Reinforced concrete Course assignments, 2018

Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 16-Feb-18
Homework assignments and solutions, 2018
All rights reserved by the author.
Foreword:
This educational material includes assignments of the course named CIV-E4040 Reinforced
Concrete Structures from the spring term 2018. Course is part of the Master’s degree programme
of Structural Engineering and Building Technology in Aalto University.
Each assignment has a description of the problem and the model solution by the author. Description
of the problems and the solutions are in English. European standards EN 1990 and EN 1992-1-1 are
applied in the problems.
Questions or comments about the assignments or the model solutions can be sent to the author.
Author: MSc. Janne Hanka
janne.hanka@aalto.fi / janne.hanka@alumni.aalto.fi
Place: Finland
Year: 2018
Table of contents:
Homework 1. Principles
Homework 2. Design of slab-beam structure in ULS
Homework 3. Analysis of beam structure in SLS
Homework 4. Design for biaxial bending and normal force in ultimate limit state
Homework 5. Strut and tie design in ultimate limit state

CIV-E4040 Reinforced Concrete & CIV-E4050 Prestressed concrete structures 23.1.2018
Design aids for EN 1990 and EN 1992-1-1 1(14)
Load combinations and partial factors, Service limit state (SLS) [EN 1990]:
Initial combination for actions pini = SW + 1.15PT
Quasi-permanent combination of actions: pqp= SW + ∑Gj + ∑ψ2Qi + PT
Frequent combination of actions: pf= SW + ∑Gj + ψ1Q1 + ∑ψ2,i+1Qi+1 + PT
Characteristic combination of actions: pc= SW + ∑Gj + Q1 + ∑ψ2,i+1Qi+1 + PT
Example of liveload combination factors: ψ0=0,7; ψ1=0,5; ψ2=0,3 (EN 1990 Class G, garages)
Selfweight of structure SW
Imposed dead load(s) G
Imposed Live load(s) Q
Prestressing force PT
Load combinations and partial factors, Ultimate limit state (ULS) [EN1990]:
STR Strenght pEd=∑ KFI γG Gj + γPHYP
pEd=∑ KFI ξγG Gj + KFIγQQ1 + ∑KFIγQψ0,i+1Qi+1 + γPHYP
Partial factor for live loads: γQ=1,5
Partial factor for dead loads: γG=1,35 ; ξγG=1,15
Partical factor for prestressing loads γP=1,0
Load factor: KFI=1 (Normally)
Hyperstatic action due to prestressing force HYP (Prestressed concrete structures)
Design values and partial factors for concrete and steel. Ultimate limit state [EN 1992-1-1]:
Design compressive strength for concrete fcd= αccfck/γC
Design tensile strength for concrete fctd= αctfctk.0.05/γC
Design bond strength for concrete fbd= 2,25η2η1fctd
Design strength for reinforcing steel fyd= fyk/γS
Design strength for tendons fpd= fp.0.1k/γS
Partial factor for concrete: γC=1,50 (Normally)
γC=1,35 (Prestressed concrete structures)
Partial factor for rebar and tendons: γS=1,15 (Normally)
γS=1,10 (Prestressed concrete structures)
Reduction factor for concrete, compression αcc=0,85
Reduction factor for concrete, tension αct=1,00
Compressive strength of concrete: fck
Tensile strength of concrete: fctk.0.05
Yield strength of reinforcement fyk
Yield strength of tendons fp0.1k
Bond strength factor related to bond condition η1 = 1 for good bond conditions
(see EC2 figure 8.2) η2 = 0,7 for poor bond conditions
Bond strength factor related to bar diameter (ϕ) η2 = 1 for ϕ ≤ 32 mm

Material properties for A500HW and B500B reinforcing steel:
Yield strength: fyk=500MPa
Ultimate strength fuk=kfyk=1.08fyk
Modulus of elasticity: Es= 200 000 MPa
Class: B
Maximum strain at maximum force εuk=5,0%
Strain limit (for inclined branch) εud=1,0%
Material properties for prestressing steel fp0,1k/fpk=1640/1860
0,1 proof strength: fp0,1k=1640 MPa
Ultimate strength: fpk = 1860 MPa
Modulus of elasticity: EP= 195 000 MPa
Maximum strain at maximum force εuk= 3,5%
Strain limit (for inclined branch) εud= 2,0%
Relaxation class and value for ρ1000 Class 2 (wire or strand – low relaxation) ρ1000=2,5%

Material properties for concrete (EN1992-1-1 table 3.1):
Concrete selfweight ρc=25kN/m3.
Coefficient of thermal expansion for concrete 10⋅⋅⋅⋅10-6 K-1
Simplified stress distribution for concrete in ultimate limit state (EN1992-1-1 3.1.6(3)):

ULS & SLS Effective flange width according to EN 1992-1-1:
Minimum clear spacing between pre-tensioned tendons and post-tension ducts according to EC2:

ULS Calculation model for ultimate flexural strength with reinforcement and prestressed tendons
where:
εcu3 = Ultimate strain in concrete UB/BT = UnBonded tendons / Bonded Tendons
AS = Area of reinforcing steel AP.BT/UB = Area of bonded/unbonded tendons
εS = Strain in reinforcing steel εP.BT/UB = εP(∞) +∆εP.BT/UB = Total strain in tendons
σS = min(fyd ; εSES) ∆εP.UB = ∆σP.UB / EP = Additional strain in unbonded tendons
= Stress in reinforcing steel ∆εP.BT = εcu(dP.BT-x)/x= Additional strain in bonded tendons
FS = σSAS ∆σP.UB = 50 MPa = Additional stress in unbonded tendons
= Force in reinforcing steel Pd.t = Pm.t / γP = Design force in tendons
σP.(∞) = Pd.t / AP.BT/UB = Design stress in tendons after all losses
FC = Force in concrete Pm.t = Force in tendons after initial and long term losses
σP.UB = min(fpd ; ∆σP.UB + σP.(∞)) = Ultimate stress in unbonded
tendons
MRd = Ultimate moment capacity σP.BT = min(fpd ; EP εP.BT)= Ultimate stress in bonded tendons
FP.BT/UB = σP AP =ULS Force in tendons

ULS Calculation model for ultimate shear strength with shear reinforcement in cracked regions
Asw = cross-sectional area of the shear reinforcement s = spacing of the stirrups
fywd = design yield strength of the shear reinforcement
ν1= 0,6 = strength reduction factor for concrete cracked in shear
αcw = coefficient taking account of the state of the stress in the compression chord
αcw = 1 (normally for non-prestressed structures)
αcw = 1 + σCP/fcd (0 < σCP < 0,25fcd)
θ = angle between the concrete compression strut and the beam axis perpendicular to the shear force.
Angle is limited to cot θ = 1 … 2.5 [recommended value in design is cot(θ)=1]
bw = minimum width between tension and compression chords
z = the inner lever arm. Approximate value z = 0,9d may normally be used
ULS Calculation model for ultimate shear strength without shear reinforcement in cracked regions
EN 1992-1-1 6.2.2
VRd,c =
, ∗ ∗ ∗
, ∗ ∗ ∗
pL = Rebar ratio = min{ ASL/(bwd) ; 0,02 }
ASL = area of the tensile reinforcement anchored beyond the section considered
bw = smallest width of the cross-section in the tensile area [mm]
k = 1+(200/d)0,5
, where d is in [mm]
fck = is in [MPa]
CRd,c = 0,18/yC , where yc is the partial material factor for concrete

SLS Crack width calculation for flexure [EN1992-1-1 §7.3.4]
Crack width: wk = sr,max * ∆ε
∆ε = εsm - εcm = max
{ − , , [ + #
$ %
$
& , ]}/$
, * /$
Crack spacing: sr.max=3,4*c+0,8*0,5*0,425*ϕ/ρp,eff
Reinforcement ratio ρp,eff = As / Ac,eff
Ac,eff = min+
, ∗ , −
, − - . /
,/
Factor kt = 0,4 (for long term loading)
Concrete tensile strength fct.eff = fctm
Concrete cover to rebar c
Diameter of rebar: ϕ
Stress in reinforcement σs
Effective heightd
Cross section height h
ycr
SLS Cracked deflection calculation [EN1992-1-1 §7.4.3]
Members which are not expected to be loaded above the level which would cause the tensile strength (fctm) of the
concrete to be exceeded anywhere within the member should be considered to be uncracked.
Members which are expected to crack, but may not be fully cracked, will behave in a manner intermediate between the
uncracked and fully cracked conditions and, for members subjected mainly to flexure, an adequate prediction of
behaviour is given by equation:
α = ξαII + (1- ξ)αI
α = Deformation parameter (strain, curvature, or rotation). As a simplification, α may be taken as a
deflection
αI = Deformation parameter calculated for the uncracked section
αII = Deformation parameter calculated for the cracked section
ξ =1 – β*(Mcr/M)2
= Distribution coefficient that considers the condition between fully cracked and uncracked section
Note! ξ = 0 for uncracked sections (M<Mcr)
β = coefficient taking account of the influence of the duration of the loading.
β=0,5 for long term loading. β=1,0 for short term loading.
Mcr = Cracking moment of the section
M = Bending moment effecting the section
Effective area of concrete
in tension surrounding
the reinforcement
Mean strain in the
reinforcement
Height of cracked
neutral axis in the
compressive zone

