INTERNAL FORCES IN A BEAM - Vocational Training Counciltycnw01.vtc.edu.hk/cbe2027/SA1_BEAM.pdf · CBE2027 Structural Analysis I Chapter 2 – Analysis of Determinate Beams HD in Civil

CBE2027 Structural Analysis I Chapter 2 – Analysis of Determinate Beams

HD in Civil Engineering Page 2-1

INTERNAL FORCES IN A BEAM

Before a structural element can be designed, it is necessary to determine the

internal forces that act within the element. The internal forces for a beam

section will consist of a shear force V, a bending moment M, and an axial

force (normal force) N. For beams with no axial loading, the axial force N is

zero.

Sign Convention (considering a small segment of the member):

Shear Force V:

Positive shear tends to rotate the segment clockwise.

Moment M:

Positive moment bends the segment concave upwards.

(so as to ‘hold water’)

Axial Force N:

Tension is positive.

An important feature of the above sign convention (often called the beam

convention) is that it gives the same (positive or negative) results regardless

of which side of the section is used in computing the internal forces.

V N

M M

N

V

V VN N

M M

Axial Force

Shear Force Bending Moment

POSITIVE SIGN CONVENTION



PROCEDURES FOR FINDING V, M AND N AT A BEAM SECTION

1. Identify whether the beam is a determinate or an indeterminate

structure. (This chapter focuses on the analysis of determinate beam

only. Indeterminate structure requires the consideration of

compatibility condition, i.e. the deformation of the structure).

2. Compute the Support Reactions

Make use of the equilibrium equations and the equations of condition if

any.

3. Draw a Free-Body Diagram of the Beam Segment

Keep all external loading on the member in their exact location.

Draw a free-body diagram of the beam segment to the left or right of the

beam. (Although the left or right segment could equally be used, we

should select the segment that requires the least computation).

Indicate at the section the unknowns V, M and N. The directions of

these unknowns may be assumed to be the same as their positive

directions.

4. Use the Equations of Equilibrium to Determine V, M & N.

If the solution gives a negative value for V, M or N, this does not mean

the force itself is NEGATIVE. It tells that actual force or moment acts

in the reversed direction only.

5. Check the Calculations using the Opposite Beam Segment if

necessary.



Example 1

Determine the shear force V and the bending moment M at the section P of

the overhanging beam shown.

4m

A B C DP

10 kN 15 kN

4 kN/m

2m 10m 3m

HBVVB C

Solution:

No. of reactions = no. of equations of equilibrium

⇒ This is a determinate beam.

Determine reactions HB, VB and VC.

ΣX = 0, HB = 0

Take moment about B,

ΣM = 0, 4*10*(10/2) + 15*13 – 10*2 – VC*10 = 0

VC = 37.5 kN

ΣY = 0, VB + VC = 10 + 4*10 + 15

VB = 27.5 kN



Determine V and M at P (using left free-body)

HBVB

4 kN/m

10 kN

AB P H

V

M

P

P

P

2m 4m

∑X = 0, since HB = 0, ∴ HP = 0

∑Y = 0, VB + VP = 10 + 4*4

27.5 + VP = 26

∴ VP = -1.5 kN

(This implies that Vp acts in downwards direction ↓)

Take moment about P,

10*6 + 4*4*4/2 + MP = VB*4

MP = 27.5*4 – 60 – 32 = 18 kNm

Determine V and M at P (using right free-body)

4 kN/m

15 kN

6m 3mVC

HP

V

MP

P

P



∑X = 0, HP = 0,

∑Y = 0, VC + VP = 15 + 4*6

37.5 + VP = 39

∴ VP = +1.5 kN

(This implies that Vp acts in upwards direction ↑ as assumed)


15*9 + 4*6*6/2 + MP = VC*6

MP = 37.5*6 – 135 – 72 = 18 kNm

(This implies that MP acts in the direction as indicated in the

free-body diagram.)



Example 2

Determine the shear force V and the bending moment M at section P of the

cantilever beam.

