equilibrium of two-dimensional systemspeople.tamu.edu/~mhaque/cosc321haque/pdf_a3.pdf · dr....

Dr. Mohammed E. Haque, P.E. Lecture Notes

COSC321Haque 1 PDF_A3 Equilibrium of Two-Dimensional Systems

Equilibrium of Two-Dimensional Systems

1. Equilibrium of a Particle

2. Equilibrium of Rigid Bodies

3. Equilibrium involving distributed loadings

4. Equilibrium involving Pulleys and Sheaves



Equilibrium of a Particle When the resultant of all concurrent forces acting on a particle is zero, the particle is in a state of equilibrium.

That means: ∑Fx = 0 ∑Fy = 0

or

Summation of Horizontal Forces ∑H = 0

Summation of Horizontal Forces ∑V = 0



Example 1:

∑Fx = 0 CAx = CBx � (CA/√2) = CB(3/5) � CA = CB(3/5)x√2 ….(1) ∑Fy = 0 CAy + CBy – 5 = 0; � (CA/√2) + CB(4/5) –5 =0 ………(2) Solve for CA and CB using Eqs 1 & 2: CB(3/5)x√2 /√2 + CB(4/5) = 5 � CB(3/5 + 4/5) = 5 � CB = 25/7 = 3.5714 k (Tension)

CA = (25/7)x (3/5)x√2 = 3.03046 k (Tension)

5 k

1

1 4

3

B

C

A

CB

C

CA

5 k

FBD of concurrent point C

CB

C

CA

5 k

FBD of concurrent point C

CBx

CBy

CAy

CAx



Equilibrium of Rigid Bodies A rigid body is in a state of equilibrium when the resultant of all external forces and moments acting on it is equivalent to zero.

That means: ∑Fx = 0 ∑Fy = 0 ∑M i = 0 (where i = any point on the rigid body)

or

Summation of Horizontal Forces is ∑H = 0

Summation of Horizontal Forces is ∑V = 0 Summation of Moments at any point (i) is ∑ Mi = 0 Example:

The cantilever beam is in the state of equilibrium, because ∑H = 0 � Support horizontal reaction, RH balances the horizontal component of Force, Fx ∑V = 0 � Support vertical reaction, Rv balances the vertical component of Force, Fy ∑M = 0 � Support Moment, M balances the moment due to Force, F

RH

RV M

Zero Rotation (Angle before bending remains unchanged after

θ

Fixed Support F

Fx

Fy



SUPPORT TYPES The three common types of support systems are

1. Roller 2. Pinned (Hinged) 3. Fixed.

1. Roller Support and Reaction The reaction force is always a single force that is perpendicular to the surface on which the roller rests. Roller supports are free to rotate and translate along the surface. The surface can be horizontal, vertical, or sloped. 2. Pinned (Hinge) Support and Reactions A pinned support can provide both vertical and horizontal reactions but not a moment. They will allow the structural member to rotate, but not to translate. 3. Fixed Support and Reactions Fixed supports can provide vertical and horizontal reactions as well as a moment. Fixed support is known as Rigid Support because it restrains both rotation and translation.



Rotation free (Angle before bending changed after bending; without support at B, the beam is unstable)

RH(A)

RV(A) RV(B)

A B θ

Simply supported beam

F

Pinned or Hinge Support

Roller Support

Pinned or Hinge Connection



RH

RV M

Zero Rotation (Angle before bending remains unchanged after bending)

θ

Fixed Support

Fixed Connection

F



Assume the following Sign convention:

• Horizontal Force (Reaction) towards Right is Positive

• Upward Vertical Force (Reaction) is Positive

• Counterclockwise Moment is Positive Example 2:

5k

10 ft

A

∑Fx = 0 � R A(H) = 0 ∑Fy = 0 � R A(V) –5 = 0 � R A(V) = 5 k (Upward) ∑M = 0 � MA – (5 x 10) = 0 � MA = 50 k-ft (Counterclockwise)

MA

RA(V)

RA(H)



Example 3:

