Modul XIV
7. Shear and Bending Moment in a Beam.
Consider a beam AB subjected to various concentrated and distrib uted loads (Fig.
7.8a). We propose to determine the shearinc force and bending moment at any point of
the beam. In the example considered here, the beam is simply supported, but the
method used could be applied to any type of statically determinate beam.
First we determine the reactions at A and B by choosing the entire beam as a free body
(Fig. 7.8fo); writing SM^ = 0 and ZMB = 0, we obtain, respectively, RB and RA.
To determine the internal forces at C, we cut the beam at C and draw the free-body
diagrams of the portions AC and CB of the beam (Fig. 7.8c). Using the free-body
diagram of AC, we may determine the shearing force V at C by equating to zero the sum
of the vertical components of all forces acting on AC. Similarly, the bending moment M
at C may be found by equating to zero the sum of the moments about C of all forces and
couples acting on AC. We could have used just as well, however, the free-body diagram
of CBt and determined the shearing force V and the bending moment M' by equating to
zero the sum of the vertical components and the sum of the moments about' C of all
forces and couples acting on CB. While this possible choice of alternate free bodies may
facilitate the computation of the numerical values of the shearing force and bending
moment, it makes it necessary to indicate on which portion of the beam the internal
forces considered are acting. If the shearing force and bending moment, however, are to
be computed at every point of the beam and efficiently recorded, we should not have to
specify every time which portion of the beam is used as a free body. We shall adopt,
therefore, the following convention:
Fig.7-8
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Note:
[The force and couple representing the internal forces acting on CB will now be denoted
by V and M', rather than by —V and — M as done earlier, in order to avoid confusion
when applying the sign convention which we are about to introduce.]
To determine the shearing force in a beam, we shall always assume that the internal
forces V and V are directed as shown in Fig. 7.8c. A positive value obtained for their
common magnitude V will indicate that this assumption was correct and that the
shearing forces are actually directed as shown. A negative value obtained for V will
indicate that the assumption was wrong and that the shearing forces are directed in the
opposite way. Thus, only the magnitude V, together with a plus or minus sign, needs to
be recorded to define completely the shearing forces at a given point of the beam. The
scalar V is commonly referred to as the shear at the given point of the beam.
Similarly, we shall always assume that the internal couples M and M' are directed as
shown in Fig. 7.8c. A positive value obtained for their magnitude M, commonly referred
to as the bending moment, will indicate that this assumption was correct, and a negative
value that it was wrong. Summarizing the sign convention we have presented, we state:
The shear V and the bending moment M at a given point of a beam are said to be
positive when the internal forces and couples acting on each portion of the beam are
directed as shown in Fig. 7.9a.
This convention may be more easily remembered if we note that:
1. The shear at C is positive when the external forces (loads and reactions') acting on
the beam tend to shear off the beam at C as indicated in Fig. 7.9k.
2. The bending moment at C is positive when the external forces acting on the beam
tend to bend the beam at C as indicated in Fig. 7.9c.
It may also help to note that the situation described in Fig. 7.9, and corresponding to
positive values of the shear and of the bending moment, is precisely the situation which
occurs in the left half of a simply supported beam carrying a single concentrated load at
its midpoint. This particular example is fully discussed in die following section.
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7.1. Shear and Bending-Moment Diagrams.
Now that shear and bending moment have been clearly defined in sense as well as in
magnitude, we may easily record their values at any point of a beam by plotting these
values against the | distance x measured from one end of the beam. The graphs
obtained in this way are called, respectively, the shear diagram and the bending-moment
diagram. As an example, consider a simply supported beam AB of span L subjected to a
single concentrated load P applied at its midpoint D (Fig. 7.1()o). We first determine the
reactions at the supports from the free-body diagram of the entire beam (Fig. 7.10b); we
find that the magnitude of each reaction is equal to P/2.
Next we cut the beam at a point C between A and D and draw the free-body diagrams of
AC and CB (Fig. 7.10c). Assuming that shear and bending moment are positive, we
direct the internal forces V and V and the internal couples M and M' as indicated in Fig.
7.9a. Considering the free body AC and writing that the sum of the vertical components
and the sum of the moments about C of the forces acting on the free body are zero, we
find V = +P/Z and M = +Px/2. Both the shear and the bending moment are therefore
positive; tJiis may be checked by observing that the reaction at A tends to shear off and
to bend the beam at C as indicated in Fig. 7.9Z? and c. We may plot V and M between A
and D (Fig. 7.10e and /); the shear has a constant value V = P/2, while the bending
moment increases linearly from M = 0 at x = 0 toM = PL/4 at x = L/2.
Cutting, now, the beam at a point E between D and B and considering the free body EB
(Fig. 7.l0d), we write that the sum of the vertical components and the sum of the
moments about E of the forces acting on the free body are zero. We obtain V = -P/2 and
M = P(L - x)/2. The shear is therefore negative and the bending moment positive; this
may be checked by observing that the reaction at B bends the beam at E as indicated in
Fig. 7.9c but tends to shear it off in a manner opposite to that shown in Fig. 7.9c. We can
complete, now, the shear and bending-moment diagrams of Fig. 7 and f; the shear has a
constant value V -P/2 between D and B, while the bending moment decreases linearly
