Application-Support Reaction in beams

MODULE – 3 SUPPORT REACTIONS

G. Ravindra Kumar, Associate Prof, Govt Engineering College, Chamarajanagar of 16

Application-Support Reaction in beams:

Types of Loads and Supports, statically determinate beams, Numerical

problems on support reactions for statically determinate beams with Point load

(Normal and inclined) and uniformly distributed and uniformly varying loads and

Moments.

Beam is a structural member resting on supports to carry vertical loads.

Generally, beams are placed horizontally. The amount and extent of external load

which a beam can carry depends upon

(a) The distance between supports and the overhanging lengths from

supports;

(b) The type and intensity of loading;

(c) The type of supports, and

(d) The cross-section and elasticity of the beam.

TYPES OF BEAMS:

Beams are classified as below:

i) Cantilever beam: One end is fixed and

the other is free.

ii) Simply supported beam: A beam supported

or resting freely bon the walls or columns at its

both ends is known as simply supported beam.

iii) Overhanging beam:

a) Simply supported beam with

single overhanging: A beam in

which a certain span length is

extended beyond the support on

one of its sides is called as beam with single overhanging.

b) Simply supported beam

with double

overhanging: A beam in

which a certain span length is extended beyond the support on both

sides is called as beam with double overhanging.

iv) Fixed beam: A beam whose ends are

rigidly fixed or built – in walls as shown in

fig is known as rigidly fixed beam or a built

in beam.

A B

L

A B

L1 L2

C

A B

L2 L3

C

L1

D

A B

L

A B

L



v) Continuous beam: A beam supported over more than 2 supports is

known as continuous beam.

vi) Propped cantilever beam: The cantilever

beam supported at free end.

TYPES OF LOADING:

Generally the beams are loaded with following types of loading.

Point Load or Concentrated Load

These loads are usually considered to be

acting at a point on the beam as shown in

Fig. is known as point load or a concentrated

load, and are denoted by Newton (N), Kilo-

Newton (KN) or ton (T), etc.,

Note: Practically point load cannot be placed

on a beam. When a member is placed on a beam it covers some space or width.

But for calculation purpose we consider as the load is transmitting at the central

width of the member.

Uniformly Distributed Load or U.D.L.

if the load on a beam is equally

distributed over a length of the beam in such a

way that rate of loading is uniform along the

length (i.e., load per unit length is a constant).

The rate of loading (udl) is as w Newton /

metre ( w N/m), w Kilo-Newton / metre

(w kN/m).

Gradually Varying Load or triangular

load:

If the spread load is uniformly varying

along the length from zero intensity at one

end to the designated intensity at the

other end. A triangular block of brickwork

practically imposes such a loading on a

beam.

w w

w/m

Gradually varing load

A B

L

A B C D



Trapezoidal Load or Uniformly varying g (UVL) :

This load is one which is spread over a beam

in a such a manner that rate of loading varies from

point to point along the beam as shown in fig.

TYPES OF SUPPORTS:

i) Simple support or knife edge support

ii) Roller support

iii) Hinged support

iv) Fixed support

v) Smooth surface support

Simple support or knife edge support:

If the beam rests simply on a support

then the support is simple support. A simple

support exerts reaction only in the direction

perpendicular to the axis of the beam as shown

in fig. VA and VB are the support reactions.

Roller support:

If the beam is supported on rollers, then

such a support is called as roller support. In

roller support the reaction always acts at right

angles or perpendicular to support surface. In

roller support the beam is free to roll left and

right or up and down.

Gradually varing load

A

w w

L

B

VA VB

Beam

Roller support

Support reaction

Beam

Roller support

Support

reaction

Beam



Hinged Support or Pin jointed support:

If the beam is supported on

hinge or pin then such a support is

called as Hinged or Pinned support. In

hinged support the beam cannot move

in any direction. In hinged beam the

reaction may be vertical (VA),

horizontal (HA) or inclined (R)

depending upon the type of loading.