HA

VA

MA

A

B P C

3m 3m

4m

40 kN

5 kN/m

Solution:

Determine the support reactions

∑X = 0, HA= 0,

∑Y = 0, VA = 5*6 + 40 = 70 kN

Take moment about A,

40 * 3 + 5*6*6/2 – MA =0

MA = 210 kNm

Determine V and M at P (using left free-body)

A

B

P

4m

40 kN

5 kN/m210 kNm

70 kN

HP

V

MP

P

∑X = 0, HP= 0,

∑Y = 0, VP + 70 = 40 + 5*4

VP = -10 kN




40*1 + 5*4*4/2 + 210 – 70*4 - Mp = 0

MP = 10 kNm

Determine V and M at P (using right free-body)

5 kN/mHP

VP

MP

P C

2m

∑X = 0, HP= 0,

∑Y = 0, VP = 5*2 = 10 kN


5*2*1 – Mp = 0

MP = 10 kNm

*Both the left free-body and the right free-body can be used to obtain the

results. However, it is noted that by using the right free-body will greatly

simplified the calculations. This shows importance of choosing the

appropriate free-body.



SHEAR FORCE AND BENDING MOMENT DIAGRAMS

By the methods discussed before, i.e. by using free-body diagrams, the

magnitude and sign of the shear forces and bending moments may be

obtained at many sections of a beam. When these values are plotted on a

base line representing the length of a beam, the resulting diagrams are called,

respectively, the shear force diagram and the bending moment diagram.

Shear force and bending moment diagrams are very useful to a designer, as

they allow him to see at a glance the critical sections of the beam and the

forces to design for. Draftmanlike precision in drawing the shear force and

bending moment diagrams is usually not necessary, as long as the

significant numerical values are clearly marked on the diagram.

The most fundamental approach in constructing the shear force and bending

moment diagrams for a beam is to use the procedure of sectioning. With

some experience, it is not difficult to identify the sections at which the shear

force and bending moment diagrams between these sections are readily

identified after some experience and can be sketched in.



Example 3

Draw the shear force and the bending moment diagrams for the beam

shown.

HA V

A VC

B

5m 3m

8 kN

2 kN/m

A C

Solution:

∑X = 0, HA = 0


2*8*8/2 + 8*5 – VC*8 = 0

VC = 13 kN

∑Y = 0, VA + VC = 2*8 + 8

⇒ VA + 13 = 2*8 + 8

⇒ VA = 11 kN

2 kN/m

11 kNx

HX

V

MX

X

For 0 ≤ x ≤ 5m

∑X = 0, HX = 0

∑Y = 0, VX + 2x = 11, ⇒ VX =11 - 2x



Take moment about the “cut”,

11x – 2x(x)/2 – Mx = 0, ⇒Mx = 11x – x2

B

5m

8 kN

2 kN/m

11 kN

x

HX

V

MX

X

For 5 ≤ x ≤ 8m

∑X = 0, HX = 0

∑Y = 0, VX + 11 = 2x + 8, ⇒ VX = 2x - 3

Take moment about X,

11x – 2x(x)/2 – 8*(x-5) - Mx = 0, ⇒Mx = 11x – x2 –8x + 40

⇒Mx = 3x – x2 + 40

X (m) 0 1 2 3 4 5 5 6 7 8

V

(kN)

11 9 7 5 3 1 -7 -9 -11 -13

M

(kNM)

0 10 18 24 28 30 30 22 12 0



B

5m 3m

8 kN

2 kN/m

11 kN 13 kN

Shear Force (kN)

119

75

31

-7-9

-11-13

00

10

1824

28 30

12

22

Bending Moment (kNm)

A C

A

BC

CB

A

+ve

+ve



Example 4

Draw the shear force and bending moment diagrams for a cantilever beam

carrying a distributed load with intensity varies linearly from w per unit

length at the fixed end to zero at free end.

HA

VA

MA

B

w kN/m

X

wxl

Solution:

At any section distance x from the free end B,

Vx = l

wxx

l

wx

22

2

=

Mx = l

wxx

l

wx

632

32

−=

−



HA

VA

MA

B

w kN/m

X

wxl

wxl

Hx

V

Mx

xx

wl/2

V = wx /2lx2

Shear Force Diagram

-wl /6M = -wx /6lx

3 Bending Moment Diagram



Example 5

Draw the shear force and bending moment diagrams for the beam subjected

to a concentrated moment M* at point C.

