Strength of Materials - 2 - Internal Forces in Statically Determined Members

7/29/2019 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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Strength of Materials

22

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

23

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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24

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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25

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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26

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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27

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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28

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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29

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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30

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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31

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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32

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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33

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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34

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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35

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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36

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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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


18/36


38

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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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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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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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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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


23/36


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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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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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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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.


27/36


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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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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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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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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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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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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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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54

k.l.

m. n.

o.p.

Fig. P.2.1 (continued)


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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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a. b.

c. d.

e. f.

Fig. P.2.3

Strength of Materials - 2 - Internal Forces in Statically Determined Members

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