EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM

When a body is in equilibrium, it has neither translatory nor rotatory motion in any direction. Thus the resultant force R and the resultant couple M are both zero, and we have the equilibrium equations for two dimensional force system

Fx = 0; Fy = 0 Eq(1)

  M = 0These requirements are both necessary and sufficient conditions for equilibrium.

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Supports: A structure is subjected to external forces and transfers these forces through the supports on to the foundation. Therefore the support reactions and the external forces together keep the structure in equilibrium.

There are different types of supports. Some of them are  a) Roller Support b) Hinged or pinned support c) Fixed or built in support  Some supports are shown in the figure along with the reactions that can be mobilised.

Types of supports

 

Types of Supports Action on body

(a) Flexible cable ,belt ,chain, rope

BODYBODY

T

Force exerted by cable is always a tension away from the body in the direction of cable

(b) Smooth surfacesContact forces are normal to the surfaces

FF

(c) Roller support Contact force is normal to the

surface on which the roller moves. The reaction will always be perpendicular to the plane of the roller . Roller support will offer only one independent reaction component.(Whose direction is known.)

( d )pinned Support / hinged support

This support does not allow any translatory movement of the rigid body. There will be two independent reaction components at the support. The resultant reaction can be resolved into two mutually perpendicular components. Or it can be shown as resultant reaction inclined at an angle with respect to a reference direction.

RvR

Rh

θ

Rv

M

RH

(e) Fixed or Built-in Support

M

(contd .)This type of support not only prevents the translatory movement of the rigid body, but also the rotation of the rigid body. Hence there will be 3 independent reaction components of forces. Hence there will be 3 unknown components of forces, two mutually perpendicular reactive force component and a reactive moment as shown in the figure.

TYPES OF BEAMS A member which is subjected to predominantly transverse loads and supported in such a way that rigid body motion is prevented is known as beam. It is classified based on the support conditions. A beam generally supported by a hinge or roller at the ends having one span(distance between the support) is called as simply supported beam. A beam which is fixed at one end and free at another end is called as a cantilever beam.

(a) Simply supported beam

span

MA

VA

B

span

AHA

(b) Cantilever beam

If one end or both ends of the beam project beyond the support it is known as overhanging beam.

A cantilever with a simple support anywhere along its length is a propped cantilever.

(c) Overhanging beam (right overhang)

MA

VA

B

span

A

HA

(d) Propped Cantilever beam

A beam which is fixed at both ends is called a fixed beam.

A beam with more than one span is called continuous beam.

VC

(f) Two Span continuous beam

VAVB

HA HB

MA

VA

span

VB

(e) Fixed beam

MB

HA

Statically determinate beam and statically indeterminate beam:Using the equations of equilibrium given in Eq(1) ,if all the reaction components can be found out, then the beam is a statically determinate beam ,and if all the reaction components can not be found out using equations of equilibrium only, then the beam is a statically indeterminate beam.In the above fig (a),(b)and (c) are statically determinate beams, where as (d),(e) and (f) are statically Indeterminate beams.

If the number of reaction components is more than the number of non-trivial equilibrium equations available then such a beam is a statically indeterminate beam.

If the number of reaction components is equal to the number of non-trivial equilibrium equations available then such a beam is a statically determinate beam

If the number of reaction components is less than the number of non-trivial equilibrium equations available then such a beam is an unstable beam.

Determination of Beam reactions

Since three equilibrium equations are available, for a planar structure a maximum of three unknown independent reaction components can be determined using these equations. 

Step I: Draw the free body diagram of the structure showing the given loadings and the reactions at the supports. 

Step 2: Apply the equations Fx = 0, Fy = 0, M = 0.

Assuming some directions and senses for unknown forces and moments.  

Step 3: solve for unknown reactions. If any of them is positive, it is along the sense initially assumed while drawing the FBD. If it is negative, it is opposite to the initially assumed sense

Problems

A beam AB of span 12m shown in the figure is hinged at A and is on rollers at B. Determine the reactions at A and B for the loading shown.