HB VB VA

A BC M*

L

a b

Solution:

∑X = 0, HB = 0


VB * L = M*, ⇒ VB = M*/L (↑)

∑Y = 0, VB – VA = 0, ⇒ VA = M*/L (↓)

Take the left free-body (for 0 ≤ x < a)

A

x

M /L*

HX

V

MX

X

∑X = 0, HX = 0

∑Y = 0, VX = M*/L

Take moment about the cut section,

(M*/L)*(x) = MX (hogging moment)



Take the left free-body (for a < x ≤ L)

A C M*

a

x

V

MX

XM /L*

∑Y = 0, VX = M*/L

Take moment about the cut section,

(M*/L)*(x) + MX = M*

MX = M* - (M*/L)*(x) ⇒ MX = M*(L – x)/L



A BC M*

L

a bM /L* M /L*

M b/L

-M a/L

AB

C

-M /L*Shear Force Diagram

Bending MomentDiagram

*

*



Example 6

Draw the shear force and bending moment diagrams for the beam shown. C

is an internal hinge of the beam.

VE

HA

VA

MA

W W

L L L L

A B C DE

Solution:

This beam has four support reactions. We can use the equations of

equilibrium (3 nos.) together with the equation of condition (1 no. – internal

hinge at C) to find the support reactions. Therefore this is a determinate

beam.

Cut the beam into the left and right free-body diagrams.

VE

HA

VA

MA

W W

L L L L

A B C DE

HC

VC

H

V

C

C

Remember:

1. The internal forces at the hinge of the left and right free-body

diagrams should be equal but opposite in direction.

2. There is no bending moment at the internal hinge.



Consider the right free-body,

∑X = 0, HC = 0

Take moment about C,

VE*(2L) – WL = 0, ⇒ VE = W/2

∑Y = 0, VE + VC = W, ⇒ VC = W/2

Consider the left free-body,

∑X = 0, HC = HA = 0

∑Y = 0, VA = VC + W, ⇒ VA = 3W/2


MA = W*(L) + VC*(2L), ⇒ MA = 2WL



W W

L L L L

A B C DE

W/23W/2

2WL

3W/2

W/2 W/2

-W/2

A B

C

D E

AB

C

D E

-2WL

-WL/2

WL/2

Shear ForceDiagram

Bending Moment Diagram

3W/2



RELATIONSHIPS BETWEEN LOAD, SHEAR FORCE AND

BENDING MOMENT

Consider a beam element subjected to distributed load as shown below.

M

V

V+dV

M+dM

dx

q

∑Y = 0, V = q(dx) + (V+dV)

∴ qdx

dV−= ⇒ ∫ ∫−=

B

A

B

A

qdxdV

⇒ VB – VA = ∫−B

A

qdx

= - (area of load-intensity diagram between points A and B)

∑M = 0, -M + q(dx)(dx)/2 + (M+dM) - V(dx) = 0

Ignore the higher order terms,

we get Vdx

dM= ⇒ ∫ ∫=

B

A

B

A

VdxdM

⇒ MB – MA = ∫B

A

Vdx

= area of shear force diagram between points A and B.



SUMMARY OF THE RELATONSHIPS BETWEEN LOADS, SHEAR

FORCE AND BENDING MOMENTS

Slope of shear force diagram at a

point

=

Intensity of distributed load at

that point

Change in shear between points

A and B

=

Area under the distributed load

diagram between points A and B.

Slope of bending moment

diagram at a point.

=

Shear at that point

Change in bending moment

between points A and B.

=

Area under the shear force

diagram between points A and B.

Concentrated Loads

Change in shear at the point of

application of a concentrated

load.

=

Magnitude of the load.

Couples or Concentrated Moments

Change in bending moment at the

point of application of a couple.

=

Magnitude of the moment of the

couple.



SHAPES OF SHEAR FORCE AND BENDING MOMENT

DIAGRAMS

A. Beams under Point Loads only

1. Shears are constant along sections between point loads.

2. The shear force diagram consists of a series of horizontal lines.

3. The bending moment varies linearly between point loads.

4. The bending moment diagram is composed of sloped lines.

B. Beams under Uniformly Distributed Loads (UDL) only

1. A Uniformly Distributed Load produces linearly varying shear

forces.

2. The shear force diagram consists of a sloped line or a series of

sloped lines.

3. A UDL produces parabolically varying moment.

4. The bending moment diagram is composed of 2nd-order

parabolic curves.