Tip: Bending moment diagrams for 1-span beams.
/1 = /2 = 34
5637 = 0.532
4
5 7 =
;
<
=
<
> /17 −
472
2
; 7 ≤ 3
0.532
4; B − 3 > 7 > 3
/1 B − 7 −
4 B − 7 2
2
; 7 ≥ B − 3
E637 =
32
4 3B2
− 232
48IJ
/1 = /2 = 0.5B4
5637 =
4B2
8
5 7 = 47/2 B − 7
E637 =
54B2
384IJ
/1 = /2 = 0
5 7 = 5 = 5637
E637 =
5B2
8IJ
/1 = /2 = K
5637 = 3K
5 7 = +
K7; 7 ≤ 3
3K; B − 3 > 7 > 3
K B − 7 ; 7 ≥ B − 3
E637 =
K3
24IJ
3B2
− 432
/L = /2 = 4M/2
5NOP =
BM4
4
−
4M2
8
5 7 =
;
=
>
/L7; 7 ≤ 3
/17 −
472
2
; B − 3 > 7 > 3
/L B − 7 ; 7 ≥ B − 3
ENOP =
M4
384IJ
8BQ
− 4BM2
+ MQ

Tip: Bending moment diagrams for 3-span continuous beam.

Note: Equal span lengths. L=L1=L2…=L5
Bending moment at critical sections: Mp=k*p*L2
(moment due to uniform load “p”)
MP=k*P*L (moment due to point load “P” at midspan)
M1 = bending moment at span1
MB= bending moment at support B (second support)
MC= bending moment at support C (third support)
MD= bending moment at support D (fourth support)
Support reactions: Qp = k’ * p * L (support reaction due to uniform load “p”)
QP = k’ * P (support reaction due to point load “P” at midspan)

CIV‐E4040 Reinforced Concrete Structures 2018 3.1.2018
Homework 1, Principles 1(2)
Return to MyCourses in PDF‐format.

Goal of this assignment is to form simplified calculation models for one casting area in a typical parking
structure.
a) Form the calculation model of the beam at Module D‐E/3
b) Calculate the effect of actions (Bending moment and Shear force in Ultimate Limit State) at critical
sections for the beam
c) Form the calculation model of the slab (supported by the beams) between modules 1‐6/E‐D
d) Calculate the effect of actions (Bending moment and Shear force in Ultimate Limit State) at critical
sections for the slab
e) Sketch a principle drawing showing the placement of tensile reinforcement in the beam (see fig. 2)
f) Sketch a principle drawing showing the placement of tensile reinforcement in the slab (see fig. 2)

Figure 1. Plan view of a parking structure. TS=Construction joint. LS=Movement joint.
Deflection curve and support reactions Bending moment curve: Placement of reinforcement:

Figure 2. Example for a one‐span cantilever beam calculation model and placement of rebar:
LS
TS TS

Sustainable engineering and design
PERUSTASO = Plan view of a typical floor
Consequence class = CC2
LOADS
qk=2.5 kN/m2 (Class F)
gk=0,25 kN/m2
sw=25 kN/m3 (selfweight of concrete)
Structure will be pos-tensioned. However in
this case you disregard any post-tensioning
effects and consider the structure as RC-
structure.
CONSTRUCTION
JOINT
MOVEMENT JOINT
CONSTRUCTION
JOINT
BEAM UNDER
CONSIDERATION

SLAB
thicnkness=180mm
BEAM UNDER
CONSIDERATION
height=750mm
width=800mm

17.1.2018, 1(4) CIV-E4040 RCS, 2018 Janne Hanka, HW01
a - b) Calculation model and loads for the BEAM
Beam cross section
Slab height hL 180mm Beam total height h 750mm
Beam width Bw 800mm Beam design
(loading) width
bf 7500mm
Span lenght Leff 16.8m
Loads
Dead load gk 0.25
kN
m
2
 Live load qk 2.5
kN
m
2

pc 25
kN
m
3

Selfweight
Basic load cases to the beam
Dead load gk.b pc Bw h bf Bw  hL  gk bf gk.b 47.025
kN
m

qk.b qk bf qk.b 18.75
kN
m

Live Load
Design ULS load to the
beam pEd.b max 1.35 gk.b  1.15 gk.b  1.5 qk.b   pEd.b 82.204
kN
m

Effects of actions in ULS
MEd
1
8
Leff
2
 pEd.b MEd 2900 kN m
Bending moment (at midspan)
Shear force (at support)* VEd 0.5 Leff pEd.b VEd 690.5 kN
*Note. Shear force used in design could be reduced acc. to EC2
C:UsersfijanhDesktopRak-43.3111

e ) Rebar sketch

a - b) Calculation model and loads for the SLAB
Width of the calculation model: bslab 2m Lets assume that the ramp is loading a 2 meter wide strip of
the slab
Typical span lenght of the slab Leff.s 7.5m
Basic load cases to the beam
Dead load gk.s pc bslab hL  gk bslab gk.s 9.5
kN
m

qk.s qk bslab qk.s 5
kN
m

Live Load
Slab is supporting the ramp at module 1-2 & 5-6
Span and thickness of ramp: Lramp 10m hramp 0.25m
Dead load from ramp to the
slab strip gk.ramp
Lramp
2
hramp pc gk  gk.ramp 32.5
kN
m
 gk.ramp gk.s 42
kN
m

