaches
such a value.
It will be more convenient to us to cover the last portion of the beam from the right to the
left (i.e. fromB to 2):
Portion B - 2
;5,7 kNYT B ==
==
====
.5,7;1
;0;0:5,7
2mkNMmx
MxxxYM
i
iBi
B
This time,2i
M = - 7,5 kN m tells us that close to the end 2 of the portion B - 2, the bending
moment reaches such a value. In this way we have obtained two values for the bending-moment at
point 2: one for the portionA - 2, close to the point 2 to the left and one for the portionB - 2 close to
the same point 2 but to the right. Since at point 2 there is a concentrated moment acting on the beam
(equal to 15 kNm), the sudden change in the bending-moment diagram at point 2 is correct.
2.3.3 TORQUE DIAGRAMS
In the preceding sections we have discussed about axial forces, shear and
bending-moment diagrams. Here we shall consider members which are in torsion.
More specifically we shall learn to draw internal forces diagrams for members
subjected to twisting couples or torques.
We say that a member is subjected to torsion if at any cross section of the
member, the internal forces are represented by a torque vector directed along the axis
of the member.
To sketch the torque diagrams the method of sections may be used, as
presented in the preceding sections.
Draw the torque diagram for a member fixed at one end and subjected to
concentrated and uniform distributed torques as shows in Fig. 2.29.
Considering the entire
member as a free body we obtain
the reactions atA.
= 0xM
00
00 42 MM
MMMA =++= ll
;
Using the method of sections
and covering the member from 1 to
A we have:
1 - 2
0MMt = ;2 - 3
000 32 MMMMt =+= ;Fig. 2.29
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Internal forces in statically determinate members
41
3 - A
xM
MMxmMMMt
l
00000 22 ++=++= ;
==
==
.4;
;3;0
0
03
MMx
MMx
Al
We note that, in such a case, the sign used for torques is not so important. As
soon as a certain sign has been adopted for the first met torque, the signs for the
other torques have to be adopted consequently. Therefore, the torque diagrams may
be sketched above or below the reference line. Like in the preceding examples, if a
concentrated torque acts at a certain section of the member, at that point a sudden
change in the torque diagram occurs (the change being equal to that concentrated
torque). The torque diagrams are usually hachured as shown in Fig. 2.29.
2.4 SUPERPOSITION METHOD
The superposition principle is a consequence of the material linear-elasticbehaviour: the effect at any point of a linear-elastic mechanical structure subjected
to several loads represents the summation of the effects produced by each of these
loads acting separately.
Using the superposition method, a complicated problem may be solved
through a summation of simple problems. For example, the shear and bending-
moment diagram for the beam shown in Fig. 2.30a, may be sketched as an
algebraical summation of the three diagrams of Fig. 2.30b.
Important remark
The drawing of the internal forces
diagrams (axial forces, shear, bending-
moment or torque diagrams) may be
performed in a unique, simple and logical
manner: the involved member is cut at an
arbitrary point, the internal force of a
certain type representing the summation of
all corresponding external loads (ormoments) acting to the left or to the right
of the cross section considered (and using the adequate sign convention).
b.
a.
Fig. 2.30
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Strength of Materials
42
In case of shear and bending moment diagrams, a particular case may also
arise. The presence of one, two or more intermediate pin connections between
different segments of a beam, offers one, two or more additional conditions for the
computation of the external reactions.
Draw the shear and bending-moment diagrams for the beam shown in Fig.
2.31.
Due to the presence of the intermediate pin connection at point 2, the bending
moment (as internal force) at that section is zero.
On the other hand, the
bending moment at point 2,
represents the summation of all
bending moments given by the
loads acting to the left side ofsection 2. We obtain therefore:
= 02i
M
.5,10324 apYaapaY AA ==
From now on the shear and
bending moment diagrams may be
sketched as if the support A and
the intermediate pin connection
would have not existed, the beambeing subjected atA by an upward
vertical concentrated external
force equal to 1,5ap. We finally
obtain the shear and bending moment diagrams shown in Fig. 2.31.
2.5 MOVING LOADS
A truck or other vehicle rolling across a beam or girder constitutes a system of
concentrated loads at fixed distance from each other. For beams carrying onlyconcentrated loads the maximum bending moment occurs under one of the loads.
