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### Transcript of Equilibrium of Forces and Supports

EQUILIBRIUM OF FORCESA 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. Fx=0, Fy=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. Dx=0, Fy=0, m=0 LAMIS THEOREM The Lamis 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 (1800- ) (1800-) a (1800-) 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

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Applying the sine rule for the triangle ab______________

bc_____________

ca______________

= Sin(1800-)

= Sin(1800-)

Sin(1800-)

From triangle abc we can find ab=F1, bc=F2 and ca=F3 F F F3______________

1 ______________

=

2 _____________

Sin

= Sin

Sin

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 450 C B 4KN 300 p 600 Using Lamis Theorem

TAB 1050

TBC TCD

4K

TBC P

F.B.D at B

F.B.D at C

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Lamis Theorem is given by F F F

1 _____________

=

2 _____________

=

3 ______________

Sin

Sin

Sin

Apply to point BAB ___________

T

= Sin 1200

BC _____________

T

4______________

= Sin1350

Sin1050

TAB = 3.58 k N TBC = 2.92 k N Apply to point CBC ___________

T

=

CD _____________

T

P =______________

Sin 1500 TCD = 5.05 k N P = 2.92 k N

Sin600

Sin1200

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 60 TBO0

TAO

A 450 1500 150N Using Lamis Theorem1 ___________

T

= Sin 1500

2 _____________

T

150 =______________

Sin1350

Sin750

150 T1 =_____________

* Sin 1500

Sin750 T1 = 77.65 N 150

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T1 =

_____________

* 1350

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 N 200 N B D A C F.B.D 300 RB 200 N RD 600 RC 200N 600 RA RB I Assume radius of roller = 200mm

FIG I Fx = 0, RBCos300 RACos600 = 0 Cos 600 __________ RB = RA , RB = 0.577RA ------------ 1 Cos 300 Fy = 0 , RBSin300 RASin600 - 200 = 0 -------------- 2

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1 in 2 0.577RA(Sin 300) + RA(Sin600) = 200 RA = 173 N RB = 99.8 N FIG II Fx = 0, RD RBCos300 RCCos600 = 0 RD 99.8Cos300 RCCos600 = 0 RD 86.4 RCCos600 -------------- 3 0 0 Fy = 0, RCSin60 RBSin30 200 = 0 RC = 288 N -------------- 4 4 in 3 RD = 86.4 + 288 * Cos600 RD = 230.4 N RA = 173 N RB = 99.8 N RC = 288 N RD = 230.4 N

SUPPORTS AND BEAMSBeam: 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 loads The 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 1. Simple Support RA B

A 2. Roller Support RA

B

A

B

A RA

B RA

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HA 3. Hinge or Pinned Support

A RA

B HA

A RA

B HA A RA

B

HA 4. Fixed Support

A

B

MA RA TYPES OF BEAMS: Depending on types of supports, beams may be classified as A 1. Cantilever Beam B

A 2. Simple Supported Beam

B

A 3. One End Hinged and other One Roller Support Beam

B

A 4. Overhanging Beam A 4. Overhanging Beam A 4. Overhanging Beam Y A B C D

B B

B

E

F

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. Overhanging Beam 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 Fx = 0, Fy = 0 and MA or MB = 0 TYPES OF LOADS W 1. Point Load HA AL/2

B

L RA RA + RB = W HA = 0 WL RB * L =______

RB

2

W RB =______

2 W RA = ______ 2

2 Uniformly Distributed Load (UDL) Sw KN/m or N/mm

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HA RA L RB

W=W*L HAL/2

RA RA + RB = W W RB =______

RB

2 W RA = ______ 2

3

L W HA2/3 L

RB

RA

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1 W=______

*L*W

2 RA + RB = W 2 RB * L W * 2 RB =______ ______

*L=0

3 *W

3 W RA =______

3

Determine the reactions at A and B for the loaded beam shown in fig141.4N 450 10N/m 100N

A

B

2m

2m 141.4Sin 450

2m

2m

10 * 2

100 N

HA

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RA2m 2 + 2 +2 / 2 = 5 6m 9m

RB

Equations of Equilibrium Fx = 0 Fy = 0 MA = 0 Fx = 0, Fy = 0, MA = 0, HA 141.4Cos450 = 0 HA = 99.98 N RA + RB 141.4Sin450 20 100 = 0 RA + RB = 219.98 N 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 N8m

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 N 60K N 30k N/m

HA A

600

B

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1m 3m 1m 1m

RA W.K.T Fx = 0 Fy = 0 MA = 0 Fx = 0,

RB

Equilibrium Equation

HA 6Cos600 = 0 HA = 30k N

1 Fy = 0, RA ______

* 3 * 30 40 60Sin600 RB = 0

2 RA + RB = 136.96k N 1 MA = 0,_____

* 2 * 30 *

______

2 RB = 84.9 k N RA = 52.06 k N.

2 * 3 + 40(4) + 60Sin600(5) - RB6 3

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