from M = PL/4 at x = L/2 to M = 0 at x = L.
We shall note that, when a beam is subjected only to concentrated loads, the shear is of
constant value between loads and the bending moment varies linearly between loads.
On the other hand, when a beam is subjected to distributed loads, the shear and
bending moment vary quite differently (see Prob. 14.3).
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SAMPLE PROBLEM 14.1
Draw the shear and bending-moment diagram for the beam and loading shown.
Solution.
The reactions are determined by considering the entire beam as a free
body; they are
RB = 46 kN
RD = 14 kN
We first determine the internal forces just to the right of the 20-kN load at A. Considering
the stub of beam to the left of section 1 as a free body and assuming V and M to be
positive (according to the standard convention), we write
F
M
 0 : -20kN – V1 = 0 - V1 = -20kN
y
1
 0:
(20kN)(0 m) + M1 = 0
+ M1 = 0
We next consider as a free body the portion of beam to the left
of section 2 and write
F
M
 0 : -20 kN - V2 = 0
y
2
V2 = -20 kN
 0 : (20kN)(2.5 m) + M2 = 0
M2 = -50kN • m
The shear and bending moment at sections 3, 4, 5, and 6 are determined in a similar
way from the free-body diagrams shown. We obtain
V3 = +26 kN
V4 = +26kN
V5 = -14kN
V6 = - 14 kN
M3 = -50 kN • m
M4 = + 28kN-m
M5 = + 28kN-m
M6 = 0
For several of the latter sections, the results may be more easily obtained by considering
as a free body the portion of the beam to the right of the section. For example,
considering the portion of the beam to the right of section 4, we write
F
M
 0:
y
4
 0:
V4 - 40kN + 14 kN = 0
V4 = +26kN
-M4 + (14kN)(2m) = 0
M4 = +28kN-m
We may now plot the six points shown on the shear and bending-moment diagrams. As
indicated in Sec. 14.1., the shear is of constant value between concentrated loads, and
the bending moment varies linearly; we obtain therefore the shear and bending-moment
diagrams shown.
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SAMPLE PROBLEM 14.2
Draw the shear and bending-moment diagrams for the cantilever beam AB. The
distributed load of 3 kips/ft extends over 8 ft of the beam and the 10-kip load is applied at
E.
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Solution. The 10-kip load is replaced by an equivalent force-couple system acting on
the beam at point D. The reaction at B is determined by considering the entire beam as
a free body.
From A to C. We determine the internal forces at a distance x from point A by
considering the portion of beam to the left of section 1. That part of the distributed load
acting on the free body is replaced by its resultant, and we write
F
M
y
1
 0 : -3x - V = 0
V = -3xkips
 0 : 3x(1/2x) + M = 0
M = - l.5x2 kip –ft
Since the free-body diagram shown may be used for all values of x smaller than 8 ft, the
expressions obtained for V and M are valid in the region 0 < x < 8 ft.
From C to D. Considering the portion of beam to the left of section 2 and again
replacing the distributed load by its resultant, we obtain
F
M
 0:
y
2
 0:
-24 - V = 0
24(x - 4) + M = 0
V= -24 kips
M = 96 – 24x (kip • ft)
These expressions are valid in the region 8 f t <C x <C 11 ft.
From D to B. Using the portion of beam to the left of section 3, we obtain for the region
11 ft < x < 16ft:
V = -34 kips
M = 226 – 34x (kip • ft)
The shear and bending-moment diagrams for the entire beam may now be plotted. We
note that the couple of moment 20 kip • ft applied at point D introduces a discontinuity
into the bending-moment diagram.
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Contoh soal dari penulis:
14-3.Sebuah batang yang berat massanya dapat diabaikan dibebani gaya F
ditengahnya, dan akan kita tinjau distribusi SFD dan BMD nya.
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Jawab:
Persamaan keseimbangan static untuk setiap FBD memberikan:
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0  x  L / 2 : FBD..I
Keseimbangan..Gaya.
F
 F / 2  Vx  0
y
Vx  F / 2
 V x  kons tan
KeseimbanganMoment.
M
P
 Mx  x F /2  0
M x  x  F / 2  grs..linier
x  0  Mx  0
x  L / 2  M x  L / 2 F / 2 
FL
4
FBD..II
L/2  x  L
Keseimbangan..Static
F
y
  F  Vx  0
Vx   F / 2
 V x  con tan t
Keseimbnagan..Moment
M
P
 M x  ( x  L / 2) F  x  F / 2  0
M x  1 / 2 xF  L / 2 F  (
x  L/2  Mx 
1
L
x  )F
2
2
L
F
4
x  L  Mx  0
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14-4.Sketlah SFD dan BMD untuk batang yang tergambar dibawah ini.
Jawab.
Step I
Dari FBD utama ,persamaan keseimbangannya adalah:
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M
F
F
 30  10(6)  BY (15)  0
AZ
X
 Ax  0
Y
 AY  BY  10  0
 BY  2kN , AX  0, AY  8kN
Step.II
Anda harus dapat memahami bahwa batang dapat dipotong menjadi 3 bagian,yakni:
i. antara ujung A dan C →FBD I
ii. antara titik C dan D,dan →FBD II
iii. antara titik D dan ujung B→FBD III
i. Batang antara ujung A dan titik C,dari FBD I:
0  x  6:
Keseimbangan..Gaya
F
Y
 V  8  0
 V x  8kN  cons tan t.....i.
 untuk..0  x  6  V x  cons tan t
Keseimbangan..Moment
M
P
 M x  x  AY  0
M x  8 xkN  m....ii
x  0  Mx  0
x  6  M x  48kN  m
ii.Batang antara titik C dan titik D →FBD II:
6m≤x≤9m
Keseimbangan Gaya:
F
Y
 V  10  8  0
V  2kN  kons tan ta....iii
Keseimbangan Moment:
M
 M x  10( x  6)  8( x)  0
PZ
M x  60  2 xkN  m....iv.
Keseimbangan Gaya:
F
Y
 V  10  8  0
V  2kN  kons tan ta...v.
Keseimbangan Moment:
M
PZ
 M x  30  10( x  6)  8( x)  0
M x  30  2 xkN  m....vi.
14-5.Sketlah SFD dan BMD untuk batang over hanging dengan berat sama rata
(homogin) = w per unit panjang.
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Jawab:
Step I.
Gambarkanlah FBD utama,untuk
persamaan keseimbangan didapat:
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X
2L
L
)  wL( )  0
3
2
 Ax  0
Y
 AY  BY  wL  0
M
F
F
AZ
 BY (
 AX  0, BY 
3
1
wL, AY  wL
4
4
Karena batang berbeban uniform dan hanya ada satu tumpuan dititik B,maka
cukup dibagi menjadi 2 bagian saja,yakni:
i. antara ujung A dan tumpuan B→FBD I
ii. antara tumpuan B dan ujung C →FBD II
Step.II.
i. 0≤x≤2L/3
Keseimbangan Gaya.
F
Y
 V  wx 
V  w(
wL
0
4
L
 x)....i.
4
Keseimbangan Moment.
x
wL
 M x  wx( ) 
( x)  0
2
4
w
M x   (2 x 2  Lx)....ii.
4
M
PZ
Step III.
ii. 2L/3≤x≤L
Keseimbangan Gaya.
F
Y
 V  wx 
wL 3wL