C. Beams under General Loading

1. Section with No Load:

Shear force diagram is a Horizontal Straight Line.

Moment Diagram is a Sloping Straight Line.

2. At a Point Load:

There is a Jump in the Shear Force Diagram.

3. At a Point Moment:

There is a Jump in the Bending Moment Diagram.

4. Section under UDL:

Shear Force Diagram is a sloping straight line (1st order)

Bending Moment Diagram is a Curve (2nd order parabolic)

5. Section under Linearly Varying Load

Shear Force Diagram is a Curve (2nd order)

Bending Moment Diagram is a Curve (3rd order)



6. The Curve of the Bending Moment Diagram is 1 order above

the Curve of the Shear Force Diagram.

7. Maximum and Minimum Bending Moments occur where the

Shear Force Diagram passes through the X-axis (i.e. at points of

zero shear) (This characteristics is very useful in finding Max.

and Min. bending moment.)



Example 7

Draw the shear force and bending moment diagrams for the beam shown.

HA V

A VF

AB C D E

F

1.5m 3m1.5m 1.5m 1.5m

20 kN 40 kN

4 kN/m

Solution:

∑X = 0, HA = 0 kN


20*1.5 + 4*3*(3 + 3/2) + 40*7.5 – VF*9 = 0

⇒ VF = 42.7 kN

∑Y = 0, VA + VF = 20 + 40 + 4*3

⇒ VA = 29.3 kN



Shear Force and Bending Moment

AB C D E

F

20 kN 40 kN

4 kN/m

29.3 42.7

29.3 29.3

9.3

-2.7

-42.7 -42.7

0

2.325m

Shear Force (kN)

+44.0

+14.0 +10.8

-63.9

-4.0-0.9

44

58

68.8 67.9 63.9

0


A B C

D

E F

A B C D E F



Example 8

Draw the shear force and bending moment diagrams for the beam shown.

HB V

B VD

A B C D E

10 kN 20 kN

4 kN/m

2m 2m 4m 2m

Solution:

∑X = 0, HB = 0 kN


20*2 + 4*8*4 – 10*2 – VD*6 = 0

⇒ VD = 24.7 kN

∑Y = 0, VB + VD = 10 + 20 + 4*8

⇒ VB = 37.3 kN



Shear Force and Bending Moment

37.3

A B C D E

10 kN 20 kN

4 kN/m

24.7

-10 -10

27.3

19.3

-0.7

-16.7

8

0

Shear Force (kN)

-20

+46.7

-34.7

+8

1

4

1

4

1

4

-20

+26.7

-8

0

10

1


A B C D

E

E D

C

BA



Deflected Shape of a Beam

The qualitative deflected shape (also called “elastic curve”) of a beam is

simply an approximate and exaggerated sketch of the deformed beam due to

the given loading. The deflected shape is useful in understanding the

structural behaviour.

Sketching the Deflected Shape of a Beam

1. The deflected shape must be consistent with the support conditions:

(a) At a Roller Support, the vertical deflection is zero but the beam

may rotate freely.

(b) At a Pin Support, the vertical and horizontal deflections are zero

but the beam may rotate freely.

(c) At a Fixed Support, the vertical and horizontal deflections are zero

and there is no rotation.

2. The deflected shape must be consistent with the Bending Moment

Diagram.

(a) Where the moment is positive, the deflected shape is concave

upwards ( ∪ ).

(b) Where the moment is negative, the deflected shape is concave

downwards ( ∩ ).

3. The transition points between positive and negative moment regions are

points of zero moment. These points are called “point of inflection” or

“point of contraflexure”.

4. The deflected shape must be a smooth curve except at internal hinges.

Normally, the vertical deflection at an internal hinge is not zero.

5. Quite often it is possible to sketch the deflected shape of a structure first

and then to infer the shape of the bending moment diagram from the

sketch. This is useful for checking whether a bending moment diagram

obtained through calculations is correct.



Deflected Shape

A B C

AB

C

Bending Moment

Deflected Shape

+ve moment -ve moment

point of inflection



Examples of Beam Deflected Shape












Example 9

Construct the complete shear force and bending moment diagrams, and

sketch the deflected shape for the beam shown.