Fixed or Built – in or Encastered Support:

If the end of the beam is fixed or

built – in, then such a support is called

as fixed support. In fixed support the

reaction may be vertical (VA), horizontal

(HA) or inclined (R) and in addition

there will be a moment (MA) acting at

fixed end as shown in fig.

VA

Support

reactions

Beam W1 W2

MA A

HA

R

Hinge or pin

support

Support

reactions

Beam W1 W2

VA

HA

R



Eg: Column supporting beam and slab in buildings. Here beam and slab forms a

joint and column is a support

Smooth Surface Support:

If the body is supported or in contact with a smooth surface, then such a

support is called as smooth surface support. In case cases, the reaction is

perpendicular to the support. In case of spheres the reaction acts perpendicular to

the surface and passes through the centre of sphere c as shown in fig.

Note:

1. If uniformly distributed and uniformly varying load are given convert them

into Concentrated (point) load. The converted concentrated load will act at

the centre of its length.

Concentrated load = 6 x 4 = 24 kN

2. If couple is given and moment about any point is to be calculated, don’t

multiply the magnitude with its distance from that point.

Smooth

surface

W

R

C

R

C

R

Smooth

surface

2m

5kN-m

A

10kN-m

15kN-m

1.5m 3m

B A

4 m

6 kN/m

B A

4 m

6 x 4 = 24kN

2 m

B A

6 x 4 = 24kN

2 m



M@A = + 5 – 10 – 15 = – 20 kN - m

Problem: ®

Draw SFD and BMD for the cantilever beam as shown in

fig.

Solution:

Reaction at support A = VA = 10 kN

Problem: ®

Draw SFD and BMD for the cantilever beam

as shown in fig.

Solution:

Reaction at support A = VA = 10 + 5 +15 = 30 kN

Problem: ®

Draw SFD and BMD. Locate the point of contra flexure for the cantilever beam

shown in fig.

Solution:

Reaction at support A = VA = 10 + 30 –

25 = 15 kN

Problem: ®

Draw SFD and BMD for the cantilever beam shown in

fig.

Solution:

Reaction at support A = VA = 6 x 4 = 24 kN

Problem:®

Determine the reactions at support for the

cantilever beam subjected to moment M at

free end.

A

10 kN

B 5m

B A 2 m 1 m 1.5 m

15 kN

C D

5 kN 10 kN

B A 2 m 2 m 2 m

25 kN

C D

30 kN 10 kN

B A

4 m

6 kN/m

B A

L M



May / June / 2010 – 06CV33 –10

marks ®

Determine the reactions at support A.

Solution:

Reaction at support A = VA = 2 + 3 x 1 = 5

kN

0.5 m B A

1 m 1.5 m

2 kN

C D

3 kN-m

3 kN/m



SIMPLY SUPPORTED BEAM WITH A POINT LOAD AT CENTRE

Reaction calculation at supports A and B

Let VA and VB be the reactions at supports A and B.

Taking moment about support A

kN 2

W V

2

LWLV

B

B

kNL

W

L

W W V

kN WVV

A

BA

Problem:

Determine the support reactions at A and B. Solution: Reaction calculations at supports A and B. Let VA and VB be the reactions at

supports A and B.


kN 13.58

108 V

6104.58268V

B

B

Problem: ®

Draw SFD and BMD for the simply supported beam as shown in fig.

Solution:

Reaction calculations at supports A and B. Let VA and VB be the reactions at supports A and B.


kN 33.58

268 V

2682

885

6104.58268V

B

B

kN 30.533.564 V

kN 64851086VV

A

BA

A B

6 kN

C

8 kN 10 kN

D E

2.5 m 2 m 1.5 m 2 m

L/2

A B

W kN

L

L/2 C

5 kN/m

C D E A B

6 kN 8 kN 10 kN

2.5 m 2m 1.5 m 2 m



Problem: ®

Determine the reactions of the

simply supported beam subjected to

clockwise moment ‘M’ at mid span as

shown in fig.

