Theory Of Structures Problems Solved By Dk Mamonai

Theory of Structures

A. Supports :

Sir Dr.Zubair

Supports are provided to resist the loads.

There are three types of Supports 1. Hinge Support: It is always in clockwise or Anti-clockwise not horizontal or vertical

2. Fixed Support: Horizontal or Vertical or Clockwise

3. Circular or Roller Support: Always normal to the surface.

B. Loads :

Dead Loads

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Live Loads

Uniformity Disturbed Load

Moving Load

Uniformly Varying Load

Wind Load

Point Load

Snow Load

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Theory of Structures

Beams

Sir Dr.Zubair

:

(i) Determinate Beam or Simply Supported Beam:

(ii) Continuous Beam : One or more span is said as Continuous Beam.

(iii) Cantilever Beam : In Sindhi it is said as CHAJA!

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Theory of Structures

Sir Dr.Zubair

Problem:1 :

Determinate Forms

P

L

Free Body Diagram

P

Ma

L

Ra

Ha

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – P = 0 Ra= P

Moment, +ve Clockwise ∑ M = 0 Ma+ Ra * L =0 Ma + PL = 0 Ma = - PL

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Theory of Structures

Sir Dr.Zubair

Problem:2 :

Determinate Forms

P= 80lb

L = 10’

Free Body Diagram

P = 80 lb

Ma

L = 10’

Ra

Ha

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – P = 0 Ra= 80 lb

Moment, +ve Clockwise ∑ M = 0 Ma+ Ra * L =0 Ma + PL = 0 Ma = - 800 lb-ft ( Anti-clockwise )

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Theory of Structures

Sir Dr.Zubair

Problem:3 :

Uniformly disturbed Load

W v/v

L WL Free Body Diagram

L/2

A

L/2 C

Ma

B

L

Ra

Ha

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – WL = 0 Ra= WL

Moment, +ve Clockwise ∑ M = 0 at point C Ma+ Ra * L/2 =0 Ma = - ( Ra * L / 2 ) Ma = - ( WL * L / 2) Ma= - W L 2 / 2

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Theory of Structures

Sir Dr.Zubair

Problem:4 :

Uniformly disturbed Load

3 kip/m

2m 6 kip Free Body Diagram

1.5

A

1.5 C

Ma

B

2m

Ra

Ha

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – WL = 0 Ra= 6 Kip

Moment, +ve Clockwise ∑ M = 0 at point C Ma+ Ra * L/2 =0 Ma= - W L 2 / 2 Ma= - 3 (2)2 / 2 Ma= - 6 kip. m

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Theory of Structures

Sir Dr.Zubair

Problem:5:

60k

3

60k

4

6’

6’

Free Body Diagram

6’

60k

Fx

60k Fy

A Ma

C

D

53.30

Ha Ra

6’

B Rb

6’

6’

First of all changing Tangent into Degree Ф = Tan-1 4/3 = 53.30 Now, finding X-Y co-ordinates of the Inclined Force Fx= F CosФ = 60 Cos 53.3 = 60 (0.59)= 36K

;

Fy = F Sin Ф = 60 Sin 53.3 = 60 (0.801)=48.10K

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha – Fx = 0,

Ra + Rb – P - Fy = 0

Ha=Fx ,

Ra + Rb -60 -48 =0

Ha= 36 K,

Ra + Rb = 108 -------------------------(i)

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Theory of Structures

Sir Dr.Zubair

Moment, +ve Clockwise ∑ M = 0 at point A 60 * 6 + Fy * 12 – Rb *18 = 0 60 * 6 + 48 * 12 = 18 Rb Rb = 52 K Putting Value of Rb in Equation (i) Ra + Rb = 108 Ra + 25 = 108 Ra = 56 K

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Theory of Structures

Sir Dr.Zubair

Problem:6: 12 k

16k

16k

3 k / ft

2 K / ft

8’

4’

4’

4’

4’

6’

Free Body Diagram 16 k A Ra

C

12 k

16 k

D

E

16 k

9k

F

G

B

Ha 4’

Rb 4’

4’

4’

4’

4’

2’

2’ 2’

As there is V.D.L = 1/2 * 6 * 3 = 9 k

;

X = L / 3 = 6 / 3 = 2’

For U.D.L = 2* 8 = 16 K

;

X= L / 2 = 8 / 2 = 4’

Equilibrium, →+ve ∑ Fx = 0, Ha = 0,

↑+ve ∑ Fy = 0 Ra – 16k – 12k – 16k – 16k + Rb – 9k = 0 Ra + Rb = 69 --------------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point A (16 * 4) + (12 * 12) + (16 *16) + (16 * 20) - (Rb * 24) + (9 * 26)= 0 1018 – 24Rb = 0 Rb = 42.41 K Putting Value of Rb in Equation (i) Ra + Rb = 69 → Ra + 42.41 = 69 → Ra = 26.6 K 09CE37

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Theory of Structures

Sir Dr.Zubair

Problem # 7

Point Load

40 K

50K

30 K

3/4 A

E

C

4’

5’

F

D

3’

3’

G 8’

3’

H

B

3’

(PART 1 ) We Disturb this Diagram into two A Point to C Point Free Body Diagram

40 k

A

E

C

Ha

Hc 4’

Ra

5’ Rc

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha - Hc = 0,

Ra – 40 + Rc = 0

Ha= Hc

Ra + Rc = 40 --------------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point A -

Rc * 90 + 40 *4 = 0

0 = -9 Rc + 160 Rc = 18 K Putting Value of Rc in Equation (i) Ra + Rc = 40 → Ra + 18 – 40 = 0 09CE37

→ Ra = 22 K 10 | P a g e

Theory of Structures (Part 2)

Sir Dr.Zubair 50k

C

Hc

32 K

F 3’

30 K

D 3’

G 4’

4’

H 3’

B 3’

Ф = Tan-1 3/4 → 36.86 Fx = F CosФ → 50 * Cos 36.86 → 40K Fy= F SinФ → 50 * Sin 36.86 → 30 K

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Hc + Fx = 0,

- Rc – Fy + Rd – 32 -30 + Rb = 0

Hc = - 40 k

Rb + Rd = 110 --------------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point A -

Rc * 18 - Fy * 15 + Rd *12 – 32*8 – 30*2 = 0

12 Rd = 1090 Rd = 91 K Putting Value of Rd in Equation (i) Rb + Rd = 110 → Rb + 91 – 110 = 0

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→ Rb = 19 K

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Theory of Structures

Sir Dr.Zubair

Problem # 8 : Rk

8’

6k

6’

Free Body Diagram

48k

3 k /ft

2’ 6k

Rk

Ma

Ra

Ha

8’ 8’

8’ 6’

2’

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra -48 – 37 - 6 = 0 Rb + Rd = 143 --------------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point A Ma + Ra *16 – ( 48 + 37 ) * 8 - 6 * 2= 0 0 = Ma + 143 *16 – 1096 - 12 Ma = -1180 K. ft Hence the direction of Moment A should be taken as reversed.

