Reinforcement for Plates

Course CT4150

Lecture12

5 Jan. 2010

Normal forces in concrete plates

• Only rebars in the x and y directions

• Equilibrium of a plate part

• Equations

• Design

Choose such that nsx+nsy is minimal

(lower bound theorem)

2cos

sin cos

xx sx c

yy sy c

xy c

n n n

n n n

n n

• Solution

Example 1• Situation

nxx=1200, nyy=-200, nxy=-400 kN/m

fy = 500 N/mm², f’c= 30 N/mm²Thickness = 100 mm• Reinforcement

nsx= 1200+400 = 1600 kN/m

nsy= -200+400 = 200• Concrete

nc= -2x400 = -800 kN/m

Asx=1600/500 = 3.20 mm = 3200 mm²/m

212–280 = 2/4x12²x1000/280 = 3231 OK

Asy=200/500 = 0.40 mm = 400 mm²/m

26–500 = 2/4x6²x1000/500 = 452 OK

c = 800/100 = 8 N/mm² < f’c OK

(safety factors omitted)

Example 2• Deep beam 4.7x7.5 m, 2 supports, opening 1.5x1.5 m,

point load 3000 kN, fcd = 16.67 N/mm², fyd = 435 N/mm²

• Forces nxx, nyy, nxy (linear elastic analysis)

• Principal stresses

• Reinforcement requirements (software)

• Reinforcement (engineer)

Moments in concrete plates• Only rebars in the x and y directions• Equilibrium of a plate part

• Result (Yield contour)2 min{( )( ), ( )( )}xy px xx py yy px xx py yym m m m m m m m m

Example 3

• Moments in a point

mxx = 13, myy = -8, mxy = 5 kNm/m• Moment capacities

mpx = 17, mpy = 0, m’px =0, m’py = 10• Is the capacity sufficient?

• Yes

25 ?min{(17 13)(0 8), (0 13)(10 8)}

25 ?min{32,26}

Design of moment reinforcement

• Carry the moments with the least amount of reinforcement.

• So, minimize mpx+ mpy+ m’px+ m’py

• 5 constraints• mpx, mpy, m’px, m’py ≥ 0•

Solution 1 (Wood-Armer moments)•

• Crack direction 45º to the reinforcing bars

px xx xy px xx xy

py yy xy py yy xy

m m m m m m

m m m m m m

2 min{( )( ), ( )( )}xy px xx py yy px xx py yym m m m m m m m m

Solution 2 (when mpx would be < 0)

Solution 3 (when mpy would be < 0)2

0

xypx xx

yy

py

mm m

n

m

2

0px

xypy yy

xx

m

mm m

m

Solution 4 (when m’px would be < 0)

Solution 5 (when m’py would be < 0)

2

0px

xypy yy

xx

m

mm m

m

2

0

xypx xx

yy

py

mm m

m

m

Solution 6 (when mpx and mpy would be < 0)

Solution 7 (when m’px and m’py would be < 0)

0

0

px

py

m

m

0

0

px

py

m

m

Example 4

• Moments in a point (as in example 1)

mxx = 13, myy = -8, mxy = 5 kNm/m• Moment capacities

mpx = 13+5²/8 = 16.13 m’px = 0

mpy = 0 m’py = 8+5²/13 = 9.92• Amount of reinforcement is proportional to16.13+0+0+9.92 = 26• Amount of reinforcement in example 317+0+0+10 = 27 (larger, so not optimal)

Example 5

• Plate bridge, simply supported

• 4 x 8 m, point load 80 kN, thick 0.25 m

Example 5 continued

• Decomposition of the load

Example 5 Torsion moment

12

18

10 kNm/m

xy

xy

m a Va T

m F

xyV mxyV m

1

2T

1

2T

1

2T Fa

x

Example 5 All moments

• Moments in the bridge middle

• Moments at the bridge support

14 1

2

18

240 kNm/m

0

10 kNm/m

xx

yy

xy

F am F

am

m F

18

0

0

10 kNm/m

xx

yy

xy

m

m

m F

Example 5 FEM moments

Example 5 Reinforcement

Middle Support Designed

2

40 10 50 kNm/m 0 10 10 50

0 10 10 0 10 10 10

40 10 0 0 10 10 10

100 2.5 0 10 10 10

40

px px px

py py py

px px px

py py py

m m m

m m m

m m m

m m m

Example 5 Upper bound check

Result

> 1OK

4 4 506 6 1.22

5 5 80pxF m

Computed requirements

ConclusionsThe design procedure used is1 Compute the force flow linear elastically2 Choose the dimensions plastically

The reason for the linear elastic analysis in the firststep is that it shows us how an as yet imperfectdesign can be improved. A plastic (or nonlinear)analysis in step 1 would shows us how thestructure would collapse; but that is not what wewant to know in design.

This procedure is applied to design many types ofstructure for the ULS.