CM01 03 [režim kompatibility]) - cvut.cz

1st task:Frame Structure

Design of reinforcement

• Design the reinforcement of the frame in the 1st floor

• Bending reinforcement of the beam

• Shear reinforcement of the beam

• Use the maximum values of internal forces from the „envelope“ of internal forces

• Design the tensile reinforcement in 3 cross-sections

• In supports, maximum values from FEM calculation should be reduced to values in the face of the column:

Bending reinforcement

sup

Ed,red Ed,FEM Ed,FEM2

bM M V= −

bsup

Bending reinforcement

Rd Ed

c s

M M

F F

=

=

• Design of reinforcement:

Bending reinforcement

s

s,rqd yd

Ed Eds,rqd s,prov s,rqd

yd B yd

Propose 0.9

Rd Ed

Ed

Ed

M M

F z M

A f z M

M MA A A

zf d f

=

=

=

= = ⇒ ≥

MEd,red in supports, MEd,FEM in midspan

Effective height of beam

B B sw2

d h c∅

= − −∅ −

Bending reinforcement, 16-25 mm (more only if necessary)

Stirrups, 6-12 mm

Example: DESIGN: 3x Ø16 (As,prov = 603 mm2)

• Check of the design:

Bending reinforcement

c cd s yd

cd s,prov yd

s,prov yd

cd

0.8

0.8

c sF F

A f A f

xbf A f

A fx

bf

=

=

=

=Width of compressed part of the

cross-section, bB in supports, beff in midspan

B 0.4z d x⇒ = −

Rd s,prov yd EdM A f z M⇒ = ≥

• Midspan – T-shaped section

Effective width beff

eff eff,i B eff,i i 0 0 eff,i i

i

where 0.2 0.1 0.2 and b b b b b b l l b b= + ≤ = + ≤ ≤∑Distance between zero moments on the beam,

for outer span of the beam l0 ≈ 0.85lBfor inner span of the beam l0 ≈ 0.7lB

B

( )( )

bal,1

B yd

ctms,prov s,min B B B B

yk

s,prov s,max B B

a a,max B

c c,min

700min ;0, 45

700

max 0.26 ;0.0013

0.04

min 2 ; 250 mm

max 20 mm; 1,2

x

d f

fA A b d b d

f

A A b d

s s h

s s

ξ ξ

= ≤ = +

≥ =

≤ =

≤ =

≥ = ∅

Detailing rules

Axial spacing of rebars

Clear spacing of rebars

• If hB ≥ 500 mm, torsion reinforcement is

necessary (add two 12 mm rebars to the middle of the beam)

Mean tensile strength of concrete, see table with

properties of concrete classes from 1st class

Shear reinforcement

• Resistance of compressed concrete struts was already checked in preliminary design (VRd,max ≥ VEd,max)

• Shear reinforcement = stirrups

Shear reinforcement – principle• The higher the shear force, the denser the stirrups

Shear reinforcement – principle• Force imposed on the structure: VEd

• Load-bearing capacity of one stirrup: Aswfyd

• Spacing of stirrups: s

• Horizontal projection of shear crack: ∆l

Ed Rd

Ed sw yd

V V

lV A f

s

=

∆=

VEd

∆l

s

Number of stirrups required on the

length of ∆l

Shear reinforcement – practice• Direct support => we can reduce theoretical maximum

shear force to the value in the distance dB from the face of the column (VEd,1)

• Up to the distance ∆l behind the support, we will design the stirrups in spacing s1 (design force is VEd,1)

• In middle part of the beam, shear force is low => stirrups will be designed with maximum possible spacing smax

• Intermediate part - ???

Spacing of stirrups:

VEd,max,FEM

VEd,1

VRd,mid

0.5bsupdB

∆l

s1smax

∆l

???∆l = z*cotgθ

see 1st HW

see bending

• Design shear force – similar triangles

• Spacing of stirrups:

Stirrups near the support (s1)

2

swsw

4

nA

π∅=

sw yd

1

Ed,1

1 B

1

1

and 0,75

and 400 mm

and 100 mm (recommended)

A fs l

V

s d

s

s

≤ ∆

≤

≤

≥

sup

B

Ed,1 Ed,max,FEM

2

bv d

V Vv

− + =

Number of legs of each stirrup, n=2

DESIGN: Stirrup øsw mm per s1 mm

Cross-sectional area of 1 stirrup

v – see slide 16

• Check shear resistance:

• Check shear reinforcement ratio

• If not checked, increase øsw or decrease si

Stirrups near the support (s1)

sw yd

Rd,sw,1 Ed,1

1

A fV l V

s= ∆ ≥

ckswsw,1 sw,min

B 1 yk

sw cdsw,1 sw,max

B 1 yd

0,08

0,5

fA

b s f

A f

b s f

ρ ρ

νρ ρ

= ≥ =

= ≤ =

Coefficient expressing effect of shear cracks and transversal deformations

ck0,6 1250

fν = −

• Design the spacing according to the condition:

• Check shear reinforcement ratio

• If not checked decrease smax

Stirrups in the middle part (smax)

ckswsw,2 sw,min

B max yk

sw cdsw,2 sw,max

B max yd

0,08

0,5

fA

b s f

A f

b s f

ρ ρ

νρ ρ

= ≥ =

= ≤ =

( )max Bmin 0,75 ;400 mms d≤

• Shear force for which smax is sufficient

• Position of VRd,mid – similar triangles:

• Length of the area reinforced by stirrups with spacing smax:

Stirrups in the middle part (smax)

sw yd

Rd,mid

max

A fV l

s= ∆

VEd,max

VRd,mid

u

Rd,mid Ed,maxV Vu

u v= →

w u l= + ∆

v

VEd,sup

y

B

Ed,sup Ed,max

y v l

V Vv

y v

+ =

= →

lB

• Theoretically, we could calculate VEd,2 using similar triangles and design spacing s2 for the stirrups in the intermediate part (s1 < s2 < smax)

• Position of VEd,2: 2*∆l from the face of the column

• BUT: this makes sense only for really long beams or beams with point forces

• In our case, we will use s1 in the intermediate part

Stirrups in the intermediate part

Layout of stirrups

Spacing of stirrups in the

given area: s1smaxs1

• Draw the scheme above with YOUR numerical values in your homework !!!

wLength of the area:

wy-w v-w

Shear force