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Theory of Structures

Sir Dr.Zubair

Problem # 9 :

R K /ft

A

B 20’

Free Body Diagram R*20 k Ha 6.66’

6.66’

Ra

6.66’ Ra

Point Load = 1/2 * 37 * 20 → 890 k Equilibrium, →+ve ∑ Fx = 0, Ha = 0,

↑+ve ∑ Fy = 0 Ra – 890 + Rb = 0 Ra + Rb = 890 --------------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point A 890 * 6.66 - Rb * 20 = 0 → 20 Rb = 5927.4 → Rd = 296.37 K Putting Value of Rd in Equation (i) Ra + Rb = 890 → Ra + 296.37 – 890 = 0

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→ Rb = 593.63 K

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Theory of Structures

Sir Dr.Zubair

Problem # 10: Find Unknown Reactions

Rk

8k

2.5 k/ft

4 k/ft

A

C

D

4’

Free Body Diagram

8’

10 k

Ra

A

Rk

E

Ma

B 6’

32 k

C

8k

D

B

Ha 2’

2’

4’

4’

6’

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – 10 - 37 – 32 -8 = 0 Ra = 139 k

Moment, +ve Clockwise ∑ M = 0 at point A Ma + Ra * 18 – 10* 16 – 37 * 14 – 32 * 10 = 0 → 0 = Ma + 2502 -1724 → Ma = -778 K . ft The direction should be reversed

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Theory of Structures

Sir Dr.Zubair

Problem # 11: 8K

RK

20 k

(0.2+R ) K

4’ 8’

5’

5’

9’

5’

Free Body Diagram: 8k

(R +0.2 )K

C

RK

20k

A

B

Rc

Rb Hc

Ra

Hb

4’ 8’ 9’

5’

9’

5’

32’

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Hb = 0,

– 8 – 160 – 37 + Ra – 20 + Rb = 0 Ra + Rb = 1722.6 -----------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point A - 8 * 28 - 1605.6 * 23 – 20 * 5 + Ra -24 = 0 → 24 Ra = 27252.8 → Rd = 1552.2 K Putting Value of Rd in Equation (i) Ra + Rb = 890 → Ra + 1552.2 – 1722.6 = 0 09CE37

→ Rb = 170.4 K 15 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem # 12: ( 10+ R/2) K/ft

20’ We break this triangle into two parts PART : 1

(10+R/2) K

Free Body Diagram (1/2 * 10 * [10+R/2] ) A Ra

V.D.L = 1/ 2 * 10 * 54.5 = 272.5 k

Ha 3.33’

3.33’

3.33’

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – 272.5 = 0 Ra = 272.5

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Theory of Structures

Sir Dr.Zubair

Part 2:

( 1/2 * 10 * [10+R/2] ) B

Rb 3.33’

3.33’

3.33’

V. D.L = 1/2 * 10 * 54.5 = 272. 5 k

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Fx = 0,

Ra – 272.5 = 0 Ra = 272.5 k

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Theory of Structures

Sir Dr.Zubair

Problem # 13:

10 k

20 k

Rk 1/2

A

B 8’

E

4’

6’

C 4’

5’

F 5’

D 4’

6’

Disturbing into 3 parts

Part 1

10 k

Free Body Diagram

10 k

8’

4’

6’

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha - He = 0,

Ra -10 + Rb + Re = 0

Ha = 79.60 k

Ra + Rb = 12.12 -----------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point A 10 * 8 - Rb * 12 – Re * 18 = 0 → -12 Rb = - 2.12 *18 → Rb = 3.65 K Putting Value of Ra in Equation (i) Ra + Rb = 12.12 → Ra + 3.65 – 12.12 = 0 09CE37

→ Rb = 8.47 K 18 | P a g e

Theory of Structures

Sir Dr.Zubair

Part 2 :

Free Body Digram: 20 k

20 k

4’

5’

5’

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

He - Hf = 0,

Re + -20 +Rc -Rf = 0

He = 79.60 k

Re + Rc = 43.874 -----------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point E 20*4 – Rc * 9 + Rf * 14 = 0 → -9 Rc = - 414.236 → Rc = 46 K Putting Value of Rc in Equation (i) Re + Rc = 43.874 → Re + 46 – 43.874 = 0

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→ Re = -2.124 K

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Theory of Structures

Sir Dr.Zubair

Part 3:

Free Body Diagram: Rk

½

4’

6’

Rk

4’

½

6’

Ф = tan -1 1 / 2 → 26.56 Fx = F Cos Ф → 37 Cos 26.56 → 79.60 k Fy = F Sin Ф → 37 Sin 26.56 → 39.79 k Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Hf - Fx = 0,

Rf - Fy + Rd = 0

Hf = 7960 k

Rf + Rd = 39.79 -----------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point F Fy * 4 – Rd * 10 =0

→

39.79 * 4 – 10 Rd = 0

→

Rd = 15.916 K

Putting Value of Rd in Equation (i) Rf + Rd = 39.79

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→

Re + 15.916 – 39.79 = 0

→

Re = 23.874 K

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Theory of Structures

Sir Dr.Zubair

Problem # 14:

0.6 k / ft

2.4 k /ft

16’

It is disturbed like that

16’

Free Body Diagram: 14.4 k 14.4 9.6 k A Ha

8’

Ra

09CE37

B 5.33

Rb

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Theory of Structures

Sir Dr.Zubair

Point Load of U.D.L = 16 * 0.6

→

9.6 k

Point Load of V.D.L = 1/2 * 16 * ( 2.4 – 0.6 )