(1)

AB

20kN 25kN 30kN

30 45

4m 3m 3m 2m

Problems

Fx = 0 HA – 25cos 30 – 30cos45 = 0

Fy = 0 VA – 20 – 25 sin30 – 30sin45 +VB = 0

MA = 0 -20×4 - 25 sin30×7 - 30 sin 45×10+ VB ×12=0

Solution

HA B

20kN 25kN 30kN

30 45

4m

3m

3m 2m

VA VB

FBDFBD

Problems

RA= 48.21 kN

Solution(contd.)

VA

HA

RA

22

AAA VHR

HA=42.86kN, VA=22.07kN, VB=31.64kN 

= 27.25

A

A

H

V1tan

(2) Find the Support reactions for the given beam loaded as shown in the figure.

60°

2m

40kN/m

A

60kN0

.5m

5m

1 m

B

Solution

[Ans: RB=140kN VA=10

HA=61.24 RA= 62.05kN

= 9.3]

60°2m

40kN/m

A60kN

1m

B

RBH=RBCos30

RB

RBv = RBCos60

C

HA

VA

Fx = 0 HA + 60 – RB Cos30 = 0

Fy = 0 VA + RB Cos60 – 40 x 2 = 0

MA = 0 -30 - 40×2×4 + RB Cos60×5 = 0

30kNm

HA

VA

Solution

FBD

2m

RA

(3) Find the Support reactions for the given beam loaded as shown in the figure.

80kN/m

100kN

3m1m

30kN 0.5m

2m

AB

Solutions

80kN/m

100kN

3m1m

30kN

2m

A

B

VA

HA

VB

15kNm

FBD

[ Ans: VB= 112.5kN VA =37.5kN HA= – 100kN

RA= 106.8kN = 20.56]

1m

120kN

A

6 m

C B

15kNm

30kN

2m

RA

HA

VA

Fx = 0 HA + 100 = 0

Fy = 0 VA + VB – 30 –120 = 0

MA = 0 - 30×2 - 15 - (120)x5 + VBx6 = 0

HA

VA

VB

100kN

FBD

(4) Find the Support reactions for the beam loaded as shown in the figure.

3m 2m 2m

20kN23kN30kN15kN/m

Fx = 0 HA = 0

Fy = 0 VA –45 –30 –23 –20 = 0

MA = 0 MA –45x1.5 –30x3 –23x5 –20x7=0 [ Ans: VA = 118kN

MA =412.5kNm]

Solution

2m 2m

20kN23kN30KN45kN

MA

VA

HA

1.5m 1.5mFBD

A

;

(5)

2m 3m 1m 2m

AC B

D

10KN/m

20KN/mFind reactions at A,B,C and D

Solution

2m 3m 1m 2m

A

C

B

D

10kN/m10kN/m

10kN/m

Rc

3m 1m 2m

Solution

2m 3mVA

C

Rc

40kN20kN

1.33m

VD

VB

2.0m

FBD of top beam

FBD of bottom beam

C D

A B

0.67m2m

Solution

For top Beam :

Fy = 0 Rc –40 –20+VD=0

MD = 0 -Rc × 6 +40 × 4 +20 × 3.33=0

Solving the above eqns

RC=37.77kN; VD=22.23kN

20kN

2m

0.6

7

RC VD

3.33m

40kN

(Contd.)For bottom beam :

Fy = 0 VA –37.77–VB=0

MB = 0 -VA× 5 +37.77 ×3=0

Solving the above eqns

VA=22.66kN; VB=15.10kN

2m 3m

RC=37.77kN

VAVB

(6) A ladder of length 5m has a weight of 200N. The foot of the ladder rests on the floor and the top of it leans against the vertical wall. Both the wall and floor are smooth. The ladder is inclined at 60 with the floor. A weight of 300N is suspended at the top of the ladder. Find the value of the horizontal force to be applied at the foot of the ladder to keep it in equilibrium.

FBD OF LADDER

300N

HB

VA

HA

200N2.

5m

2.5m

600

Solution

Fy = 0 VA – 200 – 300=0 ::VA=500N

MA = 0

HB x 5 sin60 – 200 ×2.5 cos 60 – 300 ×5cos60=0

:: HB=230.94N

Fx = 0 HA –HB=0

HA=230.94N(Ans.)

