09 Deflection-Virtual Work Method Beams and Frames

Deflection: Virtual Work

Method; Beams and

Frames

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2

Contents



Method of Virtual Work:



Beams



Frames

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3

Virtual unit load

C

A B

w C

A B

Method of Virtual Work : Beams and Frames

Cv



    L Cv dx EI M m 0 1 Real load 1 r_A r_B x₁ x₂ R_B R_A x₁ x2 B r_B x₂ v_2 m_2 x₁ r_A v_1 m_1 • Vertical Displacement w B x₂ R_B V₂ M₂ x₁ R_A V₁ M₁

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Virtual unit couple

C A B w C A B • Slope Real load x₁ x₂ R_B R_A w B x₂ R_B V₂ M₂ x₁ R_A V₁ M₁ B r_B x₂ v_2 m_2 r_A v_1 m_1 r_A x_{1 1} x₂



  C L dx EI M m 0 1   C r_B

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Example 8-18

The beam shown is subjected to a load Pat its end. Determine the slope and displacement at C. EI is constant.

2a a

A B C

P

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•Real Moment M

A B C 2a a P A B C 2a a SOLUTION

•Virtual Moment m

_

Displacement at C

1 kN

x

₁

x

2 -a m m_2 = -x₂

x

₁

x

2 -Pa M M₂ = -Px₂



   L C dx EI M m kN 0 ) )( 1 ( _{2 2} 0 2 2 0 1 1 1 ₎ 1 _{( )( )} 2 )( 2 ( 1 dx Px x EI dx Px x EI a a      





, ) 3 )( ( ) 3 8 )( 4 )( 1 ( ) 3 )( ( ) 3 )( 4 )( 1 ( 3 3 3 3 2 3 1 0 2 0 EI Pa a EI P a P EI x EI P x P EI a a C       2 3 2 1 2 1 1 Px M  2 3P 2 P 2 1 1 x m_  m_2 = -x₂ M₂ = -Px₂ 2 1 1 Px M  2 1 1 x m_ 

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A B C 2a a P A B C 2a a

•Virtual Moment m

_

•Real Moment M

Slope at C

x

₁

x

2 -1 m

x

₁

x

2 -Pa M M₂ = -Px₂



  L C dx EI M m m kN 0 ) )( 1 (   ₂ 0 2 2 0 1 1 1 ₎ 1 _{( 1)( )} 2 )( 2 ( 1 dx Px EI dx Px a x EI a a





      ), ( 6 7 ) 2 )( 1 ( ) 3 8 )( 4 )( 1 ( ) 2 )( 1 ( ) 3 )( 4 )( 1 ( 2 2 3 2 2 3 1 0 2 0 EI Pa Pa EI a a P EI Px EI x a P EI a a C       1 kN•m a 2 1 a 2 1 a x m 2 1 1    1 2    m 2 1 1 Px M  2 3P 2 P M₂ = -Px₂ a x m 2 1 1    1 2    m 2 1 1 Px M 

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Example 8-19

Determine the slope and displacement of point B of the steel beam shown in the figure below. Take E = 200 GPa, I = 250(106_{) mm}4_.

A

5 m B

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SOLUTION

•Virtual Moment m

_ A 5 m B 1 kN x 1 kN x v m_ -1x = x

•Real Moment M

A 5 m B 3 kN/m x 3x 2 x V M   2 3_x2

Vertical Displacement at B



  B L dx EI M m kN 0 ) )( 1 ( EI m kN x EI x EI dx x x EI 3 2 4 5 0 3 5 0 2 375 . 234 ) 8 3 ( 1 2 3 1 ) 2 3 )( ( 1 5 0       





) 10 250 )( 10 200 ( 375 . 234 4 6 6 3 m m kN m kN B       = 0.00469 m = 4.69mm, -1x =  3₂  2 x

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SOLUTION

•Virtual Moment m

_ A 5 m B -1 = x

•Real Moment M

A 5 m B 3 kN/m   2 3_x2

Slope at B



  B L dx EI M m m kN 0 ) )( 1 ( _  EI m kN x EI x EI dx x EI 3 2 3 5 0 2 5 0 2 5 . 62 ) 6 3 ( 1 2 3 1 ) 2 3 )( 1 ( 1 5 0       





) 10 250 )( 10 200 ( 5 . 62 4 6 6 2 m m kN m kN B       = 0.00125 rad, x _{1 kN•m} x v m_{ 1 kN•m} x 3x 2 x V M -1 =  3₂  2 x

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Example 8-20

Determine the slope and displacement of point B of the steel beam shown in the figure below. Take E = 200 GPa, I = 60(106_{) mm}4_.