→ 14.4 k

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra - 9.6 – 14.4 + Rb = 0 Ra + Rb = 24 -----------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point A 9.6 * 8 + 14.4 * 10.66 - Rb * 16 = 0

→

0 = 76.8 + 153.59 – 15 Rb → Rb = 14.39 K

Putting Value of Rb in Equation (i) Ra + Rb = 24

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→

Ra + 14.39 – 24 = 0

→ Re = 9.61 K

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Theory of Structures

Sir Dr.Zubair

Problem # 15 :

RK

200 lb /ft

100 lb/ft A

C

4’

4’

D

4’

8’

Solving this equation by parts . Part 1:

600 k 1200 k C

D 6’

2’

4’

Solution: By solving the above figure: First convert the U.D.L and V.D.L into single point loads. For U.D.L = 12 * 100 = 1200 k

For V.D.L = 1/2 * 12 * 100 = 600 k

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Rc -1200 – 600 + Rd = 0 Ra + Rb = 1800 -----------------------------(i)

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Theory of Structures

Sir Dr.Zubair

Moment, +ve Clockwise ∑ M = 0 at point C 1200 * 6 + 600 * 8 – Rd * 12 = 0

→

0 = 7200 + 4800 – 12 Rd → Rd = 1000 K

Putting Value of Rb in Equation (i) Rc + Rd = 1800

→

Part 2:

Ra + 1000 – 1800 = 0

R k

→ Re = 800 K

800 k

4’

4’

4’

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra - 89 – 800 + Rb = 0 Ra + Rb = 889 -----------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point A 89 * 4 + 800 * 8 – Rb * 12 = 0

→

0 = 356 + 6400 -12 Rb → Rb = 563 K

Putting Value of Rb in Equation (i) Ra + Rb = 889

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→

Ra + 563 – 889 = 0

→ Re = 326 K

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Theory of Structures

Sir Dr.Zubair

Shear Force and Bending Moment Shear Force : The Shear force at a cross- sectional of a beam is the unbalanced vertical force to the tight or left of the section.

Bending Moment: The bending moment at a cross-section of a beam is the algebraic sum of the moments of the forces to the right or left of the section. OR Algebraic sum of the clockwise and anti-clockwise moment is called Bending Moment.

Shear Force & Bending Moment Diagram: The shear force and bending moment will be different at different cross –sections depending upon the loading on the beam. The variation in the there can be shown graphically by plot the S.F and B.M diagram.

Shear Force Diagram for Different Types of Loading: CASE-I : For any concentrated / Point Load shear force is straight line.

CASE-II : For U.D.L the shear force diagram will be inclined line. Or Triangular Form

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Theory of Structures

Sir Dr.Zubair

CASE - III : For U.V.L uniform varying load 20 Curve will be there in its diagram

20

SIGN CONVERSION : An upward force to the right of a section contributes to a Positive Shearing force in upward L.H.S +VE

-VE

R.H.S

+VE

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-VE

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Theory of Structures

Sir Dr.Zubair

Bending Moment Diagram for Different Loads CASE-I For Concentrated Force ( Point Load ) the bending moment diagram will be inclined line.

CASE-II For UDL the shape of B.M diagram will be parabolic at 20 curve

20

CASE-III The shape for V.D.L of B.M will be at 30 curve.

30

SIGN CONVERSION: A clockwise moment to the right of then section or an anticlockwise moment to the left of the section contributes to positive bending moment. L.H.S:

+VE

R.H.S :

-VE

-VE

+VE

Positive Bending Moment is also known as Hogging Bending Moment and Negative B.M is also known as sagging B.M

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Theory of Structures

Sir Dr.Zubair

Problem : 1

2.5 k

0.6 m

1.5 k

0.6 m

2k

0.6 m

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

- 2.5 + Ra – 1.5 - 2 = 0 Ra = 6 k -----------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point B Ma + Ra * 0.18 - 2.5 * 0. 12 – 1.5 * 0.6 = 0

→ 0 = Ma + 10.8 – 3 – 0.9 → Ma = - 6.9 K

Hence the direction of moment will be reversed to which we have taken .

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Theory of Structures Shear force

Sir Dr.Zubair 1

2 2.5k

A

Ra

3 1.5 k

C

2k

D

B

Ha 0.6 m

0.6 m

0.6 m

X1 X2 X3

Section ( 1-1)

Vχ1 = Ra

→

+ 6 t ----------- (i)

If χ = 0 @ A → + 6 t If χ = 0.6 @ ( A-C) → + 6 t Section ( 2-2)

Vχ2 = Ra – 2.5

→

6 – 2.5

→

+ 3.5 t ----------- (ii)

If χ = 0.6 @ C → + 3.5 t If χ = 1.2 @ ( A -D ) → + 3.5 t Section ( 3-3)

Vχ3 = Ra – 2.5 – 1.5

→

6 – 2.5 – 1.5

→

+ 2 t ----------- (iii)

If χ = 1.2 @ D → + 2 t If χ = 1.8 @ ( A - B ) → + 3.5 t

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Theory of Structures

Sir Dr.Zubair

Bending Moment :

1

2

3

2.5 k

Ma

1.5 k

2k

Ra 0.5 m

0. 6 m

0.6 m

X1 X2 X3

Section ( 1-1)

B.Mχ1 = Ra * χ + Ma 1

If χ1 = 0 @ A

→ 6 χ1 – 6.9

→

----------- (i)

6 ( 0) – 6.9

→

- 6.9 t.m

Section ( 2-2)

B.Mχ2 = Ra * χ

2

+ Ma – 2.5 * ( χ2 – 0.6) → 6 χ2 – 69 – 2.5( χ2 – 0.6 ) → 6 χ2 – 6.9 – 2.5 χ2 – 1.5 –(ii)

If χ2 = 0.6 @ C → 6 * 0.6 – 6.9 – 2.5 * 0.6 + 1.5

→

- 3.3 tm

Section ( 3-3)

B.Mχ3 = Ra * χ + Ma – 2.5 * ( χ – 0.6) – 1.5 ( χ – 1.2) 3

…

3

3

→6 χ3 – 69 – 2.5( χ3 – 0.6 ) – 1.5 ( χ3 – 1.2 ) ----------- (iii)

If χ3 = 1.2 @ D → 6 * 1.2 – 69 – 2.5 ( 1.2 – 0.6 ) – 1.5 ( 1.2 – 1.2 ) → If χ3 = 1.8 @ B

