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EQUILIBRIUM OF FORCES “A body subjected to a force system whose resultant is zero, is said to be under equilibrium condition” Condition of static equilibrium for different force system. TYPES: COPLANAR CONCURRENT FORCE SYSTEM: The sum of components of forces along each of any two mutually perpendicular axes should be zero. σFx=0, σFy=0 COPLANAR NON-CONCURRENT FORCE SYSTEM: Equilibrium condition of a non-concurrent force system acting on a body requires the arrest of possible translational and rotational motion of the body. The resultant rand resultant moment m of the system of forces should be zero for the coplanar non-concurrent for system to be in equilibrium. σDx=0, σFy=0, σm=0 LAMI’S THEOREM “The Lami’s theorem states that, if a body is in equilibrium condition under the influence of three concurrent forces acting in a plane, the magnitude of each force is proportional to the sine of angle between the other two forces.” Fig a Fig b c (180 0 - α) (180 0 -β) (180 0 -γ) 15 α β γ

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### Transcript of Equilibrium of Forces and Supports

EQUILIBRIUM OF FORCES

“A body subjected to a force system whose resultant is zero, is said to be under equilibrium condition”

Condition of static equilibrium for different force system.

TYPES:COPLANAR CONCURRENT FORCE SYSTEM:

The sum of components of forces along each of any two mutually perpendicular axes should be zero.

σFx=0, σFy=0

COPLANAR NON-CONCURRENT FORCE SYSTEM:Equilibrium condition of a non-concurrent force system acting on a body requires the arrest

of possible translational and rotational motion of the body.The resultant rand resultant moment m of the system of forces should be zero for the

coplanar non-concurrent for system to be in equilibrium.

σDx=0, σFy=0, σm=0

LAMI’S THEOREM“The Lami’s theorem states that, if a body is in equilibrium condition under the influence of

three concurrent forces acting in a plane, the magnitude of each force is proportional to the sine of angle between the other two forces.”

Fig a Fig bc

(1800- α) (1800-β) (1800-γ) a b

PROOF:Consider a body subjected to three concurrent forces F1,F2 and F3 as shown in fig . The body

is in equilibrium condition under the action of three forces.Draw the forces F1,F2 and F3 in magnitude and direction, taken in order.

The diagram closes resulting in a triangle as shown in fig a

Applying the sine rule for the triangle

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α

βγ

ab bc ca ______________ = _____________ = ______________

Sin(1800-α) Sin(1800-β) Sin(1800-γ)

From triangle abc we can find ab=F1, bc=F2 and ca=F3

F1 F2 F3 ______________ = _____________ = ______________

Sin α Sin β Sin γ

A string is subjected to the forces 4KN and p as shown in fig. Determine the magnitudes of p and the tensions induced in various portions of the string.

D

A 600

450 Using Lami’s Theorem

CB 300

p4KN

TAB TBC

1050 TCD

4K TBC P

F.B.D at B F.B.D at C

Lami’s Theorem is given by

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F1 F2 F3 _____________ = _____________ = ______________

Sin α Sin β Sin γ

Apply to point B

TAB TBC 4 ___________ = _____________ = ______________

Sin 1200 Sin1350 Sin1050

TAB = 3.58 k N TBC = 2.92 k N

Apply to point C

TBC TCD P___________ = _____________ = ______________

Sin 1500 Sin600 Sin1200

TCD = 5.05 k NP = 2.92 k N

An electric fixture weighing 150N is supported between wall and roof by two wires as shown in fig. Determine the tensile stress developed in the wires.

C TAO

600 TBO

A 450

1500

150N

Using Lami’s Theorem

T1 T2 150 ___________ = _____________ = ______________

Sin 1500 Sin1350 Sin750

150 T1 = _____________ * Sin 1500

Sin750

T1 = 77.65 N 150

T1 = _____________ * 1350

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Sin750

T2 = 109.8 N

Two identical rollers each weighing 200 N are placed in a trough as shown in fig. Assuming all contact surfaces are smooth, find the reactions developed at contact surface A, B, C and D.

