UTM Yield Line Test

7/30/2019 UTM Yield Line Test

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FAKULTI KEJURUTERAAN AWAM

Faculty of Civil EngineeringTel:(+603) 5543 6153 Fax: (+603) 5543 5275

REINFORCED CONCRETE DESIGN (ECS478)

- TEST No. 2 Solution Page1

Question No. 1

The slab of figure below is fixed at one edge and simply supported at the other twoedges. The ultimate slab load is Wu = 10 kN/m

2. Assume the slab is reinforced in bothdirections x and y. Use yield line method to;

i. determine the expected yield line of the slab (sketch the yield line)ii. write an expression for the external and internal energyiii. write the final moment capacity equation

Given,

2.1=x

y

M

M; 8.0

'=

x

x

M

M

Mx

8 m

8 m

Mx

My


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Question No. 2

The biaxially-bent column shown in figure below is a part of a braced frame. It has toresist an ultimate axial load of 750 kN and ultimate moment about x and y axes asshown. Design the column for the first iteration (K=1.0). State your K value and Ascfrom the column design chart. You need also to calculate value of K using clause

3.8.3.1 of BS8110 : Part 1

GivenFloor to floor height = 6.5 m

cu = 30 N/mm2y = 460 N/mm2Cover = 30 mmLink = 10 mm

Longitudinal bar = 20 mm

X

X

Y

Y

35

25

30

30

20

15

27

32

Mx= 40 kNm

Mx= 30 kNm

My= 25 kNm

My= 15 kNm


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Solution # 1

2.1M

M

x

y= ThereforeMy = 1.2Mx 8.0

'M

M

x

x = ThereforeMx = 1.25Mx

External energy, Ee = Load x Area x Average distance moved

For Area A;

External Energy, Ee = 4)}3

(x)tan4x4x2

1(x)w{(

For Area B

External Energy, Ee = 2)}2

(x))tan48(x4(x)w{(

Internal Energy; Ei = Moment x Rotation x length of yield line

Sagging Yield lines;

Vertical line; Ei = My x 12 x ( 8 4 tan )

= 1.2 Mx x 12 x ( 8 4 tan )

Slanting line (resolve horizontally); Ei = 2 ( Mx x 2 x 4)

= 2 ( Mx x

tan

1 x 4)

2

tan 2

=

tan4

=

tan4

4 1

2 =

tan

1

4 m

BB

A

4 m

4 tan

8 - 4 tan

A

A A

tan 1 = 4

= 41

1

21


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Slanting line (resolve vertically); Ei = 2 ( My x 1 x 4 tan )

= 2 (1.2 Mx x 1 x 4 tan )

Hogging Yield lines;

Ei = Mx x

tan

1 x 8

= 1.25 Mx x

tan

1 x 8

Equating External and Internal Energy

Total External Energy

Ee,total = 4)}3

(x)tan4x4x2

1(x)w{(

+ 2)}

2(x))tan48(x4(x)w{(

Ee,total ={10 x 0.5 x 4 x 4 tan x31xx4 1 }4 + {10 x 4 x(8-4tan)x(

21xx4 1 )}2

Ee,total = 426.67 1 tan + 1280 1 - 640 1 tan = 1280 1 - 213.33 1 tan #

Total Internal Energy

Ei,total = {1.2Mxx 12 x(8 4 tan )} + {2(Mx x

tan

1 x 4)} + {2(1.2 Mx x 1 x 4 tan )} +

{1.25 Mx x

tan

1 x 8 }

Ei,total = 19.2Mx 1 - 9.6Mx 1 tan +

tan

M8 1x+ 9.6Mx 1 tan +

tan

M10 1x

Ei,total = 19.2Mx 1 +

tan

M18 1x

Therefore

Ee,total = Ei,total

1280 1 - 213.33 1 tan = 19.2Mx 1 +

tan

M18 1x

1280 213.33 tan = Mx (19.2 + tan

18

)

+

=

tan

182.19

tan33.2131280M x


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Solution # 2

Reference: Calculation Remarks

About X-X axis (Plane Y-Z)

Therefore = 0.85

About Y-Y axis (Plane X-Z)

Therefore = 0.75

Calculate the value of lexand ley

lox = 6.5 0.3 = 6.2 m

loy = 6.5 0.3 = 6.2 m

Therefore;

lex =()(lox) = (0.85)(6.2) = 5.27 m #

ley =()(loy) = (0.75)(6.2) = 4.65 m #

Condition 2

Condition 2

Condition 1

Condition 1

300

325

350lox

300

275250

loy


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1. Calculate the slenderness ratio

== 35.0

27.5

h

lex 15.057 > 15

==25.0

65.4

b

ley18.6 > 15

Therefore the column is slender.