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- 1.2 t.m

→ 0 t.m

30 | P a g e

Theory of Structures

Sir Dr.Zubair

Shear and Bending Moment Diagram 2.5 k

0.6 m

+6t

-6.9 tm

09CE37

1.5 k

0.6 m

+ 3.5t

-3.3 tm

2k

0.6 m

+ 2.1 t

-1.2 tm

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Theory of Structures

Sir Dr.Zubair

Problem : 2

W lb/ft

A

B L

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – WL = 0 Ra = WL -----------------------------(i)

Moment, +ve Clockwise ∑ M = 0 at point B Mn + W * L/2

09CE37

→ 0 = Ma + WL2 / 2

→ Ma = - WL2 / 2 K

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Theory of Structures

Sir Dr.Zubair

Free Body Diagram of Shear force

A

Ra

B

Ha L

X1 Section ( 1-1)

Vχ1 = Ra - Wχ

→

WL - Wχ ----------- (i)

If χ = 0 @ A → WL – W (0) → + WL If χ = L @ ( B ) → WL – W (L) → 0 Bending Moment

A

Ra

B

Ma L

X1

B.Mχ1 = Ma + Ra * χ – Wχ * χ/2

→ Ma + Ra * χ – Wχ2 /2

If χ1 = 0 @ A

→

Ma + 0

If χ1 =L @ B

→

- WL2 / 2 + WL2 – WL2/2

09CE37

----------- (i) - WL2 / 2 t.m

→ →

0 t.m

33 | P a g e

Theory of Structures

Sir Dr.Zubair

Shear and Bending Moment Diagram

W lb/ft

A

B L

+WL

0 0 -WL2 /2

09CE37

20

34 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem : 3 A cantilever beam AB 2m long carrying a uniform disturbed load of 2.5 Kn/m over a length of 1.6 m frin free hand . draw the S.F and B.M diagram for the beam. 2.5 Kn/m

A

B 0.4 m

1.6 m

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – 4 = 0 Ra = 4 KN

Moment, +ve Clockwise ∑ M = 0 at point A Ma = Ma + 4 * 1.6 / 2

09CE37

→ 0 = Ma + 4.8

→ Ma = - 4.8

KN . m

35 | P a g e

Theory of Structures

Sir Dr.Zubair

Free Body Diagram of Shear force 1

2 4 Kn

A B Ra

Ha 0.4 m

0.8 m

0.8 m

X1 X2 Section ( 1-1)

Vχ1 = Ra

→

4 KN ----------- (i)

If χ = 0 @ A → 4 KN

Section ( 2-2)

Vχ2 = Ra – 2.5 *( χ2 – 0.4 )

→ Ra – 2.5 χ2 + 1

If χ2 = 0.4 @ C → 4 – 2.5 ( 0.4) + 1 → 4 KN ----------- (ii) If χ2 = 2 @ ( B ) → 4 – 2.5 (2) + 1 → 0 ----------- (iii)

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Theory of Structures

Sir Dr.Zubair

Bending Moment 1

2 4 Kn

A

B

Ra 0.4 m

0.8 m

0.8 m

Ma X1

Section ( 1-1)

B.Mχ1 = Ma + χ

1

* Ra

If χ1 = 0 @ A

----------- (i)

→

Ma + 0

→

- 4.8 t.m

Section ( 2-2)

B.Mχ2 = Ra

* χ2 + Ma – 2.5 * ( χ2 – 0.4) { ( χ2 – 0.4)/2 }

→ Ma + Ra χ2 - 1.25 ( χ2 – 0.4 )2 → - 4.8 + 4 χ2 - 1.25 ( χ2 - 0.4 ) 2

If χ2 = 0.4 @ C → - 4.8 + 1.6 - 0 If χ2 = 2 m @ B →

09CE37

→

- 3.2 KN. m

0 KN. m

37 | P a g e

Theory of Structures

Sir Dr.Zubair

Shear and Bending Moment Diagram 2.5 Kn/m

A

B 0.4 m

+4 Kn

-4.80 kn.m

1.6 m

+ 4 Kn

0

-3.2 Kn.m

0

2o

09CE37

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Theory of Structures

Sir Dr.Zubair

Problem : 4 A cantilever beam of 1.8 m is loaded as shown in figure . Draw the shear force and bending moment diagram. 2 Kn

1.5 Kn

1 KN 2.5 Kn/m

A

B 0.3 m

C 0.6 m

D 0.3m

0.6 m

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – 2 – 1.5 – ( 0.8 * 0.6 ) -1 = 0 Ra = 4.98 KN

Moment, +ve Clockwise ∑ M = 0 at point A Ma = Ma + 2 * 0.3 + 1.5 * 0.9 + 0.48 * 1.5 → 0 = Ma + 0.6 + 1.35 + 0.72 + 1.8 → Ma = - 4.47

09CE37

KN . m

39 | P a g e

Theory of Structures

Sir Dr.Zubair

Free Body diagram and Shear Force 1

2

3

2 kn

A

1.5 kn

C

Ra

4 0.48 Kn

1 Kn

D

B

Ha 0.3 m

0.6 m

0.3 m

0.3 m

0.3 m

X1 X2 X3 X4 Section ( 1-1)

Vχ1 = Ra

→

+ 4.98 KN ----------- (i)

If χ1 = 0 @ A → +4.98 KN

Section ( 2-2)

Vχ2 = Ra – 2

→ 4.98 – 2 → +2.98 KN

If χ2 = 0.3 @ B → + 2.98 KN

Section ( 3-3)

Vχ3 = Ra – 2 – 1.5

→

4.98 – 3.5 → 1.48 KN

If χ3 = 0.9 @ C

→ + 1.48 KN

Section ( 4-4)

Vχ4 = Ra – 2 – 1.5 – 0.8( χ2 – 1.2 )

→ 1.48 - 0.8 ( χ2 – 1.2 )

If χ4 = 1.2 @ D → + 1.48 KN If χ4 = 1.8 @ E → 1.48 – 0.8 ( 1.8 – 1.2 )

09CE37

→ 1 KN

40 | P a g e

Theory of Structures

Sir Dr.Zubair

Bending Moment

1

2

3

2 kn

4

1.5 kn

0.48 Kn

1 Kn

Ma A

B

Ra 0.3 m

0.6 m

0.3 m

0.3 m

0.3 m

X1 X2 X3 X4

Section ( 1-1)