Live load from ramp to the
slab strip qk.ramp
Lramp
2
qk qk.ramp 12.5
kN
m
 qk.ramp qk.s 17.5
kN
m

Effects of actions in ULS
pEd.s max 1.35 gk.s  1.15 gk.s  1.5 qk.s   pEd.s 18.425
kN
m

At typical spans 2,3,4
At typical
spans 1,5
pEd.s.ramp max 1.35 gk.s
gk.ramp






 1.15 gk.s
gk.ramp






 1.5 qk.s
qk.ramp













pEd.s.ramp 74.5
kN
m


e ) Rebar sketch
Note: Only one design strip was consider in this solution. In actual design more design strips should be evalueated.

Homework 2, Design of reinforced beam‐slab structure in ULS 1(2)

You are designing a cast‐on‐situ beam‐slab parking structure (figure 1). Beam height is H and width Bw.
Slab (beam flange) thickness is hL. Beams are supported by circular columns that have diameter D600,
height=3000mm. Structure is braced with the columns. Column connection to the foundation slab is
assumed to be fixed.

‐ Beam concrete strength at final condition:  C35/45
‐ Exposure classes XC4, XF2, XD1. Design working life: 50 years. Consequence class CC2
‐ Rebar fyk=500MPa, Es=200GPa
‐ Beam span length: L1=17m. Spacing of beams (slab span lengths) L2=8,1m.
‐ Superimposed dead load: gsDL= 0,5 kN/m2. Concrete selfweight ρc=25kN/m3.
‐ Liveload qLL=5 kN/m2. Combination factors: ψ0=0,7; ψ1=0,5; ψ2=0,3 (EN 1990 Class G, garages)
‐ Concrete cover to rebar is c=40mm

a) Form calculation models of the middle beam and slab. Choose the slab thickness hL, beam height H and
beam width Bw. Calculate the effect of actions in Ultimate Limit State (bending moment MEd.beam  & MEd.slab
and shear force VEd.beam  & VEd.slab) at critical sections.

b) Design the required amount of beam flexural reinforcement AS.BEAM (diameter and amount) at critical
section for due to bending moment obtained in (a).

c) Design the required amount of slab flexural reinforcement AS.SLAB (diameter and spacing) at critical
section due to bending moment obtained in (a).

d) Design the required amount of beam shear reinforcement ASW (diameter, number of legs and spacing) at
critical section due to shear force obtained in (a).

e) Check the shear resistance without shear reinforcement of the slab (VRd.c.slab>VEd.slab) at critical section.

f) Draw schematic drawings (cross sections) of the beam and the slab with the designed reinforcement.

Homework 2, Design of reinforced beam‐slab structure in ULS 2(2)

Figure 1. Slab‐beam rc‐structure.

Homework 3, Analysis reinforced beam in SLS 1(2)

You are designing a cast‐on‐situ beam that is a part of beam‐slab structure (figure 1). Beam height is H and
width Bw. Slab (beam flange) thickness is hL. Beams are supported by circular columns that have diameter
D600, height=3000mm. Structure is braced with the columns. Column connection to the foundation slab is
assumed to be fixed.

‐ Beam concrete strength at final condition:  C35/45
‐ Exposure classes XC4, XF2, XD1. Design working life: 50 years. Consequence class CC2
‐ Rebar fyk=500MPa, Es=200GPa
‐ Beam span length: L1=17m. Spacing of beams (slab span lengths) L2=8,1m.
‐ Superimposed dead load: gsDL= 0,5 kN/m2. Concrete selfweight ρc=25kN/m3.
‐ Liveload qLL=5 kN/m2. Combination factors: ψ0=0,7; ψ1=0,5; ψ2=0,3 (EN 1990 Class G, garages)
‐ Assumed loading history for structure will be following: pk(t=28…29d)=selfweight only ;
pk(t=29…30d)=characteristic combination ; pk(t=30d…50years)=quasi‐permanent combination.
‐ Concrete cover is c=40mm

a) Choose the slab thickness hL, beam height H and beam width Bw. Form the calculation model of the
middle beam. Calculate the effects of actions at Service Limit State at critical section for the beam: *
‐ For quasi permanent combination MEk.qp
‐ For characteristic combination MEk.c

b) Calculate the cross‐section properties used in the analysis (Use transformed cross section properties): *
‐ Moment of inertia for uncracked section   IUC
‐ Cracking moment section   MCr
‐ Moment of inertia for cracked section   ICR

Check the SLS conditions for the beam critical section:
c) Calculate the concrete stress in top of section for characteristic combination.

d) Calculate the stress in bottom reinforcement for characteristic combination.

e) Calculate the crack width at bottom reinforcement for quasi‐permanent combination.