Therefore the problem here is to determine the bending moment under each
load when each load is in a position to cause a maximum moment to occur under it.
The largest of these various values is the maximum moment that governs the design
of the beam.
In Fig. 2.32, P1 ,P2, P3 and P4 represent a system of loads at fixed distances a,
b and c from each other; the loads move as an unit across the simply supported beam
with span l. Let us locate the position of P2 when the bending moment under this
load is maximum. If we denote the resultant of the loads on the span by R and itsposition from P2 by e, the value of the left reaction is:
Fig. 2.31
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Internal forces in statically determinate members
43
)( xeR
YA = ll
.
The bending moment under P2
is then:
= Li MM )(
.)( 12 aPxxeR
M = ll
To compute the value ofx that will give the maximum M2, we set the
derivative ofM2 with respect tox equal to zero:
0)2(2 == xeR
dx
dMl
l;
from which:
22
ex = l . (2.6)
This value ofx is independent of the number of loads to the left ofP2, since
the derivative of all terms of the form P1a with respect tox will be zero.
Equation (2.6) may be expressed in terms of the following rule: the bending
moment under a particular load is a maximum when the center of the beam is
midway between that load and the resultant of all loads then on the span . With this
rule we locate the position of each load when the moment at that load is a maximum
and compute the value of each such maximum moment.The maximum shearing force occurs at, and is equal to, the maximum reaction.
The maximum reaction for a group of moving loads on a span occurs either at the left
reaction, when the leftmost load is over that reaction, or at the right reaction when
the rightmost load is over it. In other words, the maximum reaction is the reaction to
which the resultant load is nearest.
2.6 INTERNAL FORCES DIAGRAMS FOR PLANE STRUCTURES
(2D structures - FRAMES) AND SPATIAL (3D) STRUCTURES
The principle presented above for sketching the straight beams internal forces
diagrams may be easily extended to the plane or spacial structures. Let us consider
for example the plane beam shown in Fig. 2.33, for which we have to draw the axial
force, shear and bending-moment diagrams.
An observer"O covering the beam from 1 toA (or fromA to 1, as it is easier
from the mathematical point of view) sees each straight portion of the beam as a
beam for which applies the rules presented in the preceding sections.
Fig. 2.32
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Strength of Materials
44
Therefore, for portion 1-2, at a
current section at distance x from 1,
we have (Fig. 2.33):
;0=N
;PT=
==
===
.;
;0;0:
2
1
PaMax
MxxPM
i
i
i
Fig.2. 34
The second straight portion of the beam may be covered from 2 toA orA to 2
as well. Let us suppose that the second case is being used.
Fig.2. 35
The effect of the concentrated force P is transmitted up to the current cross
section located at distancex from point 2, where the observer is placed. Therefore, at
that section we have:
Fig. 2.33
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Internal forces in statically determinate members
45
;PN =
;0=T
.aPMi =
We note that all internal forces corresponding to the portion 2 -A are constant.
We are now in the position to draw the internal forces diagrams (N, T,Mi). Itis to be mentioned that, in such cases the diagrams are sketched with respect to a
reference line representing thex axis of the beam (the axis directed along the beam).
Analogous to the straight beams, forNand Tdiagrams + means above the reference
line. For the bending-moment diagrams, "+"means below and "-" means above the
reference line (from the observer's point of view). With these remarks, the internal
forces diagrams have been represented in Fig. 2.36.
Fig. 2.36
The diagrams represented in Fig 2.36 tell us what does the plane beam feel (asa global effect) at each particular cross section, when subjected to the external load
P.
We also note that it was not necessary to compute the reactions XA, YA,MA for
sketching the internal forces diagrams.
SAMPLE PROBLEMS
a)Draw the axial force, shear and bendingmoment diagrams for the frame and the
loading shown in Fig. 2.37.
Considering the frame as a free body we
first determine the reactions:
;00 == Ax xF
;0 PYYF BAy =+=
= 0AM
==2
02P
YPYBB
ll
Fig. 2.37
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Strength of Materials
46
.2
PYY BA ==
Applying the above presented principle and letting an observer to cover the
beam fromA to 1 and then fromB to 1 we have:
A - 1:
;2
PYN A ==
;0== AXT
.0== xXM Ai B - 2:
;0=N
;
2
PYT B ==
===
==
==.
