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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