A C D

B 5 kN 14 kN•m

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A C D B 5 kN 14 kN•m 2 m 2 m 3 m A C _D B 2 m 2 m 3 m 1 kN

•Virtual Moment m

_

•Real Moment M

0.5 kN 0.5 kN x₃ x₂ 6 kN 1 kN



  B L dx EI M m kN 0 ) )( 1 (







    3 0 3 2 0 2 2 2 1 1 2 0 1 (0.5 )(6 ) (0)(0) 1 ) 14 ( ) 5 . 0 ( 1 dx dx x x EI dx x x EI 1 1 1 0 x.5 m_  x₁ 2 2 0 x.5 m_  m_ m_ 14 x₃ x₂ x₁ M₁ = 14 - x₁ M₂ = 6x₂ 2 0 2 0 3 ) 3 )( 1 ( ) 3 5 . 0 2 7 )( 1 ( ) 3 ( 1 ) 5 . 0 7 ( 1 2 ₁2 ₁3 ₂3 0 2 2 2 1 2 1 2 0 1 x EI x x EI dx x EI dx x x EI      





) 60 )( 200 ( 667 . 20 667 . 20    EI B = 0.00172 m = 1.72 mm, 1 1 0 x.5 m_  _{m2  0 x.5 2} M₁ = 14 - x₁ M₂ = 6x₂

Displacement at B

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A C D B 5 kN 14 kN•m 2 m 2 m 3 m A C _D B 2 m 2 m 3 m

•Virtual Moment m

_

•Real Moment M

0.25 kN x₃ x₂ 6 kN 1 kN



  B L dx EI M m m kN 0 ) )( 1 (   



 



 



3 0 3 2 0 2 2 2 1 1 2 0 1 ( 0.25 )(6 ) (0)(0) 1 ) 14 ( ) 25 . 0 ( 1 dx dx x x EI dx x x EI x₁ m_ m_ 14 x₃ x₂ x₁ M₁ = 14 - x₁ M₂ = 6x₂





    2 0 2 2 2 1 2 1 2 0 1 ( 1.5 ) 1 ) 25 . 0 5 . 3 ( 1 dx x EI dx x x EI ) 60 )( 200 ( 333 . 2 333 . 2 _  EI B  = 0.000194 rad 1 kN•m _{0.25 kN} 0.5 -0.5 m_1 = 0.25x₁ m_2 = -0.25x₂ 2 0 2 0 3 ) 5 . 1 ( 1 ) 3 25 . 0 2 5 . 3 ( 1 ₁2 ₁3 x₂3 EI x x EI     M₁ = 14 - x₁ M₂ = 6x₂ m_1 = 0.25x₁ m_2 = -0.25x₂

Slope at B

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Example 8-21

(a) Determine the slope and the horizontal displacement of point C on the frame. (b) Draw the bending moment diagram and deflected curve.

E = 200 GPa I = 200(106_{) mm}4 A B C 5 m 6 m 2 kN/m 4 kN 1.5 EI EI

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A C

•Virtual Moment m

_ A B C 5 m 6 m 2 kN/m 4 kN x₁ x₂ 1 x₂ x₁ M₂= 12 x₂ 12 kN 16 kN 12 kN m₂= 1.2 x₂ m₁= x₁ 1.2 kN 1 kN 1.2 kN

•Real Moment M

M₁= 16 x₁- x₁2



 CH L dx EI mM kN 0 ) )( 1 ( 



 



5 0 2 2 2 1 2 1 1 6 0 1 (1.2 )(12 ) 1 ) 16 ( ) ( 5 . 1 1 dx x x EI dx x x x EI





   5 0 2 2 2 6 0 1 3 1 2 1 (14.4 ) 1 ) 16 ( 5 . 1 1 dx x EI dx x x EI ) 200 )( 200 ( 1152 600 552 ) 3 4 . 14 ( 1 ) 4 3 16 ( 5 . 1 1 5 0 6 0 3 2 4 1 3 1        EI EI x EI x x EI CH = + 28.8 mm , M₂= 12 x₂ m₂= 1.2 x₂ m₁= x₁ M₁= 16 x₁- x₁2

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A C A B C 5 m 6 m 2 kN/m 4 kN x₁ x₂ x₂ x₁ M₂= 12 x₂ 12 kN 16 kN 12 kN m₂= 1-x₂/5 m₁= 0 1/5 kN 0