22;
;0;0
;2
2
lll
PPMx
Mx
xP
xYM
i
i
Bi
B
2 - 1:;0=N
;22
PP
PPYT B =+=+=
==
==+=+=.0;
;2
;0;)(
2)(
1
2
i
iBi
Mx
PMxPxx
PPxxYM
l
l
ll
The axial force, shear and bending moment diagrams have been represented in Fig.
2.38.
Fig. 2.38
b)Draw the axial force, shear and bending-moment diagrams for the frame shownin Fig.2.39.
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Internal forces in statically determinate members
47
In Fig. 2.39b the simplified form of the frame together with the reactions have
been represented. Considering the entire beam (frame) as a free body and using the
external reference coordinate system Oxy, we determine the reactions as follows:
Fig. 2.39
;5050 apXapXF AAx ===
;2020 apYYapYYF BABAy =+=+=
;6025220 apYaapaapaYM BBA ===
.4022520 apYaapaapaYM AAB ==+= We have found therefore that:
=
=
=
.6
;4
;5
apY
apY
apX
B
A
A
The second equation of equilibrium (Fy = 0) may be used as a checking equation
when the reactions YA and YB were computed from MA = 0 and MB = 0. TheN, T
and Mi diagrams are shown in Fig. 2.40.
Fig. 2.40
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Strength of Materials
48
c)Draw the axial force, shear and bending-moment diagrams for the curved beamof radius R shown in Fig. 2.41.
Although the axis of the beam is not a straight line, the principle presented
above for sketching theN, T,Mi diagrams remains valid.
The problem consists in determining the internal forces corresponding to eachparticular cross section of the beam. In order to locate the current cross section an
angular parameter must be used (instead of the linear parameter x which has
been used up the now) for each particular portion of the curved beam.
Fig. 2.42 Fig. 2.43
Let us now consider the first portion 12 of the built-in arch shown in Fig.
2.41. The axial force, shear and bending moment equations for segment 1 2 are
obtained similarly by passing a cross section aa anywhere between 1 and 2. As
discussed above, the cross section is located by the parameter . When varies
between 0 and 90 the whole portion 1 - 2 of the curved beam is covered. The
concentrated load P which acts at section 1, transmits its effect through the segment
BCDEup to the current cross sectionD'E located by parameter (Fig. 2.42).
This means that a vertical downward load P will act at the centroid O of the
current cross sectionDE. It is in fact the internal force exerted on the current cross
section and may be resolved into two components: one component perpendicular to
the current cross section
DE and the other one contained within the plane of thecross section. The first component represents the current axial force (N) while the
second component represents the corresponding shearing force (T).AnobserverO
placed at the current cross section, using the proper positive sign convention will see
that:
:cos= PN
:sin= PT
==
==
==
==
.;2
;0;
2
;0;0
;;0
PT
N
T
PN
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Internal forces in statically determinate members
49
The bending moment exerted by the concentrated load P, acting at point 1,
with respect to the centroid O of the current cross section is (Fig. 2.43):
:)cos1()cos( == PRRRPMi
==
==
.;2
;0;0
2
1
PRM
M
i
i
The same reasoning may be
applied when the second portion of the
curved beam is to be covered. This time
it will be more convenient to us to cover
the beam from A to 2. But in this case
we have to compute the reactions YA andMA at first. This will be done by considering
the entire curved beam as a free body and using the corresponding equations of
equilibrium, Fig.2.42.
PYPPYF AAy 3020 === ;
PRMRPRPMM AAA 40220 +=== .
We write:
;cos3cos PYN A ==
==
==
.0;2
;3;0
2N
PNA
;sin3sin PYT A ==
==
==
.3;2
;0;0
2 PT
TA
)cos1(34)cos( +=+= PRPRRRYMM AAi ;
==
===
.;2
;4;0
2PRM
PRMM
i
AiA
The axial force, shear and bending moment diagrams have been represented in Fig.2.45. It is to be noted that all the established properties of the N, T,Mi diagrams of
straight beams remains valid.