B.Mχ1 = Ra * χ If χ1 = 0 @ A

1

+ Ma

----------- (i)

→

Ma + 0

→

-4.47 Kn.m

Section ( 2-2)

B.Mχ2 = Ra

* χ2 + Ma – 2 * ( χ2 – 0.3 )

→ 4.98 χ2 – 4.4 - 2 χ2 + 0.6 If χ2 = 0.3 @ B → 4.98 ( 0.3) - 4.4 – 2 ( 0.3) + 0.6

09CE37

→

- 2.97 KN. m

41 | P a g e

Theory of Structures

Sir Dr.Zubair

Section ( 3-3)

B.Mχ3 = Ra

* χ3 + Ma – 2 * ( χ3 – 0.3 ) – 1.5 (χ3 – 0.9 )

→ -4.47 + 4.98 χ3 - 2 χ3 + 0.6 – 1.5 χ3 +1.35 → 1.48 χ3 – 2.52 If χ3 = 0.9 @ C → 1.48 ( 0.9 ) - 2.52

→

- 1 .188 KN. m

Section ( 4-4)

B.Mχ4 = Ra

* χ4 + Ma – 2 * ( χ4 – 0.3 ) – 1.5 (χ4 – 0.9 ) – 0.8 (χ4 – 1.2) (χ4 -1.2 / 2)

→ -4.47 + 4.98 χ3 - 2 χ3 + 0.6 – 1.5 χ3 +1.35 – 0.4 (χ4 – 1.2)2 → - 2.52 + 1.48 χ4 – 0.4 (χ4 – 1.2) 2 If χ4 = 1.2 @ D → -2.52 + 1.48 (1.2 ) – 0.4 ( 1.2 – 1.2 ) 2 → If χ4 = 1.8 @ E

09CE37

→ -2.52 + 1.48 (1.8 ) – 0.4 ( 1.8 – 1.2 ) 2 →

- 0.744 KN. m 0 KN. m

42 | P a g e

Theory of Structures

Sir Dr.Zubair

Shear and Bending Moment Diagram 2 Kn

1.5 Kn

1 KN 2.5 Kn/m

A

B 0.3 m

+4.98 kn

- 4.47 kn.m

C 0.6 m

+2.98 kn

D 0.3m

+1.48 kn

0.6 m

+1.48 kn

-2.97 kn.m -1.18 kn.m

- 0.744 knm

1kn

0

20

09CE37

43 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem : 5 A 5m long cantilever bream carries a point load of R t at the free hand and a UDL of ( 1+ 0.2 R ) t /m over a length of 3m from the free end . Draw the shear force and bending moment diagram. Rt 18.8 t

A

B 2m

C 3m

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – 56.4 – 89 = 0 Ra = 145.4 t

Moment, +ve Clockwise ∑ M = 0 at point A 0 = Ma + 56.4 * 3.5 +89* 5 → 0 = Ma + 197.4 + 445 → Ma = - 642.4 t . m 09CE37

44 | P a g e

Theory of Structures

Sir Dr.Zubair

Free Body diagram and Shear force

1

2

(1+0.2 R )*3

A

Rt

B

Ra

C

Ha 2m

1.5 m

1.5 m

X1

X2

SHEAR FORCE

Section ( 1-1)

Vχ1 = Ra

→

+ 145.4 t ----------- (i)

If χ1 = 0 @ A → +145.4 t

Section ( 2-2)

Vχ2 = Ra – 18.8 (χ2 -2 )

09CE37

→ 145.4 – 18.8 (χ2 -2 )

If χ2 = 2 @ B →

145.4 – 18.8 ( 2 - 2 )

→

+ 145.4 t

If χ2 = 5 @ C →

145.4 – 18.8 ( 5 - 2 )

→

+ 89 t

45 | P a g e

Theory of Structures

Sir Dr.Zubair

Bending Moment : 1

2

(1+0.2 R )*3

Rt

Ma A

B

C

Ra 2m

1.5 m

1.5 m

X1

X2

Section ( 1-1)

B.Mχ1 = Ra * χ If χ1 = 0 @ A

1

+ Ma → 145 χ1 - 624.4

→

----------- (i)

Ma + 0

→

- 642.4 t.m

Section ( 2-2)

B.Mχ2 = Ra

* χ2 + Ma + 18.8 * ( χ2 – 2 ) ( χ2 – 2 / 2 )

→ 145.4 χ2 – 624.4 - 9.4 ( χ2 + 2) 2 If χ2 = 2 @ B → 145.4 * 2 – 642.4 – 9.4 ( 2- 2 ) 2

→

If χ2 = 5 @ C → 145.4 * 5 – 642.4 – 9.4 ( 5- 2 ) 2

→

09CE37

- 351.6 t. m 0 t. m

46 | P a g e

Theory of Structures

Sir Dr.Zubair

Shear and Bending Moment Diagram:

Rt 18.8 t

A

B

C

2m

+145.4 t

3m

+145.4t

89t

0 -642.4 tm

09CE37

20

-351.6 tm

47 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem : 6 A cantilever beam of 150 cm span carries a load of 2t at 60 cm from the fixed end , a load of 1t at the free end and a load of 4 ton UDL over 60 cm from 2 ton load towards the free end . Draw S.F and B,M diagram

2t

1t 6 t /m

A

B 0.6 m

C 0.6 m

0.3 m

Load intensity = Load / Length = 4 / 0.66 = 6 t/m

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – 2 – (4) - 1 = 0 Ra = 7 t

Moment, +ve Clockwise ∑ M = 0 at point A 0 = Ma + 2 * 0.66 + 4 * 0.99 + 1 * 1.5 → 0 = Ma + 1.33 + 3.99 + 1.5 → Ma = - 6.33 t . m

09CE37

48 | P a g e

Theory of Structures

Sir Dr.Zubair

Free Body diagram and Shear force

1

2 2t 4t

A

1t

B

Ra

C

Ha 0.6 m

0.3 m

0.3 m

0.3 m

X1

X2

Section ( 1-1)

Vχ1 = Ra

→

+ 7 t ----------- (i)

If χ1 = 0 @ A → + 7 t

Section ( 2-2)

Vχ2 = Ra – 6 ( χ2 – 0.66 )