•Real Moment M

•Virtual Moment m

_

M₁= 16 x₁- x₁2



  C L dx EI mM m kN 0 ) )( 1 (  



 



 5 0 2 2 2 1 2 1 1 6 0 ) 12 )( 5 1 ( 1 ) 16 ( ) 0 ( 5 . 1 1 dx x x EI dx x x EI



   5 0 2 2 2 2 ) 5 12 12 ( 1 0 x x dx EI ) 200 )( 200 ( 50 50 ) 3 5 12 2 12 ( 1 5 0 3 2 2 2      EI x x EI C  = + 0.00125 rad , 1 kN•m 1/5 kN M₂= 12 x₂ m₂= 1-x₂/5 m₁= 0 M₁= 16 x₁- x₁2

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+ + 60 60 M , kN•m 16 -12 -+ V , kN 4 A B C 5 m 6 m 2 kN/m 4 kN 12 kN 16 kN 12 kN CH= = 28.87 mm _C = 0.00125 rad ,

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Example 8-22

Determine the slope and the vertical displacement of point C on the frame. Take E = 200 GPa, I = 15(106_{) mm}4_. 5 kN 3 m 60o 2 m A B C

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•Virtual Moment m

_

•Real Moment M

3 m B C 30o

Displacement at C

3 m B C 30o x₁ 1 kN x₁ 1 kN C 30o n_1 v_1 m_1 = -0.5x₁ 1.5 m 1.5 kN•m 1 kN m_1 = -0.5x₁ x₁ 5 kN 1.5 m x₁ 5 kN C 30o N₁ V₁ 7.5 kN•m M1 = -2.5x1 x₂ 2 m A 1.5 kN•m m_2 = -1.5 x₂ 5 kN 2 m A 7.5 kN•m x₂ 1.5 kN•m 1 kN v_2 n_{2 m} 2 = -1.5 x₂ 7.5 kN•m 5 kN N₂ V₂ M₂ = -7.5



   L Cv dx EI M m kN 0 ) )( 1 ( 



  



  2 0 2 1 1 3 0 1 ( 1.5)( 7.5) 1 ) 5 . 2 ( ) 5 . 0 ( 1 dx EI dx x x EI ) 15 )( 200 ( 75 . 33 75 . 33 ) 25 . 11 ( 1 ) 3 25 . 1 ( 1 2 0 3 0 2 2 3 1      EI x EI x EI Cv = 11.25 mm ,

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•Virtual Moment m

_

•Real Moment M

3 m B C 30o x₁ 1.5 m 1 kN•m x₂ 2 m A 1 kN•m m_1 = -1 m_2 = -1 2 m A 3 m B C 30o x₁ 5 kN 1.5 m 7.5 kN•m x₂ 7.5 kN•m 5 kN M₁ = -2.5x₁ M₂ =- 7.5 x₁ 5 kN C 30o N₁ V₁



  L C dx EI M m m kN 0 ) )( 1 (   



  



  2 0 2 1 1 3 0 ) 5 . 7 )( 1 ( 1 ) 5 . 2 ( ) 1 ( 1 dx EI dx x EI ) 15 )( 200 ( 25 . 26 25 . 26 ) 5 . 7 ( 1 ) 2 5 . 2 ( 1 2 0 3 0 2 2 1     EI x EI x EI C  = 0.00875 rad,

Slope at C

1 kN•m x_{1 C} 30o n_1 v_1 1 kN•m m_1 = -1 M₁ = -2.5x₁ x₂ 1 kN•m n_2 v_2 m_2 = -1 x₂ 7.5 kN•m 5 kN N₂ V₂ M₂ = -7.5

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Virtual Strain Energy Caused by Axial Load, Shear, Torsion, and Temperature

• Axial Load



 L n dx AE nN U 0 Where

n = internal virtual axial load caused by the external virtual unit load N = internal axial force in the member caused by the real loads L = length of a member

A = cross-sectional area of a member E = modulus of elasticity for the material • Bending



 L b dx EI mM U 0 Where

n = internal virtual moment cased by the external virtual unit load M = internal moment in the member caused by the real loads L = length of a member

E = modulus of elasticity for the material

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• Torsion



 L t dx GJ tT U 0 Where

 t = internal virtual torque caused by the external virtual unit load T = internal torque in the member caused by the real loads G = shear modulus of elasticity for the material

J = polar moment of inertia for the cross section, J = c4_{/2, where c is the}

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• Shear



 L s dx GA vV K U 0 ) ( Where

v = internal virtual shear in the member, expressed as a function of x and caused by the external virtual unit load