f) Calculate the beam deflection for quasi‐permanent combination. **

Condition # Combination EN1990 Limitation EC2 Clause
Final
I Max concrete compression Characteristic σcc.c < 0,6*fck 7.2(2)
I Max rebar tension Characteristic σs.c < 0,8*fyk 7.2(2)
II Max concrete compression Quasi‐permanent σcc.c < 0,45*fck 7.2(3)
III Max deflection Quasi‐permanent
Creep factor = 2
Δ < Span / 250 7.4.1(4)
IV Max crack width Quasi‐permanent wk.max < 0,3mm 7.3.1(5)

*You can use the same dimensions and rebar chosen in HW1. Or you can use the following:
hL=300mm ; bw=1800mm ; H=1500mm. Stirrups: Φsw=16mm. Bottom rebar: 20pcs–Φ32mm in one row

**Consider the loading history

Homework 3, Analysis reinforced beam in SLS 2(2)

Figure 1. Slab‐beam rc‐structure.

AALTO UNIVERSITY CIV-E4040 RCS 2018 Janne Hanka
2. PALKIN MITOITUS
Kaikki tässä kappaleessa annettavat viittaukset ovat EN 1992-1-1:een [1] ellei toisin mainittu.
2.1 Materiaaliominaisuudet
2.1.1 Betoni
Lieriölujuuden ominaisarvo fck 35MPa
Betonin lieriölujuuden ominaisarvo jännityshetkellä fck.i 25MPa
Betonin taivutusvetolujuuden keskiarvo fctm 3.2 MPa
Betonin vetolujuuden 5% fraktiili fctk.0.05 2.2 MPa
Betonin taivutusvetolujuus jännityshetkellä fctm.i 2.6 MPa
Kimmokertoimen keskiarvo Ecm 34 GPa
Kimmokertoimen keskiarvo jännityshetkellä Ecm.i 31 GPa
Betonin murtopuristuma murtorajatilassa εcu3 0.35 %
2.1.2 Betoniteräkset
Vetolujuuden ominaisarvo ja kimmokerroin fyk 500MPa
Betoniterästen kimmokerroin Es 200GPa
Betoniterästen murtovenymä εuk 5% (Sitkeysluokka B)
2.1.4 Osavarmuusluvut
Betonin osavarmuusluku ja
pienennyskerroin murtorajatilassa
αcc 0.85 αct 1
γS 1.15 γC 1.5
Betoniterästen osavarmuusluku
2.1.4.1 Suunnittelulujuudet
Betonin puristuslujuuden suunnitteluarvo fcd
αcc fck
γC
 fcd 19.8 MPa
Vetolujuuden mitoitusarvo fctd
αct fctk.0.05
γC
 fctd 1.5 MPa
Betoniterästen myötölujuuden suunnitteluarvo fyd
fyk
γS
 fyd 434.8 MPa
HW02 (a)
2.2 Poikkileikkausominaisuuksien laskenta
2.2.1 Palkin mittatiedot
Palkin kokonais- ja laipan korkeus h 1400mm hf 280mm
Palkin uuman korkeus hw h hf hw 1120 mm
Palkin uuman leveys ja palkkiväli bw 1200mm bf 8100mm
Palkin jänneväli (yksiaukkoinen palkki) Leff 17m
Tuen pituus palkin suunnassa at
1
2
600 mm Lenght of support in beam direction

2.2.1.1 Koordinaatisto
Poikkileikkausominaisuu
ksia laskettaessa
käytetään seuraavaa
koordinaatistoa:
- Positiivinen +y suunta
on alaspäin
poikkileikkauksen
painopisteestä (PP)
- Negatiivinen -y suunta
on ylöspäin
poikkileikkauksen
painopisteestä (PP)
- Positiivinen (+) jännitys
on vetoa ja
- Negatiivinen (-) jännitys
on puristusta
σx
N
A
M
I
y=
Kuva 2-1. Poikkileikkauksen koordinaatisto jännityksiä laskettaessa.
2.2.1.2 Taivutusvetolujuus
Korotettu taivutusvetolujuus
[Kaava (3.23)]
fctm.fl max 1 1.6
h
1m






fctm fctm.fl 3.2 MPa
Laskelmissa käytettävä
taivutusvetolujuus [7.1(2)] fct.eff fctm.fl fct.eff 3.2 MPa
2.2.2 Laipan tehollisen leveyden laskenta
Kuva 2-2. Laipan toimivan leveyden parametrit [1, s.58]
Palkin jäyhyymomentin laskeminen edellyttää palkin laipan toimivan leveyden laskentaa
EC2 kappaleen 5.3.2.1 mukaan
Yksiaukkoisella rakenteella voidaan momenttien nollakohtien
välisenä etäisyytenä käyttää palkin jänneväliä
L0 1.00 Leff
T-palkin toimiva leveys
[EC2 kaavat (5.7, 5.7a ja
5.7b]
beff.1 min 0.2 bf bw  0.1 L0 0.2 L0  beff.1 3.1m
beff.2 beff.1 beff.2 3.1m
beff beff.1 beff.2 bw beff 7.4m