If χ2 = 0.66 @ B → + 7 t

Section ( 3-3)

Vχ3 = Ra – 6 (χ2 – 0.66 ) - 2 If χ3 = 1.33 @ C If χ3 = 1.5 @ D

09CE37

→ 7–4-2 →1 t

→ +1t → +1 t

49 | P a g e

Theory of Structures

Sir Dr.Zubair

Bending Moment :

1

2 2t

Ma

4t

A

1t

B

C

D

Ra 0.6 m

0.3 m

0.3 m

0.3 m

X1

X2

Section ( 1-1)

B.Mχ1

= Ra * χ1 + Ma

If χ1 = 0 @ A

----------- (i)

→

Ma + 0

→

- 6.33 t.m

Section ( 2-2)

B.Mχ2 = Ra

* χ2 + Ma – 2 * ( χ2 – 0.6 ) – 6 * ( χ2 – 0.6 ) ( χ2 – 0.6 )/2

→ 7 χ2 – 6.33 – 2 (χ2 – 0.6 ) – 3 (χ2 - 0.6 ) 2 If χ2 = 0.66 @ C → 7 * 0.66 – 6.33

→

- 2.16 t. m

Section ( 3-3)

B.Mχ3 = Ra

* χ3 + Ma – 2 * ( χ3 – 0.6 ) – 4 * ( χ3 – 0.9)

If χ3 = 0.9 @ D → 7 * 0.9 – 6.3 – 2 ( 0.9 – 0.6 ) – 0

→ 7 χ3 – 6.33 – 2 (χ3 - 0.6 ) – 4 (χ3 – 0.9 ) →

If χ3 = 1.5 @ B → 7 * 1.5 – 6.3 – 2 ( 1.5 – 0.6 ) – 4 ( 1.5 - 0)

09CE37

- 0.6 t . m →

0 t.m

50 | P a g e

Theory of Structures

Sir Dr.Zubair

Bending Moment and Shear Diagram 2t

1t 6 t /m

A

B

C

0.6 m

+ 7t

-6.31 t.m

09CE37

0.6 m

0.3 m

+7t

-21.6 t.m

1t

20

-0.6 t.m

1t

0

51 | P a g e

Theory of Structures

Sir Dr.Zubair

TRUSSES

09CE37

52 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem 1 : 20 k

15 k

B

D

60o A

600 5’

5’

60o C

10’

60o

5’

5’

E

10’

Ra

Rb

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – 20 – 15 + Re = 0 Ra + Re = 35 ---------------------- (i)

Moment, +ve Clockwise ∑ M = 0 at point A 0 = 20 * 5 + 15 * 15 – 20 Re = 0 → 0 = 325 – 20 Re → Re = 16. 25 k

09CE37

Ra = 18.75 k

53 | P a g e

Theory of Structures

Sir Dr.Zubair

B 60o

A

Joint A :

C

Ra = 18.75

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

-F ac – F ab x = 0,

Ra – F ab y = 0

F ac = - F ab x ------------------------(i) As Fx = F Cos Ф and Fy = F SinФ

F ab y = 18.75 k

F ab y = F ab Sin 60 0

→

Equation (i) will be

F ab = 21.65 ( C ) , F ab x = F ab Cos 60 0

→

F ab x = 10.82

F ac = - 10.82 ( C )

20 k

B

D

A

C

Joint B : Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

F ab x – F bd + F cb x = 0,

F ab y – 20 – F cb y = 0

10.82 + F cb x - F bd ------------------------(ii)

18.75 – 20 – F cb

F cb y = F cb Sin 60 0

→

y

F cb = 1.44 ( C ) , F cb x = F cb Cos 60 0

= 0 → F cb

→

y

= 1.25

F cb x = - 0.721

Equation (ii ) will be → 10.82 – F bd – 0. 721 → F bd = - 10.82 ( C )

09CE37

54 | P a g e

Theory of Structures

Sir Dr.Zubair

B

D C

Joint C :

60o

A

60o

E

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

F bc x – F ac - F cd x - F ce = 0,

- F bc y - F cd y = 0

Fce + F cd x = -10.1

-1.25 – F cd

------------------------(iii)

F cd y = F cd Sin 60 0

→

y

= 0 → F cd

F cd = 1.43 ( T ) , F cb x = F cb Cos 60 0

Equation (iii ) will be → F ce – 0.721 = - 10.1 →

→

y

= - 1.25

F cb x = -0.721

F ce = - 9.378 ( C ) or Fce = 9.378 ( T )

15 K

B

D

C E

Joint D : Equilibrium, →+ve ∑ Fx = 0, F bd – F cd x - F de x = 0, Fde x = - 9.379 F de x = F de Cos 60 0

09CE37

→

F de = - 18.75 k ( C )

55 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem # 2 :

2k 3k

F

3k

G

A 30o

E

C

D

60o 60o

60o 60o

3

3

30o B 3

Ra

Rb

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha = 0,

Ra – 3 – 2 – 3 + Rb = 0 Ra + Rb = 8 ---------------------- (i)

By applying Sine Law Sin 60 / AG = Sin 90 / 3 → AG = 3 Sin 60 → AG = 2.59 Cos 30 = AC ‘ / AG → AC ‘ = 2.59 Sin 60 / 3 = Sin 60 / CF → Cos 60 = CD ‘ / 3

CF = 3

→ CD ‘ = 1.5

Moment, +ve Clockwise ∑ M = 0 at point A 0 = 3 * AC’ + 2 * ( 3 + CD ‘ ) + 3 * ( 6 + DB ‘ ) – Rb * 9 → 0 = 28.46 – 9 Rb

→ Rb = 4.27 k 09CE37

Ra = 4 k 56 | P a g e

Theory of Structures

Sir Dr.Zubair

G

60o

A

C

Ra Joint A : Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

- F ag x + F ac = 0,

Ra - F ag y = 0

F ag x = F ac ------------------------(i) F ag y = F ag Sin 60 0 Equation (i ) will be

→

F ag

y

= 4k

F ag = 8 k ( C ) , F ag x = F ag Cos 60 0

→ F ac = F ag x →

→

F ag x = 6.92 k

F ac = 6.92 k ( T )

3k F

G

A

C

Here if we solve it directly it will be more different for us to get the assumes beacuase 2 unknown will come in each equation of equilibrium. So we rotate it as