V = internal shear in the member expressed as a function of x and caused by the real loads

K = form factor for the cross-sectional area: K = 1.2 for rectangular cross sections K = 10/9 for circular cross sections

K 1 for wide-flange and I-beams, where A is the area of the web G = shear modulus of elasticity for the material

A = cross-sectional area of a member

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• Temperature dx T₁ T₂ T₂ > T₁ dx y c T y d ) 2 ( ) (    dx c T d ) 2 ( ) (   



 md U_temp



  L temp dx c T m U 0 ) 2 ( T₂ c c T₁ T = T₂ - T₁ c T 2    T₁ T₂ y c y T 2  y T₂ > T₁ T₁ O d 2 2 1 T T T_m   d M M

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Example 8-23

From the beam below Determine :

(a) If P = 60 kN is applied at the midspane C, what would be the displacement at point C. Due to shear and bending moment.

(b) If the temperature at the top surface of the beam is 55 o_{C , the}

temperature at the bottom surface is 30 o_{C and the room temperature is 25} o_C.

What would be the vertical displacement of the beam at its midpoint C and the the horizontal deflection of the beam at support B.

(c) if (a) and (b) are both accounted, what would be the vertical displacement of the beam at its midpoint C.

Take = 12(10-6_)/o_{C. E = 200 GPa, G = 80 GPa, I = 200(10}6_{) mm}4 _{and A =}

35(103_{) mm}2_{. The cross-section area is rectangular.}

A C B

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A B 1 kN x x A C B 2 m 2 m P SOLUTION



  L i i bending dx EI M m 0



 /2 0 ) 2 )( 2 ( 2 L EI dx Px x /2 0 ) 3 4 ( 2 _Px3 _L EI   ) 200 )( 200 ( 48 ) 4 ( 60 48 3 3   EI PL = 2 mm,



  L i i shear dx GA V Kv 0



 /2 0 ) 2 )( 2 1 ( 2 L GA dx P K ) 35000 )( 80 ( 4 ) 4 )( 60 ( 2 . 1 4 2 2 / 0    GA KPL GA KPx L = 0.026 mm, shear bending C     _{= 2 + 0.026 = 2.03 mm,} P/2 P/2 M diagram PL/4 x x x P 2 P₂ x P/2 P/2 V diagram • Part (a) : 0.5 kN 0.5 kN m diagram 0.5x 1 0.5x 0.5 0.5 v diagram

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•Part (b) : Vertical displacement at C

SOLUTION A B 1 kN x x m diagram 0.5 kN 0.5 kN 0.5x 1 0.5x T_room = 25 oC , T₁=55o_C T₂=30o_C 260 mm Temperature profile 5 . 42 2 30 55    m T Cv = -2.31 mm ,



  Cv L dx c T m kN 0 2 ) ( ) )( 1 ( 

- Bending



  2 0 ) 5 . 0 ( 2 ) ( 2 x dx c T  2 0 ) 2 5 . 0 ( ) 10 260 ( ) 25 )( 10 12 ( 2 2 3 6 _x       A C B 2 m 2 m 55 o_C, 30 o_C 260 m

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A B 1 kN

• Part (b) : Horizontal

displacement

at B

x 0 0 T_room = 25 oC , T₁=55o_C T₂=30o_C 260 mm Temperature profile 5 . 42 2 30 55    m T BH = 0.84 mm ,



   L BH n T dx kN 0 ) ( ) )( 1 ( 

- Axial



  4 0 ) 1 ( ) ( T dx  4 0 ) )( 25 5 . 42 )( 10 12 ( _ 6 _{ x}   1 kN 1 1 n diagram BH = 0.84 mm Cv = 2.31 mm , A C B 2 m 2 m 55 o_C 30 o_C 260 m

Deflected curve

A C B

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P A C 55 B o_C 30 o_C 260 m

• Part (c) :

C = -2.03 + 2.31 = 0.28 mm, A C B C = 2.03 mm P

=

A B 55 o_C, 30 o_C _C = 2.31 mm

+

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Example 8-24

Determine the horizontal displacement of point C on the frame.If the

temperature at top surface of member BC is 30 o_{C , the temperature at the bottom}

surface is 55 oC and the room temperature is 25 oC.Take = 12(10-6)/oC, E = 200

GPa, G = 80 GPa, I = 200(106_{) mm}4 _{and A = 35(10}3_{) mm}2 _{for both members.}

The cross-section area is rectangular. Include the internal strain energy due to axial load and shear.

