Transformed-section method - KOCW [PDF]

EXAMPLE 6.18 (cont.) • Applying the flexure formula to the transformed section, the maximum normal stress in the concr

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6.6 COMPOSITE BEAMS •

Transformed homogeneous beam obtained through a transformation factor: Transformed-section method

n

E1 E2

and dF  dA   ' dA' dzdy   ' ndzdy

  n '

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Transformed-section Method E1  n E 2 F (x) 

  xdA  

1





A1

 

A1

A1



1





A2

yE 2ndA 



A2

y E1d A 

yn d A 



At



A2

yE 2dA  0 yE 2dA  0

yd A  0

y d At  0

Copyright ©2014 Pearson Education, All Rights Reserved

M ( x )    y x d A 



1





A1

y 2 E1d A 

1





=

1  2 y E2ndA     A1

=

E2  y 2ndA     A1

=

E2

=

E2It





At



A2





x1 = 



x2 = 

E1 y

= 

nM y It

E2 y

= 

My It





A2



y 2 d At

y 2 E2dA

A2

y 2 E2dA  

y 2dA  

EXAMPLE 6.17 A composite beam is made of wood and reinforced with a steel strap located on its bottom side. It has the cross-sectional area shown in Fig. 6–38a. If the beam is subjected to a bending moment of 2 kN•m, determine the normal stress at points B and C. Take Ew = 12 GPa and Est = 200 GPa.

Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 6.17 (cont.) Solutions • We will transform the section into one made entirely of steel. 12 150  9 mm bst  nbw  200

• The transformed section is as shown. • The location of the centroid (neutral axis), y

 y A  0.010.020.150  0.0950.0090.15  0.03638 m 0.020.15  0.0090.15 A Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 6.17 (cont.) Solutions • The moment of inertia about the neutral axis is 3 2 1 I NA    0.15  0.02    0.15  0.02  0.03638  0.01  12  3 2 1    0.009  0.15    0.009  0.15  0.095  0.03638   12 

 9.358 106  m 4

• Applying the flexure formula, the normal stress at B’ and C is 2  0.17  0.03638   B'



C 

9.358 10

2  0.03638 

9.358 10

6



6



 28.6 MPa

 27.87 MPa (Ans)

• The normal stress in the wood at B is  B  n B ' 

12 28.56  1.71 MPa (Ans) 200 Copyright ©2014 Pearson Education, All Rights Reserved

6.7 REINFORCED CONCRETE BEAMS

h' y  0 :bh  nAst (d  h' )  0 2 b i2  h  nAst h'  nAst d  0 2 '

EXAMPLE 6.18 The reinforced concrete beam has the cross-sectional area as shown. If it is subjected to a bending moment of M=60 kN•m, determine the normal stress in each of the steel reinforcing rods and the maximum normal stress in the concrete. Take Est =200 GPa and Econc = 25 GPa.

Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 6.18 (cont.) Solutions

• The total area of steel is Ast  2 12.52   982 mm 2 , thus

   

200 103 2   A'  nAst  982  7856 mm 25 103

• We require the centroid to lie on the neutral axis.

 yA  0 h'  7856  400  h '  0 2 h '2  52.37h ' 20949.33  0  h '  120.90 mm

300  h '

• The moment of inertia of the transformed section is computed about the neutral axis, 2 1  120 . 9   3 2 6 4 I   300 120.9   300120.9    7856400  120.9    788.67 10 mm  2  12 

Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 6.18 (cont.) Solutions • Applying the flexure formula to the transformed section, the maximum normal stress in the concrete is

 conc max  601000120.961000  9.20 MPa (Ans)

788.67 10 60100010001000400  120.9  'conc   21.23 MPa 6 788.67 10

• The normal stress in each of the two reinforcing rods is therefore

  21.23  169.84 MPa (Ans)   

 200 103  st  n 'conc   3  25 10

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6.9 STRESS CONCENTRATIONS •

Stress concentrations occur at locations where cross section suddenly changes; e.g.

 max

Mc K I

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EXAMPLE 6.20 The transition in the cross-sectional area of the steel bar is achieved using shoulder fillets. If the bar is subjected to a bending moment of 5 kNㆍm, determine the maximum normal stress developed in the steel. The yield stress is σY = 500 MPa.

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EXAMPLE 6.20 (cont.) Solutions • From the geometry of the bar, r 16   0.2 h 80

w 120   1 .5 h 80

• Thus K is 1.45 and we have  max  K

50.04 Mc  1.45  340 MPa I 1 3    0 . 02 0 . 08 12 

• This result indicates that the steel remains elastic since the stress is below the yield stress (500 MPa).

Copyright ©2014 Pearson Education, All Rights Reserved

6.10 INELASTIC BENDING •

Elastic Range Mc y

  T  2 I  2d 



h M 6M C  m ax   32   2 2 h  2 bmax b hd   12 c  0 bh 2 3  M   c m ax 6 2 max 0





Onset of Yield: MY bh 2 MY   Y  S Y 6 ܵ: Elastic section modulus Copyright ©2014 Pearson Education, All Rights Reserved

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