09CE37

57 | P a g e

Theory of Structures

Sir Dr.Zubair

y 30 k 300

A

F

x

G

E

Joint G : Fx = F Cos 60 → 3 Cos 60 → 1.5

,

Fy = F Sin 60 → 3 Sin 60 → 2.6

Equilibrium, →+ve ∑ Fx = 0, F ag ™ + F gf – F x ™ = 0, F gf = 6.5 N ( C )

09CE37

↑+ve ∑ Fy = 0 F gc - Fy = 0

Note ™ denotes the values which are already known to us

F gc = 2.6 ( C )

58 | P a g e

Theory of Structures

Sir Dr.Zubair

G

Joint C :

F

A

C

D

Equilibrium, →+ve ∑ Fx = 0, - F ac – Fcd + F cg x + F cf x = 0, ----( ii ) F cg y = F cg ™ Sin 60 0

→ F ag y = 2.251 , F cg x = F cg ™ Cos 60 0

→

F cg x = 1.3 k

Equation (ii ) will be → - F ac – Fcd + F cg x + F cf x = 0 → - 6.9 – F cd + 1.3 + F cf x = 0 → F cf x - F cd = 5.6 -------------- ( iii)

↑+ve ∑ Fy = 0 F cf y - F cg y = 0 F cf

F cf

y

y

= 4k

= F cf Sin 60 0

→ F cf = 2.59 ( T ) , F cf x = F cf Cos 60 0

→

F cf x = 1.295 k

Equation (iii) will be → 1. 295 – F cd = 5.6 → - Fcd = 4.3 ( C ) → Fcd = 4.3 ( T)

Joint F

2k

From Figure We have indentify that F fg = F bf = 6. 5 F cf = F fd = 2.59

09CE37

F G C

B D

59 | P a g e

Theory of Structures

Sir Dr.Zubair

Method of Sections Previously we have studied methods of joints for analysis of trusses In this methods we can directly analysis a truss force without calculating from initial point .

Here condition of moment is used rather than condition of statistics ( except in exceptional cases ) Need to cut a section where maximum unknown members force becomes zero or can be neglected.

PROCEDURE Cut a section Draw a free body diagram Assume compression and tension Apply co-planer applications of forces and equation of moment to determine members force

09CE37

60 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem : 1

D m =1/3 B

F m=3/2

A

12

Ra

↑+ve ∑ Fy = 0

C

12

1200

E

12

1200

G

12

1200

→ Ra + Rh – 1200 -1200 – 1200 = 0

H

Rh

→

Ra + Rh = 3600---- ( i )

Taking momet at A. Ma = 0 → 1200 *12 + 1200 * 24 + 1200 * 36 - Rh * 48 = 0 → 48 Rh = 86400 → Rh = 1800 lb D

Ra = 1800 lb

B

Ra=1800

C

1200

E

Take the point where maximum unknowns can be neglected. Taking moment at B.

→

1800 * 12 – F ce * 8 = 0

→

F ce = 2700 lb ( T )

Taking moment at E

→

1800 * 24 – 1200 * 12 – F bd y * 12 = 0

→ Fbd y = 2400 lb

Fbd y = F bd * Sin 18.45 → Fbd = 7583.4 lb ( C ) Taking moment at D

→ F be x * 12 – F ce * 12 + 1800 * 24 = 0 → F be x= 2400 lbs

F be x = F be * Cos 33.69 → F be = 2884.43 ( C )

09CE37

61 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem # 2:

Required needed forces Rce = ? ,

Rbd= ? ,

Rcd= ? , R gf= ?

For Fce , we take moment at D. F ce * 20 + Ra * 40 – 20 * 20 = 0 20 Fce =- -30 * 40 + 400 Fce = 40 k ( C ) or Fce = 40 k ( T )

For Fbd taking moment at C. Fbd * 20 + 30 * 20 = 0 Fbd = - 30 ( T ) or 30 ( C )

For Fcd we take moment at B. Fcd y * 20 + Fce * 20 + 30 * 20 = 0

→

20 Fcd y = 200

Fcd y = F cd * Sin 45 → 10 = Fcd * Sin 45

09CE37

→

→

Fcd y = 10 k ( C)

Fcd = 14.14 k

62 | P a g e

Theory of Structures

Sir Dr.Zubair

Here Equation of moment fails so we have apply direct method which is to apply equation of statics.

↑+ve ∑ Fy = 0 F gf + 30 = 0 F gf = - 30 ( C ) or 30 ( T )

09CE37

63 | P a g e

Theory of Structures

Sir Dr.Zubair

ARCHES

09CE37

64 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem 1 : 3 hinge parabolic arch of 20 ft and standardize of 5’ caring a concentrate load of 20 kips at 6’ from the left side support. Draw the bending moment diagram for the arch, position and magnitude of maximum bending moment for the arch also find the value of the reaction forces C 20 kips yc=5’ A

B 20 ft

DATA: Yc = 5’ P = 20 kips L = 20 ‘ B.M = ?

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha - Hb = 0,

Va + Vb = 20 ------ ( i )

Ha = Hb +ve Clockwise ∑ M = 0 at point A 20* 6 – Vb * 20 = 0 → Vb = 6 kips

putting this value in eq(i)

Va = 14 kips

R.H.S +ve Clockwise ∑ M = 0 at point C Vb * 10 + Hb * 5 = 0 → Hb= 12 kips

as Hb=Ha , So ,

Ha = 12 kips

RESULTANT Ra = √ Ha2 + Va 2 → 18.44 kips Rb= √ Hb2 + Vb 2 → 13.4 kips

09CE37

65 | P a g e

Theory of Structures

Sir Dr.Zubair

B.M max ( +ve ) Using Formula Y = 4 Yc χ ( L – χ ) / L2 Y = 4 * 5* 6 ( 20 – 6 ) / (20)2 Y = 4 .2 s.ft

Va * 6 – Ha * Y = 14 * 6 – 12 * 4.2

→ 33. 6 k.ft

B.M max ( -ve ) Hb * Y – Vb * χ --------------- (iii) Using Formula Y = 4 Yc χ ( L – χ ) / L2 Y = 4 * 5 * χ ( 20 – χ ) / L2 Y = χ - χ2 / 20----(iv) putting Y value in equation (iii) 12 * ( χ – χ 2 / 20) – 6 * χ 12χ – 12 χ2/20 – 6 χ 6 χ – 12 χ2/20 Differentiating w.r.t χ χ = 5 ‘ p putting this value in eq (iv) 09CE37

Y = - 15 ‘ 66 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem 2 : 3 hinge parabolic arch of 20 m and standardize of 4 m caring a concentrate U.D.L of 2 ton/m from the left side support over the length of 8 m . Draw the bending moment diagram for the arch, position and magnitude of maximum bending moment for the arch also find the value of the reaction forces

C

A

8m

DATA:

B 20 m

Yc = 4 m P = 2 to / m L = 20 m B.M = ?