30 kN/m

10 kN20 kN

2m 2m5m

A B C D

Solution:

VH

B VB C

30 kN/m

10 kN20 kN

2m 2m5m

A B C D

∑X = 0, HB = 0 kN


20*7 + 30*7*(3.5 – 2) + (30*2/2)*(5 + 2/3) - 10*2 – VC*5 = 0

⇒ VC = 121 kN

∑Y = 0, VB + VC = 10 + 20 + 30*7 + 30*2/2

⇒ VB = 149 kN



-10

-70

50

D

Shear Force (kN)

A

B C

79

-71

20

2.63m


A B C D

2.63 m

-80-60

24

30 kN/m

10 kN20 kN

A B C D

Deflected Shape

P.I. P.I.



Example 10


sketch the deflected shape for the beam shown. B is an internal hinge of the

beam.

20 kN/m

50 kN

6m 6m 2m

A B C D

Solution:

VC

HA

VA

MA

20 kN/m

50 kN

6m 6m 2m

A B C D

Consider the free-body diagram BCD,



VC

HA

VA

MA

20 kN/m

50 kN

A

B C D

B

H

V

B

B

H

V

B

B

Free-body Diagram

∑X = 0, HB = 0 kN


20*6*3 + 50*8 - VC*6 = 0

⇒ VC = 126.7 kN

∑Y = 0, VB + VC = 20*6 + 50

⇒ VB = 43.3 kN

Consider the free-body diagram of AB

∑X = 0, HA = HB = 0 kN


43.3*6 – MA = 0

⇒ MA = 259.8 kNm

∑Y = 0, VA – 43.3 = 0

⇒ VA = 43.3 kN



43.3 43.350 50

-76.7

A B C D

2.165 m

Shear Force (kN)



A B C D

2.165 m


-259.8

-100

46.9

20 kN/m

50 kN

A B C D

Deflected Shape

pt. of inflection



Example 11


sketch the deflected shape for the beam shown. Joint C is an internal hinge.

30 kN/m100 kN

10 m 5m 3m 3m

A B C DE

Solution:

VH

A VA E VB

30 kN/m100 kN

10 m 5m 3m 3m

A B C DE



Consider the free-body diagram of CDE

VH

A VA E VB

30 kN/m100 kN

A B C DEC

Hc

Vc

Hc

Vc

By symmetry, VC = VE = 100/2 = 50 kN

∑X = 0, HC = 0 kN

Consider the free-body diagram of ABC,

∑X = 0, HA = HC = 0 kN


50*15 + 30*15/2*5 – VB*10 = 0

⇒ VB = 187.5 kN

∑Y = 0, VA + VB = 50 + 30*15/2

⇒ VA = 87.5 kN



Determination of the position of zero shear force

87.5

30 kN/m

A

187.5

B C

50 kN

P P

X

2X kN/m

MM

Let P be the position where the shear force equals to zero.

Consider the vertical equilibrium of the right hand side diagram,

2X (X/2) + 50 – 187.5 = 0

∴ X = 11.73 m

Determination of the value of maximum bending moment

Consider the right hand side free body,




2X (X/2)*(X/3) + 50X – 187.5 (X – 5) + M =0

X3/3 – 137.5X + 937.5 + M = 0

M = -11.733/3 + 137.5*11.73 – 937.5

M = 137.4 kNm.

Determination of the position of zero bending moment

Consider the free-body diagram below,

187.5

B C

50 kN

Q

X

2X kN/m

Take moment about Q,

2X (X/2)*(X/3) + 50X – 187.5 (X – 5) =0

X3/3 – 137.5X + 937.5 = 0

X = 8.11 m



A B C

D E

87.5

-112.5

7550

-50 -50

Shear Force (kN)

11.73 m

-291.7

150137.4 8.11 m

11.73 m

A

B C

D E


30 kN/m100 kN

A B C DE

Deflected Shape

P.I.



Example 12


sketch the deflected shape for the beam shown. Joints D and F are internal

hinges.

AB

C

D E F GH

120 kN150 kN

10 kN/m

5m 5m 2m 2m3m 3m 6m

3

45

Solution:

Resolve the inclined external load into vertical and horizontal components.

HA V

A VH

VG

VC

AB

C

D E F GH

120 kN

10 kN/m

5m 5m 2m 2m3m 3m 6m

120 kN

90 kN



Break the beam into three free-body diagrams, namely ABCD, DEF and

FGH.

