4 - Beam - Vocational Training Counciltycnw01.vtc.edu.hk/cbe5029/4 - Beam.pdf · Draw a free-body...

Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams

Page 4-1

BEAM

A horizontal or inclined structural member that is designed to resist forces acting to its axis is called a beam INTERNAL FORCES IN A BEAM Whether or not a beam will break, depend on the internal resistances built up in the beam and on the strength of material that the beam is made of. What are the internal resistences and how to find them? If the beam can still maintain in one piece without breaking under the actions of lateral loads, then the beam as a whole, and any part of it, must be in the condition of equilibrium. By means of equations of equilibrium for the plane structure, we should be able to find the internal resistance in a beam.

ΣFx = 0, ΣFy = 0 and ΣM = 0.

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 for Internal Forces in a Beam

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.


Page 4-2

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 MN

V

V

V

N 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.


Page 4-3

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.


-4


Page 4-5

Example 1 Find the shear and moment on sections at 2 m and 4 m from the left end for the following beam.

Solution To analyze internal forces in the beam, it will first be necessary to calculate the reaction at one support.

ΣMD = 0, VA*7 = 40*4 + 20*2

VA = 28.571 kN

ΣFx = 0, HA = 0

ΣFy = 0, VA + VD = 20 + 40

VD = 31.429 KN


Page 4-6

A free-body diagram of the part of the beam to the left of the section at 2 m is drawn to analyze the shear and moment on the section. At x = 2m,

ΣFy = 0,

V=28.571 kN ↓

ΣM0 = 0,

M = 28.571*2 M = 57.142 kNm (Sagging) A new free-body diagram is drawn to shown V and M on the section 4 m from right of the left end. Again, both V and M are drawn in the positive direction for internal shear and moment.


Page 4-7

At x = 4m,

ΣFy = 0,

28.571- 40 = V V = -11.429 kN ↑ (Negative sign indicates that the shear force acts

upwards direction.)

ΣM0 = 0,

-28.571*4 + 40*1 + M = 0 M = 74.824 kNm (Sagging)


Page 4-8

Example 2 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 kN4 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


Page 4-9

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

HB VB

4 kN/m10 kN

A B P HV

M

PP

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/m15 kN

6m 3mVC

HP

V

MP

PP


Page 4-10

∑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)

Take moment about P, 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.)


Page 4-11

Example 3 Determine the shear force V and the bending moment M at section P of the cantilever beam.

HA

VA

MAA

B P C 3m 3m4m

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


Page 4-12

Take moment about P, 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

Take moment about P, 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.


Page 4-13

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. Draftsman like 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.


Page 4-14

Example 4 Draw the shear force and the bending moment diagrams for the beam shown.

HA VA VC

B

5m 3m

8 kN

2 kN/mA C

Solution:

∑X = 0, HA = 0 Take moment about A, 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 kN x

HX

V

MX

X

For 0 ≤ x ≤ 5m

∑X = 0, HX = 0 ∑Y = 0, VX + 2x = 11, ⇒ VX =11 - 2x


Page 4-15

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


Page 4-16

B

5m 3m

8 kN

2 kN/m

11 kN 13 kN

Shear Force (kN)11 9 7 5

3 1

-7-9 -11

-13

0 010

1824 28 30

12

22

Bending Moment (kNm)

A C

AB

C

CBA

+ve

+ve


Page 4-17

Example 5 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

wxxl

wx22

2=⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

Mx = l

wxxl

wx632

32−=⎟

⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−


Page 4-18

HA

VA

MA

B

w kN/m

X

wxl

wxl

Hx

V

Mx

xx

wl/2

V = wx /2lx 2

Shear Force Diagram

-wl /6M = -wx /6lx 3 Bending Moment Diagram


Page 4-19

Example 6 Draw the shear force and bending moment diagrams for the beam subject to a concentrated moment M* at point C.

HB VB VA

A BC M*

La b

Solution:

∑X = 0, HB = 0

Take moment about A, VB * L = M*, ⇒ VB = M*/L (↑)

∑Y = 0, VB – VA = 0, ⇒ VA = M*/L (↓)

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

A

xM /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)


Page 4-20

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

A C M*

ax

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


Page 4-21

A BC M*

La b

M /L* M /L*

M b/L

-M a/L

AB C

-M /L*Shear Force Diagram

Bending MomentDiagram

*

*


Page 4-22

RELATIONSHIPS BETWEEN LOAD, SHEAR FORCE AND BENDING MOMENT Consider a beam element subject to distributed load as shown below.

M

V

V+dV

M+dM

dx

q

∑Y = 0, V = q(dx) + (V+dV)

∴ qdxdV

−= ⇒ ∫ ∫−=B

A

B

AqdxdV

⇒ VB – VA = ∫−B

Aqdx

= - (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

AVdxdM

⇒ MB – MA = ∫B

AVdx

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


Page 4-23

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.


Page 4-24

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)


Page 4-25

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.)


Page 4-26

Example 7 Draw the shear force and bending moment diagrams for the beam shown.

HA VA VF

AB C D E F

1.5m 3m1.5m 1.5m 1.5m

20 kN 40 kN4 kN/m

Solution: ∑X = 0, HA = 0 kN Take moment about A, 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


Page 4-27

Shear Force and Bending Moment

AB C D E F

20 kN 40 kN4 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

4458

68.8 67.9 63.9

0


A B CD

E F

A B C D E F


Page 4-28

Example 8 Draw the shear force and bending moment diagrams for the beam shown.