Solution:

Reaction calculation at supports A and B



kN L

M V

MLV

B

B

kN L

M -V V

kN 0VV

BA

BA

Solution: check this once again

Reaction calculation at supports A

and B

Let VA and VB be the reactions at

supports A and B.


kNL

M

2

wL V

2

LLwMLV

B

B

kNL

M

2

wLV

2

wL

L

M wLVwLV

kNwL VV

A

BA

BA

A B

M kN-m

C L/2 L/2

L

B

C

w /m

A

M

L/2 L/2

kNL

M

2

wL



Problem:® Determine the reactions at the supports A and B. Solution: Bracket load of 8 kN acting at F with an offset of 0.5 m is converted into a point load = 8 kN and a anticlockwise couple of magnitude = 8 x 0.5 = 4 kN-m at ‘F’. Reaction calculations at supports A and B. Let VA and VB be the reactions at supports A and B.


kN 128

96 V

832

2256.5848V

B

B

kN 61218 V

kN 18258VV

A

BA

Problem: Calculate the support reactions for the simply supported beam as shown in fig. Solution:

The trapezoidal loading diagram is first divided into two parts.i.e., udl of area ABCD and triangular load EDC. The varying load and udl is converted into point load.

support.left from m 2.67 83

1at acting kN 28 78

2

1 load triangular of Area

span) (Mid m 4 2

8

2

Lat acting kN 40 85 loadpoint into Udl


Reaction calculations at supports A and B. Taking moment about support A

kN 29.338

234.67 V

83

128

2

8408 V

B

B

B

5 kN/m

C D A

8 kN - m

8 kN

E F

1.5 m 2 m 1 m 2 m 1 m 0.5 m

B

5 kN/m

8 m

A

12 kN/m

VA

B

5 kN/m

8 m

A

7 kN/m

5 kN/m

28 kN

40 kN

C D

E

VB

VA VB

B

8 m

A

28 kN 40 kN

2.67 m 4 m



kN 38.6729.3368 V

kN 682840VV

A

BA

June / July 2009 – 06CV33 - 14 marks ®

For the beam shown in fig, obtain the support reactions.

Solution:

Reaction calculation at supports A and D

Let VA and VD be the reactions at supports A and D.


kN 50 V

2

8850125016012016V

D

D

kN 405090 V

kN 908550VV

A

DA

Problem:

Find the support reaction at supports A and D for the force system shown in

fig.

Solution:

Reaction calculation at supports A and D

Let VA and VD be the reactions at supports A and D.


kN 42.5 V

2

22202005401008V

D

D

A

B

160 kN-m 120 kN-m

50 kN

C

D

5 kN/m

VA VD 8 m 4 m 4 m

A B

200 kN-m

100 kN-m

40 kN

C D

2 m 3 m 3 m

20 kN/m

VA VD



kN 37.542.580 V

kN 8022040VV

A

DA

PROBLEMS ON OVER HANGING BEAMS

Dec 2010 – 06CV33 – 12 marks ®

Draw the SFD and BMD for the beam loaded as shown in the fig.

Solution:

Reaction calculation at supports A and C

Let VA and VC be the reactions at supports A and C.


kN 6 V

2

33263.55V

C

C

kN 3.569.5 V

kN 9.5323.5VV

A

DA

Problems on beams with inclined loading:

Problem: ®

A horizontal beam 8 m long is hinged at support A and on is on roller at support

Band is loaded with oblique loads as shown in fig. Construct the axial thrust, Shear

force and bending moment diagrams.