Fig. 2.43
Fig. 2.44
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Strength of Materials
50
We shall now consider a simple spatial structure (a 3D structure, i.e. a three
dimensional structure) represented by a beam, fixed at one end, and subjected to two
concentrated loads P and Q at the other end, Fig. 2.46.
d)Draw the axial force, shear, bending moment and torque diagrams for the 3Dbeam shown in Fig. 2.46.
Although the problem seems to be a little bit more completed the basic
principle for sketching the internal force diagrams remains unchanged.
Before solving the problem there are still some important remarks to be done:
As it will be discussed later, the sign of the shearing force has no physicalconsequences in designing a beam or a certain mechanical structure. This is why,
in case of complicate structures, we shall give the T sign up and we shall
represent the Tdiagrams as they are convenient to us;
The sign of the bending moment does also depend upon the relative position ofthe observer, Fig. 2.47. Although the two beams represented in Fig. 2.47 are
entirely identical from the geometrical and loading point of view, the two
corresponding bending - moment diagrams have different signs. These signs are,
after all, simple conventions. On the other hand, it is to be observed that in both
cases of Fig 2.47, the bending-moment diagrams occupy the same position withrespect to the reference line. In other words, this means that the position of the
bending-moment diagrams do not depend upon the observer's position. The
Fig. 2.45
Fig. 2.46
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Internal forces in statically determinate members
51
location of the bending moment diagram with respect to the reference line
corresponds to the position of the beam fibres in tension.
For this reason, in many cases, we shall not note the sign of the bending
moment diagrams and we shall represent these diagrams on the side corresponding to
the beam fibres in tension, Fig. 2.48.
Let us now return to the original
problem regarding the simple 3D
structure shown in Fig. 2.46. Covering
the beam from 1 to A and attaching a
proper coordinate system (whose Ox
axis is usually directed along the beam)to each main portion of the beam (Fig.
2.46) we obtain the axial force, shearing
force, bending - moment and torque
diagrams shown in Fig. 2.49. It should
be noted that, for each main portion of
the beam, two shearing forces and two bending-moments could exist simultaneously
(about Oy and Oz axes).
We shall now conclude our analysis concerning the main types of internal
forces by observing that these diagrams are in fact a graphical representation of aglobal mechanical effect occured at any particular cross section of a given member
subjected to external loads.
Fig. 2.47
Fig. 2.48
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Strength of Materials
52
Fig. 2.49
While these diagrams represent a first and necessary step in the analysis of a given
structural member, they do not tell us whether the external loads may be safely
supported. Whether or not a given structural member will break under the external
loading clearly depends upon the ability of the material to withstand the
corresponding elementary forces occurred at the level of each particular point of the
member cross sections. This is why, after a short study of the moments of inertia
within the next chapter, some other chapters of the text will be devoted to the
analysis of the stresses and of the corresponding deformations in various structural
members, considering axial loading, shearing loading, torsion and bendingsuccessively. Each analysis will be based upon a few basic concepts, namely, the
conditions of equilibrium of the forces exerted on the member, the relations existing
between stress and strain in the material and the conditions imposed by the supports
and loading of the member. The study of each type of loading will be complemented
by examples, sample problems and problems to be assigned, all designed to
strengthen the students understanding of the subject.
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Internal forces in statically determinate members
53
PROBLEMS TO BE ASSIGNED
P2
P.2.1Draw the axial force, shear and bending moment diagrams for the members, frames andloading shown (Fig. P.2.1).
a. b.
c. d.
e. f.
g. h.
i. j.
Fig. P.2.1
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Strength of Materials
54
k.l.
m. n.
o.p.
Fig. P.2.1 (continued)
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Internal forces in statically determinate members
55
r. s.
t. u.
Fig. P.2.1 (continued)
P.2.2 Draw the torque diagrams for the members and loading shown (Fig. P.2.2).
Fig. P.2.2
P.2.3Draw the axial force, shear, bending moment and torque diagrams for the 3-D structures
and loading shown (Fig. P.2.3).
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Strength of Materials
a. b.
c. d.
e. f.
Fig. P.2.3
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