300N

HB

VA

HA

200N

2.5m

2.5m

600

Solution

(7) Find the reactions at the supports A and C of the bent 20

kN

/m

B C

3m

2m

A

Solution 2

0 k

N/m

B C

3m

2mVA

HA

FBD

X

Y

VC

Solution (contd.)

B C

3m

2m

VA

HA

FBD

60kN

Fx = 0 60 –HA=0

Fy = 0 VA+VC=0

MA = 0

VCx2-60 ×1.5=0 VC

Solving the above

Ans: VA = - 45kN

VC = 45 kN

HA = 60kN

FBD after finding reactns

RA=75 kN

36.90

B C

VA

HA

60kN VC

# ve sign for VA

indicates ,reaction is downwards and not upwards as assumed initially.

(8)    A roller (B) of weight 2000N rests as shown in the fig. on beam CD of weight 500N.Determine the reactions at C and D. Neglect the weight of beam AB.

C

D

BA

30°4m

1m

Solution: 2000N

RAB

R B

VD

FBD of Roller

D

30°

2.5m

1m

300

500N

Hc

Vc

1.5m

FBD of beam CD

Solution: 2000N

RAB

300 R BCD

FBD of Roller

FBD of Roller :

Fy = 0 RBCD cos 300 –2000=0

Fx = 0 RAB – RBCD sin 300 =0

Solving above eqns : RBCD=2309.4N;

RAB=1154.7N

For bottom beam :

Fy = 0 VD –500+Vc –2309.4cos30=0

MC = 0

-VD × 5cos30 + 500 × 2.5 × cos30-2309.4 × 1=0

Solving the above eqns: VD=783.33N; VC=1716.67N

Fx = 0

2309.4 sin 30 –HC =0 Hc=1154.7 N

FBD of beam CD D

30°

2.5m

1m

300

500N

Hc

Vc

1.5m

2309.4N

VD

(9)    Compute the reactions for the bent beam shown in the figure at A and F.

B C D

3m 3m4m

50 N/m

45°A

F

300Nm

4m

Solution

MF = 0 – VA × 14 +200 × 5 – 300=0

VA=50N

FX=0 HF=0

FY=0 VA +VF=200;

VF=200 – 50=150N18kN

9kN

26.570

RA=20.12kN

45°A

B C D

F

3m 3m4m 4mVA

FBD

200 N

2m

HF

VF

300Nm

(10) Determine the support reactions for the trees shown

4m 4m 4m

A 3KN

G3KN

F 3KN

B C D E

4m

4m 4m 4m

A 3KN

G 3KN

F 3KN

B

C D EHB

HA

VA

Solution

FBD

4m

MA = 0 HB × 4 – 3 × 4 – 3 × 8 – 3 × 12=0

HB=18kN

FX=0 : –HA+HB=0

HA=18kN

FY=0 VA –3 –3 –3=0;

VA=9kN

18kN

9kN26.570

RA=20.12kN

4m 4m 4m

A 3KN3KN

3KN

HB

HA

VA4m

4kN/m

12kN/m

(1)Find the reactions at A,B,C and D for the beam loaded as shown in the figure(Ans.RA=RB =34kN;RC=28.84kN;

MC=-140kNm ; θC=-33.69 ˚ )

Problems for practice

12kN/m

4kN/m

20 kN

30kN

1m 2m 1m 1m 2m 1m 1m 2m

AB

C

34

40kNm

(2)A uniform bar AB of weight 50N shown in the figure supports a load of 200N at its end. Determine the tension developed in the string and the force supported by the pin at B.(Ans. T=529.12N;RB=807.15N, θB=64.6˚)

2.5m2.5m

200N

2.5m

A

B

60˚

string

(3)Find the position of the hinged support (x),such that the reactions developed at the supports of the beam are equal..

(Ans.x=2m.)

2.0m 1.4m1.0m 3.0m0.6

15kN18kN/m

10kN/m

x

(4)A right angled bar ABC hinged at A as shown in fig carries two loads W and 2W applied at B &C .Neglecting self weight of the bar find the angle made by AB with vertical(Ans:θ =18.44˚)

0.5L2W

θ

B

L m

WC

A