Solution:

Resolving the inclined

forces both

horizontally and

vertically

Vertical components at C, D and E

VC = 3 sin 300 = 1.5 kN ↓

VD = 2 sin 450 = 1.414 kN ↓

VE = 4 sin 600 = 3.464 kN ↓

Horizontal components at C, D and E

HC = 3 cos 300 = 2.598 kN →

HD = 2 cos 450 = 1.414 kN ←

HE = 4 cos 600 = 2.0 kN ←

2 kN/m

A

3 m B

2 m 1 m

3.5 kN

C D

2 m 2 m 2 m 2 m

450 B

D A

E

3kN

C

2 kN 4 kN

600 300



Reaction calculation at supports A and B:

Let VA and VB be the vertical reactions at supports A and B.

Let HA be the horizontal reaction at support A.


kN 3.68 V

21.541.41463.4648V

B

B

kN 2.6983.686.378 V

kN 6.3783.4641.4141.5VV

A

BA

0.816 2 - 1.414 - 1.5 at A forceNet

0.816 at A reaction Horizontal

Problem:

Determine the reaction components for the loaded beam shown in fig

2 m 2 m 2 m 2 m

B D A E

1.5kN

C

1.414 kN 3.464 kN

2.598 kN 1.414 kN 2.0 kN

HA=0.816kN

VA = 2.698 kN VB = 3.68kN

1 m 1 m 2 m

B A

100kN

600

2

1

50 kN/m



Problem:

Calculate the reaction at the supports A and B. .

Solution:

Resolving the inclined force both horizontally and vertically

Vertical component at C

VC = 100 sin 600 = 86.60 kN ↓

Horizontal component at C:

HC = 100 cos 600 = 50 kN ←





kN 112.73 V

2

4430486.6906V

B

B

kN 93.87112.73206.6 V

kN 206.686.60430VV

A

BA

kN 50 at A forceNet

C

4 m 1 m 1 m

86.60 kN

50 kN HA=50 kN

VA = 93.87 kN VB =112.73 kN

A

B D

90 kN – m

30 kN/m

600

A B

90 kN – m 4 m 1 m 1 m

100 kN 30 kN/m

C

D

600 A B

C

D

90 kN – m 4 m 1 m 1 m

100 kN



kN 50 at A reaction Horizontal

Problem:

Find the reactions at the support

Solution:

Resolving the inclined force of 30 kN inclined at 300 with horizontal,

Horizontal force at D = 30 x cos 300= 25.98 kN

Vertical force at D = 30 x sin 300= 15 kN





kN 74.98 V

62

2220525.98406V

B

B

kN 974.9865.98 V

kN 65.9822025.98VV

A

BA

Horizontal reaction at A = 15 kN →

June / July 2017 – 08 marks

Find the support reactions for the beam loaded as shown in fig.

300 B

C

A

D E

30 kN 20 kN/m

2 m 2 m 3 m 1 m

40 kN-m

B

C

A

D E

20 kN/m

2 m 2 m 3 m 1 m

30 cos300=15kN

30 sin300 =25.98 kN

40 kN-m

HA = 15 kN

VA = -9 kN VB =74.98 kN

A B

C D

60 kN /m

6 m 2 m 2 m

30 kN 10 kN/m 20 kN

2 m

E



Problem:

For the beam loaded as shown in fig. find the magnitude of the load ‘P’ acting

at ‘E’ such that the reactions at supports A and B are equal.

Solution:

Reaction calculation at supports A and B and determination of point

load ‘P’.



1---- kN 135P 0.25 V

52

66205402P

2

22408V

B

B

2---- 0.5P120V

P2402V

V V

kN P24062040P240VV

B

B

BA

BA

Equating eq 1 and eq 2

0.25 P + 135 = 120 + 0.5 P

0.25 P = 135 – 120 = 15

kN 600.25

15P

kN 150 V

kN 150 60 0.5120 V

A

B

40 kN/m

A

2 m B

3 m

40 kN P = ?

C D

3 m

20 kN/m

3 m 2 m E F

8 m

Application-Support Reaction in beams

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