Equilibrium, →+ve ∑ Fx = 0,

↑+ve ∑ Fy = 0

Ha - Hb = 0,

Va + Vb = 16 ------ ( i )

Ha = Hb +ve Clockwise ∑ M = 0 at point A 16 * 4 – Vb * 20 = 0 → Vb = 3.2 t

putting this value in eq(i)

Va = 12.8 t

R.H.S +ve Clockwise ∑ M = 0 at point C Va * 16 + Ha * 4 – 16 * 6 = 0 → Hb= 8 t

as Hb=Ha , So ,

Ha = 8 t

RESULTANT Ra = √ Ha2 + Va 2 → 15.094 t Rb= √ Hb2 + Vb 2 → 8.61 t 09CE37

67 | P a g e

Theory of Structures

Sir Dr.Zubair

B.M max ( +ve ) Va * χ – Ha * Y – 2 * χ * χ/2 -------- (v) Using Formula Y = 4 Yc χ ( L – χ ) / L2 Y = 4 * 4 * χ ( 20 – χ ) / (20)2 Y = 0.8 χ – 0.04 χ2 Putting Value of Y in eq ( v) B.M= 12.8 * χ – 8 * ( 0.8 χ – 0.04 χ 2 ) - χ2 B.M = 12.8 χ – 6.4 χ + 0.32 χ2 – χ2 B.M = 6.4 χ – 0.68 χ2 ---- ( vi) Differentiating w.r.t. χ 0 = 6.4 – ( 0.68 ) ( 2 ) χ Χ = 4.7 m from left hand side support Putting value of χ in eq ( vi ) B.M = 6.4 * 4.7 – 0.68 * ( 4.7)2 B.M ( max) = 15.05 t.m

09CE37

68 | P a g e

Theory of Structures

Sir Dr.Zubair

B.M max ( -ve ) Vb * χ – Hb * Y --------------- (vii) Using Formula Y = 4 Yc χ ( L – χ ) / L2 Y = 0.08 χ – 0.04 χ 2 put in equation (vii) 3.2 * χ – 8 * ( 0.8 χ 2 – 0.04 χ 2 ) B.M = - 3.2 χ + 0.32 χ2 ---------------- ( viii) Differentiating w.r.t χ 0 = - 3.2 + 0.64 χ χ=5m putting this value in eq (viii) B.M = -3.2 * 5 + 0.32 * ( 5 )2 = - 8 t.m

09CE37

69 | P a g e

Theory of Structures

Sir Dr.Zubair

COLUMNS

09CE37

70 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem 1 : 1.5 m long column and has a circular cross-section of 5m diameter one of the ends of the column is fixed and other end is free . Find the Rakiens formula and Find the Eulers Crippling load .

Assume: Rakiens Constant = 1 / 1600 Modulus of Elasticity : 1.2 x 106 kg / cm2 Crushing Stress = 5600 kg/cm2

Rakiens Formula: P = Fc * A / 1 + a ( L/K)2

K = √ I /A

DATA: L = 1.5 m = 150 cm

,

E = 1.2 * 106 kg / cm2

d = 5 cm ,

a = 1 / 1600

fc= 5600 kg/cm2 Finding Area = π / 4 (d)2 →

π / 4 ( 5) 2 → 19.63 cm2

As one head is fixed = L = 2l

→ 2 ( 150 ) → 300 cm

Finding Inertia = π / 64 (d)4 Finding K = √ i/A

→ 3.14 / 64 ( 5) 2 → 30.67 cm4

→ √ 30.67 / 19.63 → 12.4 cm

Putting Value in formula P = 5600 * 19.63 / 1 + { 1 / 1600 ( 300 / 12.4 ) 2 } P = 2971.78 kg Finding Euler’s Formula P = π 2 E I / 4 L 2 ----------------------- As one end is fixed other is free P = (3.14)2 * (1.2 * 10 6) ( 30.67 ) / 4 ( 300 ) 2 P = 362.87 * 10 6 kg

09CE37

71 | P a g e

Theory of Structures

Sir Dr.Zubair

Problem 2 : A cast iron 3 m in length when hinged at both ends has a critical pickling load of P kilogram when the column is fixed at both the ends its critical load rises to P + 30,000 kilogram . if ratio of external diameter is 1.25 and E is equal to 1 * 10 6 . Determine the external and internal diameter of column.

L = 3000 cm

,

do/di = 1.25 → do = 1.25 di

Load when both ends are hinged : P kg Load when both ends are fixed : P + 30,000 E = 1 * 10 6 Using Euler’s theorem when both ends are hinged

L =l

P = π 2 E I / L 2 -------------------------- ( i ) Finding interia ( i) = i = π / 64 ( do2 – di2 ) Putting value of do i= π/ 64 ( 1.25 di )2 - ( di ) 2 i= π / 64 ( 0.25 di ) 4 ----------------------------- (ii) putting value of i P=π2EI/L2 →

in equation (i) P = π 2 E (π / ( 0.25 di ) 4 ) / 64 * L 2 → P = π 3 (0.25 di) 4 -------------------------- (iii)

Using Eulers theorem when both ends are fixed.

L = l / 2 → 3000 / 2 → 1500 cm

P + 30, 000 = 4 π 2 E I / L 2 Putting value of i P + 30 , 000 =

4 π 2 E { π / ( 0.25 di ) 4 } / 64 * L 2 → π 3 ( di ) 4

P + 30,000 = π 3 ( di ) 4 ------------------------------ (iv) Putting value of P from eq (iii) into eq ( iv) π 3 (0.25 di) 4 + 30,000 = π 3 ( di ) 4

30,000 = ( 31 – 0.121 ) di4 Di = 5.58 cm as we know, do = 1.25 * di

09CE37

→ do = 7.5 cm

72 | P a g e