200 N200 N

Assume radius of roller = 200mm

DA

C

F.B.D 200N

300 600

RB RA

200 N RB

RD

600

RC

FIG I

σFx = 0, RBCos300 – RACos600 = 0Cos 600

RB = RA __________ , RB = 0.577RA ------------ 1 Cos 300

σFy = 0 , RBSin300 – RASin600 - 200 = 0 -------------- 2

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B I

1 in 2

0.577RA(Sin 300) + RA(Sin600) = 200 RA = 173 N RB = 99.8 N

FIG II

σFx = 0, RD – RBCos300 – RCCos600 = 0 RD – 99.8Cos300 – RCCos600 = 0 RD – 86.4 – RCCos600 -------------- 3

σFy = 0, RCSin600 – RBSin300 – 200 = 0 RC = 288 N -------------- 4

4 in 3

RD = 86.4 + 288 * Cos600

RD = 230.4 N

RA = 173 NRB = 99.8 NRC = 288 NRD = 230.4 N

SUPPORTS AND BEAMS

Beam: A beam is a structural element which has considerable length compared to its c/s dimensions and is supported at a few points.

It is subjected to transverse loadsThe support reactions adjust themselves to keep the beam in equilibrium.If the support reactions can be determined using the equations of equilibrium, then the beam is said to be statically determinate beam.

Types of supports

4 types of supports

A B1. Simple Support

RA

A B A B A B

2. Roller Support RA

RA RA

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HA A B HA A B HA A B3. Hinge or Pinned Support

RA RA

RA

HA A B4. Fixed Support

MA RA

TYPES OF BEAMS:

Depending on types of supports, beams may be classified as

A B1. Cantilever Beam

A B

2. Simple Supported Beam

A B3. One End Hinged and other One Roller Support Beam

A B4. Overhanging Beam

A B4. Overhanging Beam

A B

4. Overhanging Beam Y

A B C D E F

. Overhanging Beam

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Except lost one all are statically determinate beam, because the unknown RA, RB, HA, HB, MA, MB are deermined by using 3 equations of equilibrium that is σFx = 0, σFy = 0 and σMA or MB = 0

L/2

L

RA RB

RA + RB = W HA = 0

WLRB * L = ______

2

W RB = ______

2

W RA = ______

2

Sw KN/m or N/mmHA

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RA RB

L

W = W * L HA

L/2

RA RB

RA + RB = W

W RB = ______

2

W RA = ______

2

WkN/m

L

WHA

2/3 L RB

RA

1 W = ______ * L * W

2

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RA + RB = W

2 RB * L –W * ______ * L = 0

3

2 RB = ______ * W

3

W RA = ______

3

Determine the reactions at A and B for the loaded beam shown in fig

141.4N 10N/mA 450 100N B

2m 2m 2m 2m

141.4Sin 450

10 * 2 100 N

HA

RA RB

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2m

2 + 2 +2 / 2 = 5

6m

9m

Equations of Equilibrium

σFx = 0σFy = 0ΣMA = 0

ΣFx = 0, HA – 141.4Cos450 = 0 HA = 99.98 N

ΣFy = 0, RA + RB – 141.4Sin450 – 20 – 100 = 0 RA + RB = 219.98 N

ΣMA = 0, 141.4Sin450(2) + 141.4Cos450(0) – 100 + 20 * 5 + 100 * 6 – RB * 8 = 0

199.96 + 0 – 100 + 100 + 600 = 8 RB

RB = 99.99N RA = 219.98 – 99.99 RA = 119.98 N

99.98 N

119.98 N 8m 99.99 N

Beam AB shown in fig, has hinged support at A and roller support at B. Determine the reactions developed at the support when the forces shown in the figure are acting.

40k N60K N

30k N/mHA 600

A B

1m

3m 1m 1m

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RA RB

W.K.T σFx = 0 ΣFy = 0 Equilibrium Equation ΣMA = 0

σFx = 0, HA – 6Cos600 = 0 HA = 30k N

1 σFy = 0, RA - ______ * 3 * 30 – 40 – 60Sin600 – RB = 0 2

RA + RB = 136.96k N

1 2σMA = 0, _____ * 2 * 30 * ______ * 3 + 40(4) + 60Sin600(5) - RB6

2 3

RB = 84.9 k NRA = 52.06 k N.

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