0
4
4
V  w( L  x)....iii .
Keseimbangan Moment.
3wL
2L
x
wL
 M PZ  M x  4 ( x  2 )  wx( 2 )  4 ( x)  0
M x   w(1 / 2 x 2  Lx  1 / 2 L2 )....iv.
PROBLEM 14.1
Draw the shear and bending-moment diagrams for the beam and loading shown.
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PROBLEM 14.2
Draw the shear and bending-moment diagrams for the beam ana loading shown.
PROBLEM 14.3
Sketch the shear and bending-moment diagrams for the cantilever beam shown.
PROBLEM 14.4
The simple beam AC is loaded by a couple of magnitude T applied at point B. Draw the
shear and bending-moment diagrams of the beam.
“SELAMAT –BELAJAR”
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Modul XIV:
7. Defleksi
Ambil :
E: modulus elastisitas
I: moment inertia
Bila Mx adalah harga bending moment disebarang posisi x, maka persamaan
diffrensial untuk lenturan beam tsb :
Gambar.7.1.
d2y
Mx

........(7.1)
2
dx
EI
Solusi:
Pengintegrasian sekali dari pers.(7.1) didapat :
Mx
dy
 
dx  c1
dx
EI
Karena persamaan diatas merupakan persamaan garis singgung yg bersudut θ dengan
sumbu // sumbu x, maka:
tan  
dy
M
   x dx  c1......(7.2)
dx
EI
disini θ disebut sudut defleksi.
Dan dari persamaan (7.2) dengan mengintegrasikannya , maka didapat lenturan y,yaitu:
y   
Mx
dxdx  c1 x  c2 .......(7.3)
EI
disini konstanta c1 dan c2 dihitung dengan ketentuan “ boundary condition “
(lihat Gambar 7.2.dan 7.3.)
Boundary Condition
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Gambar. 7.2.
Gambar.7.3.
Dari Gambar.7.2.
Dititik A untuk x=0 →y=0
}……(7.4)
Dititik B untuk x=l →y=0
Dari Gambar.7.3.
Dititik A untuk x=0,maka:
y0 

