Equilibrium of Forces and Supports

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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 NONCONCURRENT FORCE SYSTEM:Equilibrium condition of a nonconcurrent 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 nonconcurrent 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) (1800β) (1800γ) 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(1800α) Sin(1800β) Sin(1800γ)
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
TYPES OF LOADSW
1. Point LoadHA A B
L/2
L
RA RB
RA + RB = W HA = 0
WLRB * L = ______
2
W RB = ______
2
W RA = ______
2
2 Uniformly Distributed Load (UDL)
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
3 Uniformly Varying Load (UVL)
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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