HW02 (a) - BEAM
2.7 Kuormien vaikutusten laskenta poikkileikkauksittain
Rakenteen omapaino (SW) gsw 25
kN
m
3
Agr gsw 90.3
kN
m

Pysyvä hyötykuorma (DL) g1 bf 0.5
kN
m
2
 g1 4.1
kN
m

Muuttuva hyötykuorma (LL) q1 bf 5
kN
m
2
 ψ1 0.5 ψ2 0.3 q1 40.5
kN
m

Kuva 2-11. Palkin laskentamalli ja kuormitukset.
2.7.1 Kuormien yhdistelyt
pc gsw g1 q1 pc 134.8
kN
m

Kuormien ominaisyhdistelmä
pf gsw g1 ψ1 q1 pf 114.6
kN
m

Kuormien tavallinen yhdistelmä
Kuormien pitkäaikaisyhdistelmä pqp gsw g1 ψ2 q1 pqp 106.5
kN
m

Pysyvien kuormien yhdistelmä
pg gsw g1 pg 94.3
kN
m

Murtorajatilan
yhdistelmä
pEd max 1.35 gsw g1  1.15 gsw g1 
1.5 q1 







pEd 169.3
kN
m

2.7.2 Kuormien vaikutusten laskenta käyttörajatilassa
Momentin yhtälö Mp p x( )
p Leff x
2
1
x
Leff







Kuormien ominaisyhdistelmä Mc x( ) Mp pc x  Mc 0.5 Leff  4871.5 kN m
Kuormien tavallinen yhdistelmä Mf x( ) Mp pf x  Mf 0.5 Leff  4139.9 kN m
Kuormien pitkäaikaisyhdistelmä Mqp x( ) Mp pqp x  Mqp 0.5 Leff  3847.3 kN m
Pysyvien kuormien yhdistelmä Mg x( ) Mp gsw g1 x  Mg 0.5 Leff  3408.4 kN m
2.7.3 Kuormien vaikutusten laskenta murtorajatilassa
Taivutusmomentti MEd x( ) Mp pEd x  MEd.beam Mp pEd
Leff
2







 MEd.beam 6114.2 kN m
Leikkausvoima VEd x( )
x
MEd x( )
d
d
 VEd 0( ) 1438.6 kN

0 5 10 15
0
5600
4200
2800
1400
0
ULTIMATE LIMIT STATE MEd(x)
Pitkäaikaisyhdistelmä (SW+PT+DL+0.3LL)
Tavallinen yhdistelmä (SW+PT+DL+0.5LL)
Ominaisyhdistelmä (SW+PT+DL+1.0LL)
BENDING MOMENT SLS AND ULS
x [m]
M[kNm]
Kuva 2-12. Palkin taivutusmomentin kuvaaja käyttörajatilan kuormien yhdistelyille.
Alle on taulukoitu momentin arvot kuormien alkeistapauksille ja kuormien yhdistelyille.
X 0 0.1Leff Leff
Taulukko 2-2. Taivutusmomentin arvot kuormien alkeistapauksille
X
0
1.7
3.4
5.1
6.8
8.5
10.2
11.9
13.6
15.3
17
m
 Mp gsw X 
0
1174
2088
2740
3132
3262
3132
2740
2088
1174
0
kN m
 Mp g1 X 
0
53
94
123
140
146
140
123
94
53
0
kN m
 Mp q1 X 
0
527
936
1229
1405
1463
1405
1229
936
527
0
kN m

Taulukko 2-3. Taivutusmomentin arvot kuormien yhdistelyille.
X
0
1.7
3.4
5.1
6.8
8.5
10.2
11.9
13.6
15.3
17
m
 Mc X( )
0
1754
3118
4092
4677
4871
4677
4092
3118
1754
0
kN m
 Mf X( )
0
1490
2650
3478
3974
4140
3974
3478
2650
1490
0
kN m
 Mqp X( )
0
1385
2462
3232
3693
3847
3693
3232
2462
1385
0
kN m
 MEd X( )
0
2201
3913
5136
5870
6114
5870
5136
3913
2201
0
kN m


HW02 (a) - SLAB
Calculation model of the slab:
Slab span L2 bf bf 8.1m
Slab (design) width bslab 10m bslab 10 m
Maximum SHEAR force and BENDING moment happens with the following Live Load arrangement
MAX BENDING MOMENT AT SUPPORT B
MEd.slab 0.107 bslab 1.15 hf 25
kN
m
3
0.5
kN
m
2


