HB VB VD

A B C D E

10 kN 20 kN4 kN/m

2m 2m 4m 2m Solution: ∑X = 0, HB = 0 kN Take moment about B, 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


Page 4-29

Shear Force and Bending Moment

37.3

A B C D E

10 kN 20 kN4 kN/m

24.7

-10 -10

27.319.3

-0.7

-16.7

8

0

Shear Force (kN)

-20

+46.7

-34.7

+8

14

14

14

-20

+26.7

-80

101


A B C DE

E DC

BA


Page 4-30

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.


Page 4-31

Deflected Shape

A B C

A BC

Bending Moment

Deflected Shape

+ve moment -ve moment

point of inflection


Page 4-32

Examples of Beam Deflected Shape


Page 4-33



Page 4-34



Page 4-35



Page 4-36

Example 9 Construct the complete shear force and bending moment diagrams, and sketch the deflected shape for the beam shown.

30 kN/m

10 kN 20 kN

2m 2m5m

A B C D

Solution:

VHB VB C

30 kN/m

10 kN 20 kN

2m 2m5m

A B C D

∑X = 0, HB = 0 kN Take moment about B, 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


Page 4-37

-10

-70

50

D

Shear Force (kN)

AB C

79

-71

20

2.63m


A B C D

2.63 m

-80-60

24

30 kN/m

10 kN 20 kN

A B C D

Deflected Shape

P.I. P.I.


Page 4-38

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.


Page 4-39

Principle of Superposition

HA

VA

MA

P1P2w 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 bendingmoment diagram


Page 4-40


Page 4-41

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 ofbending moment (kNm)


Page 4-42

Bending Moment Diagrams by Parts

4m 2m

5 kN/m10 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


Page 4-43

Beam Deflection Beam deflection can be determined by using the following beam deflection table. The deflection of the beam is inversely proportional to the quantity EI, which is the flexural rigidity of the beam.


Page 4-44


Page 4-45


Page 4-46

Example 10 Use the methods of superposition to find the deflection at the free end of the following cantilever beam. EI of the beam is constant.


Page 4-47

Solution


Page 4-48

where WT = W1 + W2

From the deflection tables, WT = PL3/3EI + wL4/8EI WT = 2(9)3/(3EI) + 0.5(9)4/(8EI) WT = 896 /(EI) kN-m3

Example 11

Use the method of superposition to find the deflection at the middle of the following simply supported beam. EI of the beam is constant.


Page 4-49

Solution

10 kN

A B

3 m

y1

A B

y2

6 m

1.5 kN/m

Total deflection at the mid-span of the beam, yM

yM = y1 + y2

From the beam deflection tables,

yM = Pa(3L2 - 4a2) / 48EI + 5wL4 / 384 EI

= 10(3)[3(12)2 - 4(3)2] / 48EI + 5(1.5)(12)4 / 384EI

= 652/EI kN-m3


Page 4-50

Example 12

Determine the displacement at point C and the slope at the support A of the beam. EI of the beam is constant.

Solution


Page 4-51

Total deflection at the point C of the beam, yc

yc = y1 + y2

yc = PL3 / 48EI + 5wL4 / 768EI

yc = 8(8)3 / 48EI + 5(2)(8)4 / 768EI

yc = 139/EI kN-m3

Total slope at the support A of the beam, θc

θc = θ1 + θ2

= PL2 / 16EI + 3wL3 / 128EI

= 8(8)2 / 16EI + 3(2)(8)3 / 128EI

= 56/EI (Clockwise direction)


Page 4-52

Example 13

Use the method of superposition to find the deflection at the free end of

the following cantilever beam.


Page 4-53

Solution

The uniform loading produces a deflection at B and because the beam is

continuous and has a slope at B, an additional deflection occurs at C

equal to (L1-L2)θB as shown in the following figures.

LL 21

0.5 kN/m

CA B

-

y

y1

2

CA B

2 kN

y3

0

Total deflection at C = y1 + y2 + y3

= wL24 / 8EI + (L1 -L2 )*wL2

3 /6EI + PL13 / 3EI

= 0.5(6)4 / 8EI + 3*0.5*(6)3 /6EI + 2*(9)3 / 3EI

= 624/EI kN-m3


Page 4-54

Example 14

A cantilever beam is subject to a uniformly distributed load of 0.5 kN/m

over the length BC as shown in the following figure. If the flexural rigidity

(EI) of the beam is 6000 kN-m2, calculate the deflection at the free end C of

the beam.


Page 4-55

Solution

0.5 kN/m C

A

B

0.5 kN/m

CAB

0.5 kN/mC

AB

y1

y2

y3

0

Total deflection at C = y1 - y2 - y3

= 0.5(7)4 / 8EI - 0.5(2)4 / 8EI -5*0.5*(2)3 /6EI = 145.73/EI kN-m3

= 145.73/6000 m

= 0.024 m


Page 4-56


Page 4-57

4 - Beam - Vocational Training Counciltycnw01.vtc.edu.hk/cbe5029/4 - Beam.pdf · Draw a free-body...

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