5k

10 ft

A

4 3

(5k/5) x 3 = 3 k

(5k/5) x 4 = 4 k

6 ft MA

∑Fx = 0 � R A(H) – 3 = 0 � R A(H) = 3k (Right) ∑Fy = 0 � R A(V) – 4 = 0 � R A(V) = 4 k (Up) ∑M = 0 � MA – (4x 10) + (3 x 6) =0 � MA = 22 k-ft (Counterclockwise)

RA(H)

RA(V)

First, find the vertical and horizontal component of the force, 5k



Example 4:

5 k 10 ft

24 ft

A

B

6 k 5 ft

MA

RA(V)

RA(H)

∑Fx = 0 � R A(H) = 0 � R A(H) = 0 ∑Fy = 0 � R A(V) - 5 - 6 = 0 � R A(V) = 11 k (Up) ∑M = 0 � MA - (6x 5) + (5 x 10) = 0 � MA = -20 k-ft (Clockwise)



Example 5:

5 k 10 ft

24 ft

A

B

6 k 5 ft

MA

RA(V)

RA(H)

∑Fx = 0 � R A(H) + 2 = 0 � R A(H) = - 2 (Left) ∑Fy = 0 � R A(V) - 5 - 6 = 0 � R A(V) = 11 k (Up) ∑M = 0 � MA - (6x 5) - (2x24) + (5 x 10) = 0 � MA = 28 k-ft (Counterclockwise)

2 k



Example 6:

∑Fx = 0 � R A(H) = 0 � R A(H) = 0 ∑Fy = 0 � R A(V) - 5 = 0 � R A(V) = 5 k (Up) ∑M = 0 � MA + 50 = 0 � MA = - 50 k-ft (Clockwise)

5 k 10 ft

24 ft

A

B

5 k M= 5 x 10 = 50 k-ft

24 ft

A

B

MA

RA(V)

RA(H) MA

RA(V)

RA(H)



Simply Supported Beam (Loadings are symmetrical about mid-span vertical axis)

A B

P1 (kips)

L (ft)

w (kips/ft) downward

A B

L (ft)

RB2 =0.5 P1 RA2 = 0.5 P1

RB1 = wL/2

RB(V) = RB1 + RB2 + RB3 RA(V) = RA1 + RA2 + RA3

A B

L (ft)

P2 (kips)

RB3 = P2 RA3 = P2

P2 (kips) P2 (kips)

Fig 2

Fig 3

A B

L (ft)

w (kips/ft)

P1 (kips)

RA1 = wL/2

Fig 1

P2 (kips)

X X

0.5L

RA(H) = 0



Example 7: Simply Supported Beam (Loadings are symmetrical about mid-span vertical axis). Find Reactions.

A B

10 k

10 ft

2 k/ft (downward)

RB(V) RA(V)

5 k 5 k

2 ft 2 ft

5 ft

RA(H) = 0

∑Fx = 0 � RA(H) = 0 (there is no horizontal force on the beam) ∑Fy = 0 � RA(V) = RB(V) = 5 + (0.5 x 10) + (0.5 x 2 x 10) = 20 k (Up)



Simply Supported Beam (Loadings are NOT symmetrical about mid-span vertical axis) Example 8:

A B

10 ft

RB(V) RA(V)

5 k

2 ft

RA(H) = 0

∑Fx = 0 � RA(H) = 0 (there is no horizontal force on the beam) ∑MA = 0 � RB(V) x 10 – (5x2) =0 � RB(V) = 1 k (Up) ∑MB = 0 � -RA(V) x 10 + [5x(10-2)] =0 � RA(V) = 4 k (Up) Proof: ∑Fy = 0 � RA(V) + RB(V) - 5 = ? � 4 + 1 –5 =0



Example 9: Simply Supported Beam (Uniformly distributed continuous Loadings are NOT symmetrical about mid-span vertical axis) Find the reactions.

∑Fx = 0 � RA(H) = 0 (there is no horizontal force on the beam) ∑MA = 0 � RB(V) x 10 – (8x2) =0 � RB(V) = 1.6 k (Up) ∑MB = 0 � -RA(V) x 10 + [8x(10-2)] =0 � RA(V) = 6.4 k (Up) Proof: ∑Fy = 0 � RA(V) + RB(V) - 8 = ? � 6.4 + 1.6 – 8 =0

A B

10 ft

RB(V) RA(V)

2 k/ft x 4 ft = 8 k 4/2 = 2 ft

RA(H) = 0

4 ft

A B

10 ft

RB(V) RA(V)

2 k/ft

RA(H)

4 ft



Example 9: Simply Supported Beam (Uniformly distributed continuous Loadings are NOT symmetrical about mid-span vertical axis) Find the reactions.