A B C 5 m 6 m 2 kN/m 4 kN 1.5 EI,1.5AE, 1.5GA EI,AE,GA 260 mm

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B C Moment, m (kN•m) A B C Shear, v (kN) A 5 m 6 m A C B

Virtual load

1 x₂ x₁ 1.2 kN 1 kN 1.2 kN 1.2 1 1.2 1 1 1 -1.2 -1.2 6 6 1x₁ 1.2x₂ B C Axial, n (kN) A

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B C Shear, V (kN) A B C Axial, N (kN) A A B C 5 m 6 m 2 kN/m 4 kN x₁ x₂ 12 kN 16 kN 12 kN

Real load

12 4 12 4 16 4 16 - 2x₁ -12 -12 60 16x₁ - x₁2 12x₂ 60 B C Moment, M (kN•m) A

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•Due to Axial

x₁ x₂



  i i i i i CH E A L N n kN)( ) 1 ( AE 1.5AE AE AE ) 5 )( 4 )( 1 ( 5 . 1 ) 6 )( 12 )( 2 . 1 (   5 m 6 m AE m kN   77.6 2 ) 10 200 )( 10 35000 ( 6 . 77 2 6 2 6 m kN m m kN CH       = 1.109(10-5) m = 0.0111 mm, B C Virtual Axial, n (kN) A 1.2 1 1.2 1 B C Real Axial, N (kN) A 12 4 12 4

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•Due to Shear

x₁ x₂ 1.5GA 5 m 6 m GA



 CH L dx GA vV K kN 0 ) ( ) )( 1 ( 2 5 0 1 6 0 1 _1.2( 1.2)( 12) 5 . 1 ) 2 16 )( 1 ( 2 . 1 dx GA dx GA x





     GA m kN x GA x x GA      2 2 2 1 1 4 . 134 ) 4 . 14 )( 2 . 1 ( ) 2 2 16 )( 5 . 1 2 . 1 ( 5 0 6 0 ) 10 35000 )( 10 80 ( 4 . 134 2 6 2 6 _m m kN m kN CH       = 4.8(10-5_{) m = 0.048 mm,} B C Virtual Shear, v (kN) A 1 1 -1.2 -1.2 B C Real Shear, V (kN) A 16 4 16 - 2x₁ -12 -12

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Taxila, Pakistan

35

•Due to Bending





   5 0 2 2 2 1 2 1 1 6 0 1 (1.2 )(12 ) 1 ) 16 ( ) ( 5 . 1 1 dx x x EI dx x x x EI = 0.0288 m = + 28.8 mm ,



 CH L dx EI mM kN 0 ) )( 1 ( EI m kN x EI x x EI 3 2 3 2 4 1 3 1 ₎ 1152 3 4 . 14 ( 1 ) 4 3 16 ( 5 . 1 1 5 0 6 0      ) 10 200 )( 10 200 ( 1152 4 6 2 6 3 m m kN m kN CH       x₁ x₂ 1.5EI 5 m 6 m EI B C Virtual Moment, m (kN•m) A 6 6 1x₁ 1.2x₂ B C Real Moment, M (kN•m) A 60 16x₁ - x₁2 12x₂ 60

Department of

Civil

Engineering

University of

Engineering

and

Technology,

Taxila, Pakistan

36

T₁=30oC T₂=55o_C 260 mm Temperature profile

•Due to Temperature

CH = 0.0173 m = 17.3 mm ,



  CH L dx c T m kN 0 2 ) ( ) )( 1 ( 



   5  0 2 6_{)(42.5 25)} 10 12 )( 1 ( dx T_m= 42.5o_C

- Bending

- Axial

CH = 0.00105 m = 1.05 mm ,



  CH Ln T dx kN 0 ) ( ) )( 1 ( 



      5 0 2 3 6 2 ) 10 260 ( ) 30 55 )( 10 12 )( 2 . 1 ( dx x A B C 5 m x₁ x₂ 260 mm 30o_C 55o_C T_room = 25o_C B C m (kN•m) A 6 6 1x₁ 1.2x₂ B C n (kN) A 1.2 1 1.2 1

Department of

Civil

Engineering

University of

Engineering

and

Technology,

Taxila, Pakistan

37

A B C 2 kN/m 4 kN

•Total Displacement

Temp CH Bending CH Shear CH Axial CH Total CH) ( ) ( ) ( ) ( ) (         = 0.01109 + 0.048 + 28.8 + (17.3 + 1.05) = 47.21 mm CH= 47.21 mm

Department of

Civil

Engineering

University of

Engineering

and

Technology,

Taxila, Pakistan

38