D E F

120 kN

90 kN

HA V

A VH

VG

VC

AB

C

D F GH

120 kN

10 kN/mHD

VD

H

VD

D

H

VF

F

HV

F

F

Consider the free-body FGH first,

∑X = 0, HF = 0

Consider the free-body DEF,

By symmetry, VD = VF = 120/2 = 60 kN

∑X = 0, HD = HF +90 = 0 + 90 = 90 kN.

Consider the free-body ABCD,

∑X = 0, HA = HD , ⇒HA = 90 kN.


120*5 + 60*12 = VC*10, ⇒VC = 132 kN.

∑Y = 0, VA + VC = 120 + 60 , ⇒VA = 48 kN.

Consider the free-body FGH ,

Take moment about H,

60*8 + 10*8*4 = VG*6, ⇒VG = 133.3 kN.

∑Y = 0, VG + VH = 10*8 + 60 , ⇒VH = 6.7 kN.



A B

C

D E F GH

48 48

-72 -72

60 60

-60 -60 -80

53.3

-6.7

5.33 m

Shear Force (kN)

A

B

C D

E

F GH

5.33 m


240

-120

180

-140

2.2

A

B

C D

EF G

H

120 kN150 kN

10 kN/m3

45

Deflected Shape



PRINCIPLE OF SUPERPOSITION

The principle of superposition states that on a linear elastic structure, the

combined effect (e.g., support reactions, internal forces and deformation) of

several loads acting simultaneously is equal to the algebraic sum of the

effects of each load acting individually.

There are two conditions for which superposition is NOT valid.

1. When the structural material does not behave according to Hooke’s law;

that is, when the stress is not proportional to the strain.

2. When the deflections of the structure are so large that computations

cannot be based on the original geometry of the structure.



Principle of Superposition

HA

VA

MA

P1P2

w kN/m

L L

B.M. due to P1

B.M. due to P2

B.M. due to w

2P L

P L

2wL

1

2

2

m1

m2

m3

(a)

(b)

(c)

m1 + m2 + m3

2P L1 P L2 2wL2+ +

(d)

+

+

Complete bending

moment diagram



SUPERPOSITION

4 kN/m 40 kNm

5 kN

15

77.5

2.5m 2.5m

4 kN/m

10

12.5

5 kN

5

25

40 kNm40

+

+

-77.5

-12.5

-52.5

-25

-40 -40

-12.5

-12.5

+

+

Superposition of loading Superposition of

bending moment (kNm)



Bending Moment Diagrams by Parts

4m 2m

5 kN/m

10 kN

25 kN5 kN

5 kN/m

10 kN10 kN

10 kN

15 kN5 kN

+

10 kNm

-20 kNm

-20 kNm

10 kNm -20 kNm

2.5 kNm

+

Bending Moment



Tutorial 2 (Analysis of Beams)

Find the support reactions, draw the shear force and bending moment

diagrams, and sketch the deflected shapes of the beams shown below.

Q1.

8m 8m 4m

A B C D

15 kN/m

(internal hinge)

Q2.

200 kN

40 kN/m

A B C D

5m 12m5m

C is an internal hinge

Q3.

3m 3m 4m 4m 8m

100 kN 150 kN

25 kN/m

AB C D E F




Q4.

15 kN/m10 kN/m

100 kN

6m 3m 3m 3m 3m 6m

AB C D E F G

C & E are internal hinges

Q5.

150 kN 60 kN 80 kN 30 kN

4m 4m 4m 4m 4m 4m

A B

C

D E FG

E is an internal hinge

Q6.

2.5m 2.5m 2m 4m

20 kN

3 kN/m

AB

C

DE

D is an internal hinge



Q7.

3m 6m 6m3m 3m 3m

100 kN 50 kN 50 kN 20 kN

AB

C

D E

F

G

D is an internal hinge

Q8.

3m 2m 2m 2m

5 kN/m

5 kN 4 kN 3 kN

A B C D E

B is an internal hinge

Q9.

6 kN/m

3 kN/m

5m 2m 2m5m

AB C D E


Documents

INTERNAL FORCES IN A BEAM - Vocational Training Counciltycnw01.vtc.edu.hk/cbe2027/SA1_BEAM.pdf · CBE2027 Structural Analysis I Chapter 2 – Analysis of Determinate Beams HD in Civil