 L2
2

0.121 bslab 1.5 5
kN
m
2


















 L2
2


MEd.slab 1200.9 kN m
MAX SHEAR FORCE (RIGHT SIDE OF SUPPORT B)
VEd.slab 0.607 bslab 1.15 hf 25
kN
m
3
0.5
kN
m
2


















 L2
0.620 bslab 1.5 5
kN
m
2


















 L2

VEd.slab 800.7 kN

HW02 (b) - BEAM ULS
2.8 Mitoitus murtorajatilassa
Mitoitetaan tarvittavat teräkset ensin murtorajatilassa. Teräsmäärää lisätään, jos
käyttörajatilan tarkastelut niin vaativat.
Pääterästen halkaisija ϕm 32mm
Hakaterästen halkaisija ϕh 16mm
Suojabetoni hakaterästen pintaan cbot 40mm
Distance between TWO rows of rebar ccbot.rows 80mm
Centroid of BOT rebar from BOT of section (2 ROWS ASSUMED!) ebot cbot ϕh
ϕm
2

ccbot.rows
2
 ebot 112 mm
Vetoterästen tehollinen korkeus ds h ebot ds 1288 mm
Calculation for REQUIRED amount of rebar for the GIVEN BENDING MOMENT:
Required compr. height λxreq.beam ds ds
2
2
MEd.beam
beff fcd
 λxreq.beam 32.9 mm
Required amount of rebar As.req
λxreq.beam beff fcd
fyd
 As.req 11059.7 mm
2

Required number of rebar
rounded up
nbot.req ceil
As.req
0.25 ϕm
2
 π











 nbot.req 14

Calculation for BENDING MOMENT RESISTANCE with the CHOSEN amount of rebar
Kuva 2-13. Taivutuskestävyyden laskentamalli murtorajatilassa.
Laskelmissa käytettävä teräsmäärä AS.BEAM nbot.req 0.25 ϕm
2
 π AS.BEAM 11259 mm
2

Poikkileikkauksen normaalivoimien
tasapainoehto murtorajatilassa:
As fyd. λ x beff η fcd 0=
Josta ratkaistaan neutraaliakselin korkeus
otaksuen, että vain laippa on puristettu x
AS.BEAM fyd
λ beff η fcd
 x 41.9 mm
Jännitysblokin tehollinen korkeus λ x 33.5 mm
Tarkistetaan neutraaliakselin
korkeuden määrityksesä tehty
otaksuma pitää paikkansa:
x hf 1 Vain laippa on puristettu, OK!
FLANGE COMPRESSED!
Tarkistetaan, että otaksuma
vetoterästen myötäämisestä
pitää paikkansa
εs
εcu3 ds x 
x
 εs 10.4 % >
fyd
Es
0.22 % OK!
Momenttikapasiteetti MRd ds
λ x
2







AS.BEAM fyd MRd 6223 kN m
Maksimimomentti
tarkasteltavassa
poikkileikkauksessa
MEd
Leff
2






6114 kN m MRd > MEd OK! MRd MEd.beam 1
nbot.req 14 ϕm 32 mm IS REQUIRED FOR THE BEAM BOTTOM REBAR IN ULS!

HW02 (c) - SLAB ULS
Only TOP rebar calculation in critical section is shown here:
2.8 Mitoitus murtorajatilassa
Mitoitetaan tarvittavat teräkset ensin murtorajatilassa. Teräsmäärää lisätään, jos
käyttörajatilan tarkastelut niin vaativat.
Pääterästen halkaisija / Slab main rebar diamater ϕslab 16mm
Jakoterästen halkaisija / Slab installation rebar diamater ϕslab.inst 10mm
Suojabetoni laatan terästen pintaan / Slab rebar cover cslab 40mm
Centroid of SLAB TOP rebar (1 ROWS ASSUMED!) eslab.top cslab
ϕslab
2
 eslab.top 48 mm
Centroid of SLAB BOT rebar (1 ROWS ASSUMED!) eslab.bot cslab ϕslab.inst
ϕslab
2
 eslab.bot 58 mm
Vetoterästen tehollinen korkeus / Slab top rebar eff. height dslab.top hf eslab.top dslab.top 232 mm
Calculation for REQUIRED amount of rebar for the GIVEN BENDING MOMENT:
Required compr. height λxreq.slab.top dslab.top dslab.top
2
2
MEd.slab
bslab fcd
 λxreq.slab.top 27.8 mm
Required amount of rebar As.req.slab.top
λxreq.slab.top bslab fcd
fyd
 As.req.slab.top 12663.2 mm
2

Required number of rebar
rounded up
nreq.slab.top ceil
As.req.slab.top
0.25 ϕslab
2
 π











 nreq.slab.top 63
Required spacing of rebar
rounded down
ccreq.slab.top Floor
bslab
nreq.slab.top
10mm






 ccreq.slab.top 150 mm
Due to simplifiation same rebar shall be used in BOTTOM! (safe side because moments are smaller at the mid span of slab)