dy
....(7.5)
 0
dx

Contoh:
1. Hitunglah sudut defleksi θ dan defleksi y dari Gambar dibawah ini.
Jawab:
Dari pers.(7.2) dan (7.3) jelas diperlukan menghitung Mx dan gaya2 reaksi dititik A
dan B terlebih dahulu.
FBD.I:
Dari FBD I didapat:
i.Keseimbangan Gaya:
 Fy  Ay  B y  R  0
 Ay  R  B y  1 / 2 pl  B y ......(i )
ii.Keseimbangan Moment:
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2
  lR  lB y  0
3
2 R
2
 By  l  2 / 3R   1 / 2 pl  1 / 3 pl.....(ii )
3 l
3
3 2
1
Ay  1 / 2  pl  1 / 3  pl  (
) pl  pl....(iii )
6
6
M
A
FBD II:
Dari FBD II ,jumlah momen dititik P adalah:
M
P
  xAy  ( x  2 / 3 x) R x  M x  0
 1 / 6  plx  1 / 3  x  1 / 3  p x x
 1 / 6  plx  1 / 6 x 2

x
p
l
p 2
(l x  x 3 ).....(iv )
6l
Substitusikan pers.(iv) ini ke pers.(i) didapat:
M
d2y
p
 x 
( x 3  l 2 x)
2
EI
6 EIl
dx
Setelah integrasi,didapat:
dy
p
x4 l 2 x2

(

)  c1
dx
6 EIl 4
2
p
x5 l 2 x3
y
(

)  c1 x  c 2
6 EIl 20
6
Dari”boundary condition”,Gamb.7.2,konstanta2 integral c1 dan c2 dihitung sbb:
Diujung titik A untuk x=0 →y=0,maka c2=0.
Diujung titik B untuk x=l→y=0,maka:
p
l5
l 2l 3
pl 5 6  20
(

)
(
)
6 EIl 20
6
6 EIl 120
14 pl 5
7 pl 4


720 EIl
360 EI
3
7 pl
 c1 
........(v)
360 EI
c1l  
Jadi:
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dy
p x4 l 2 x2
7 pl 3

(

)
dx 6 EIl 4
2
360 EI
p 
x 2
x 4

7  30( )  15( ) ......(vi)
360 EI 
l
l 
p x5 l 2 x3
7 pl 3
y
(

)
x
6 EIl 20
6
360 EI
pl 4  x
x 3
x 5

7( )  10( )  3( ) .....(vii)
360 EI  l
l
l 
(2)Hitunglah sudut defleksi θ,dan defleksi y dari Gambar disamping ini.
Jawab.
FBD I
FBD II
Dari FBD I didapat:
F
 Ay  R  0
y
R  Ay  1 / 2 pl......(i )
M
 M A  2 / 3  lR  0
M A  2 / 3l 
1
1
pl  pl 2 ....(ii )
2
3
Dari FBD II didapat:
M
p
 M x  M A  xAy  1 / 3 xRx  0
1
1 1 x
1
pl  x x p  pl 2
2
3 2 l
3
p

( x 3  3l 2 x  2l 3 )....(iii )
6l
M x  xAy  1 / 3 xRx  M A  x
1 x3 1
 1 / 2 plx  p
 pl 2
6
l
3
Substitusikan ke pers.(i) ,didapat:
d2y
p

( x 3  3l 2 x  2l 3 ).....(iv )
2
6 EIl
dx
Dengan mengintegrasikan,didapat:
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dy
p x 4 3l 2

( 
 2l 3 x)  c1
dx 6 EIl l
2
5
2
p x
l
y
(  x 3  l 3 x 2 )  c1 x  c 2
6 EIl 20 2
Dari “boundary condition” Gambar 7.3,kondisi dititik A untuk x=0 yaitu y=0,dan
dy
 0 ,maka didapat:c1=c2=0.
dx
Thus,
dy
p
x4
3

(
 l 2 x 2  2l 3 x)
dx
6 EIl 4
2
3
pl  x
x 2
x 4

8( )  6( )  ( ) ....(v)
24 EI  l
l
l 
p
x5
l 2x 3
(

 l3x2 )
6 EIl 20
2
pl 4 
x 2
x 3
x 5

20( )  10( )  ( ) .....(vi)
120 EIl 
l
l
l 
y
Latihan soal.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
21