2. Consider moment about X-X axis

(Plane Y-Z)

Mi= 0.4 M1 + 0.6 M2 0.4 M2

Therefore;

Mi = 0.4 (-30) + 0.6 (40)

Mi = 12 kNm #

0.4 M2 = 0.4 (40) = 16 kNm

Take the greatest value, Mix=16 kNm

2.1 Calculate additional moment, Madd,

(Assume K= 1)

uadd NaM = = (N) )( Kha

Madd= (N) ))'(2000

1

(

2

b

le

(K)(h)

= )35.0)(0.1)()35.0

27.5)(

2000

1)(750(

2

= 29.76 kNm #

Mix=16 kNm

Madd= 29.76 kNm


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2.2 For the first iteration (K=1), the total

moment, Mtxis;

a) M2= 40 kNm

b) Mix+ Madd= 16 + 29.76 = 45.76 kNm

c) M1 +2addM

= 30 + 31.25 = 44.88 kNm

d) eminN = (0.05 x 0.35)(750) = 13.13 kNm

Mtx= 45.76 kNm #

3. Consider moment about Y-Y axis

(Plane X-Z)

Mi= 0.4 M1 + 0.6 M2 0.4 M2

Therefore;

Mi = 0.4 (-15) + 0.6 (25)Mi = 9 kNm #

0.4 M2 = 0.4 (25) = 10 kNm

Take the greatest value, Miy=10 kNm

3.1 Calculate additional moment, Madd,

(Assume K= 1.0)

Madd= (N) ))'

(2000

1(

2

b

le (K)(h)

= )25.0)(0.1)()25.0

65.4)(

2000

1)(750(

2

= 32.43 kNm #

Mtx= 45.76 kNm.

Miy=10 kNm

Madd= 32.43 kNm


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3.2 For the first iteration (K=1), the total

moment, Mtyis;

a) M2= 25 kNm

b) Miy+ Madd= 10 + 32.43 = 42.43 kNm

c) M1 +2addM

= 15 + 16.22 = 31.22 kNm

d) eminN = (0.05 x 0.25)(750) = 9.38 kNm

Mty= 42.43 kNm #

4. Calculate'' b

M

h

M yx or'' b

M

h

M yx to

use either equation 40 or 41

h'= h Cover - link - bar/2= 350 30 10 20/2

h'= 300 mm

b'= b Cover - link - bar/2= 250 30 10 20/2b'= 200 mm

300

76.45

'=

h

Mx = 0.153 and

200

43.42

'=

b

My

= 0.212

Therefore;

'' b

M

h

M yx < , then use equation 41

xyy MhbMM

''' +=

Mty= 42.43 kNm


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5. Calculate the value of ;

29.0

)30)(250)(350(

107503

==x

bhf

N

cu

By interpolation

= 65.0)29.03.0)(2.03.0

65.077.0( +

= 0.662

Therefore;

xyy Mh

bMM

'

'' +=

= )76.45)(300200)(662.0(43.42 +

My = 62.625 kNm #

6. Calculatebh

Nand

2bh

M;

bh

N=

)250)(350(

10x7503

= 8.57 N/mm2

2bhM =

2

6

)250)(350(10x625.62 = 2.86 N/mm2

8.0250

200

h

d== ; fcu= 30 N/mm

2; fy= 460 N/mm2

Refer to Chart 27 of Bs8110: Part 3,

bh

Asc100 = 1.2;Asc= 1050mm

2

k= 0.8

Provide 4T20 = 1257 mm2

bh

N= 8.57

2bh

M= 2.86

Asc= 1050mm2

k= 0.8


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Check on the value of Kat interpolated from

the design chart using equation 33 of BS8110

1

=

baluz

uz

NN

NNK

K=baluz

uz

NN

NN

where

Nuz = 0.45fcubh + 0.95fyAsc

= (0.45 x 30 x 350 x 250) + (0.95 x 460 x 1050)

= 1640.1 kN #

Nbal = 0.25fcubd

= 0.25 x 30 x 350 x 200

= 525 kN #

K=5251.1640

7501.1640

=0.798 #

The value of Kfrom the design chart and from

calculation is almost the same.

Shear Reinforcement;

Min. =4

20

4

bar=

= 5mm

Max Spacing = 12 bar = 12 x 20 = 240mm

Provide R10 @ 200 mm c/c

250 mm

350 mm4T20

R10 @ 200 mm c/c

Documents

UTM Yield Line Test