∑Fx = 0 � RA(H) = 0 (there is no horizontal force on the beam) ∑MA = 0 � RB(V) x 10 – [10x(4/3)] =0 � RB(V) = 1.3333 k (Up) ∑MB = 0 � -RA(V) x 10 + [10x{10-(4/3)}] =0 � RA(V) = 8.6667 k (Up) Proof: ∑Fy = 0 � RA(V) + RB(V) - 10 = ? � 8.6667 + 1.3333 – 10 =0

A B

10 ft

RB(V) RA(V)

0.5(5 x4) = 10 k (1/3)(4) = 4/3 ft

RA(H) = 0

4 ft

A B

10 ft

RB(V) RA(V)

5 k

RA(H)

4 ft



Example 10. Calculate Reaction at A and B

6 panels @ 12 ft = 72 ft

A B

12

5

5’

10’

15’ C

D

E

Ay By

Bx

2k x(5/13)

2k x(12/13)

∑Fx = 0 � - BX + 2k(5/13) +2k(5/13) + 2k(5/13) = 0 � BX = 3x(2k)x(5/13) = 2.308k (Left) ∑MA = 0 � By(72) – 2k(13) – 2k(26) – 2k(39) = 0 � By = (26+52+78)/72 = 2.1667k (Up) ∑Fy = 0 � Ay + 2.1667 – 3[2k(12/13)] = 0 � Ay = 3.372k (Up)

6 panels @ 12 ft = 72 ft

A B

12

5

5’

10’

15’ C

D

E

2k

2k

2k

Ay By

Bx

13’

13’

13’



Equilibrium involving Pulleys and Sheaves Example 1:

W P P =W

W

2W

FBD of the Pulley



Example 2:

W=3k

P C

B A

W=3k

P

P

P

P

P

P P

C

A B

B A

FBD of the Pulleys

FBD of Pulley C: ∑Fy = 0 � 3P = 3k � P=1k FBD of Pulley A: ∑Fy = 0 � A=2P � A = 2(1k) = 2k FBD of Pulley B: ∑Fy = 0 � B=2P � B = 2(1k) = 2k



Example 3: What weight, W can be lifted by a pull of P=1k.

W=?

P=1k C

B A

W

P=1k

P

P

P

P

P P

C

A B

B A

FBD of the Pulleys

FBD of Pulley C: ∑Fy = 0 � W=3P = 3(1k) = 3k



Example 4: What weight, W can be lifted by a pull of P=1k.

W=?

P=1k

A

B

C

2P

W = 4P

P=1k

A

B

C

P

P

P

2P

2P

2P

FBD of the Pulleys

FBD of Pulley A: ∑Fy = 0 � W=4P = 4(1k) = 4k



Example 5: Calculate support reactions at A and B Solution: From FBD of Pulley: T=W=2k ∑Fx = 0 � Cx = 2k; ∑Fy = 0 � Cy = 2k; From FBD of the beam: ∑Fx = 0 � Ax + T – Cx = 0 � Ax +2 –2 = 0 � Ax=0 ∑MA = 0 � By(12) – T(2) – Cy(16) =0 � By(12) –2(2) – 2(16) =0 � By =36/12 =3k (up) ∑MB = 0 � Ay(12) –T(2) – Cy(4) =0 � Ay(12) -2(2) –2(4) =0 � Ay= 1k (Downward) Proof: Look at the original figure with support reactions ∑Fy = 0 � -Ay +By –W =? -1 +3 –2 =0

A B

6 ft

W=2k

2 ft

4 ft

Cable

6 ft

2 ft radius pulley

C

A B

6 ft

W=2k 2 ft

4 ft 6 ft

C

C

T=2 k

Cx

Cy

Cy

Cx

T=2 k

Ax

Ay By

equilibrium of two-dimensional systemspeople.tamu.edu/~mhaque/cosc321haque/pdf_a3.pdf · dr....

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