Calculation for BENDING MOMENT RESISTANCE with the CHOSEN amount of rebar
Kuva 2-13. Taivutuskestävyyden laskentamalli murtorajatilassa.
Laskelmissa käytettävä teräsmäärä As.slab.top
bslab
ccreq.slab.top
0.25 ϕslab
2
 π


 As.slab.top 13404 mm
2

Poikkileikkauksen normaalivoimien
tasapainoehto murtorajatilassa:
As fyd. λ x beff η fcd 0=
Josta ratkaistaan neutraaliakselin korkeus
otaksuen, että vain laippa on puristettu xslab.top
As.slab.top fyd
λ bslab η fcd
 xslab.top 36.7 mm
Jännitysblokin tehollinen korkeus λ xslab.top 29.4 mm
Tarkistetaan, että otaksuma
vetoterästen myötäämisestä
pitää paikkansa
εs.slab
εcu3 dslab.top xslab.top 
xslab.top
 εs.slab 1.9 % >
fyd
Es
0.22 % OK!
Momenttikapasiteetti MRd.slab dslab.top
λ xslab.top
2







As.slab.top fyd MRd.slab 1266 kN m
Maksimimomentti
tarkasteltavassa
poikkileikkauksessa
MEd.slab 1201 kN m MRd > MEd OK! MRd.slab MEd.slab 1
ccreq.slab.top 150 mm ϕslab 16 mm IS REQUIRED FOR THE SLAN REBAR IN ULS!
FOR TOP AND BOT REBAR!
2.8.1.1 Taivutuksen edellyttämän vähimmäisraudoituksen tarkastelu / MINIMUM REBAR FOR SLAB
Palkin vähimmäisraudoitus
positiiviselle momentille
[9.2.1.1(1) kaava 9.1]
As.min max 0.26
fct.eff
fyk
 0.0013






bslab dslab.top As.min 3872 mm
2

Tarkistetaan ehto [9.2.1.1(1)] As.slab.top As.min 1 OK!

HW02 (d) - BEAM SHEAR ULS
2.8.2 Mitoitus leikkaukselle
Momenttivarsi leikkausmitoituksessa z 0.9 ds z 1159.2 mm
Leikkausterästen kulma αs 90deg
Betonin puristuspaarteen
kulma 21,8<θ<45
θs 45deg cot θs  1
Leikkausterästen mitoituslujuus fywd fyd fywd 435 MPa
Puristusjännitys murtorajatilassa σcp
Pd.t
Agr
 σcp 0 MPa
2.8.2.1 Maksimileikkauskestävyys leikkausraudoitettuna
Lujuuden pienennyskertoimen arvo
(kaava 6.6) υ 0.6 1
fck
250MPa







 υ 0.5
Kerroin, jonka avulla otetaan
huomioon poikkileikkauksen
puristusjännitystilan vaikutus
αcw 1
σcp
fcd
 αcw 1
Leikkauskestävyyden yläraja VRd.max
υ αcw fcd bw z
cot θs  tan θs 
 VRd.max 7118 kN
Leikkausvoima tuella VEd.max VEd 0( ) VEd.max 1439 kN
Leikkauskestävyyden ylärajan
tarkistus tuella [EC2 6.2.1(8)] VRd.max VEd.max 1 OK!

2.8.2.2 Leikkauskestävyys leikkausraudoitettuna
Leikkausvoima kriittisessä
poikkileikkauksessa etäisyydellä "d"
tuen reunasta.
VEd.d VEd ds at  VEd.d 1169.9 kN
Leikkausterästen halkaisija ϕh 16 mm
Leikkausterästen leikkeiden lukumäärä nh 4
Leikkausterästen pinta-ala Asw nh
ϕh
2
π
4
 Asw 804 mm
2

REQUIRED SPACING or
STIRRUPS rounded down cch Floor
Asw
VEd.d
z fywd cot θs  10mm






 cch 340 mm
Leikkauskestävyys (EC2 kaava 6.8) VRd
Asw
cch
z fywd cot θs  VRd 1192 kN
Leikkauskestävyyden tarkistus min VRd VRd.max  VEd.d 1 OK!
2.8.2.3 Minimileikkausterästen tarkistus
Leikkausraudoitussuhteen
vähimmäisarvo [9.2.2(5) kaava 9.4]
ρw.min 0.08
fck
MPa
fywd
MPa







ρw.min 0.109 %
Peräkkäisten hakojen suurin sallittu väli
jänteen suunnassa [9.2.2(5) kaava 9.6] smax1 0.75 ds 1 cot αs   smax1 966 mm
Peräkkäisten hakojen suurin sallittu väli
jänteen suunnassa [9.3.2(4) kaava 9.4] smax2
Asw
ρw.min bw sin αs 
 smax2 616 mm
Leikkaushakojen leikkeiden suurin
sallittu jakoväli [9.3.2(4) kaava 9.8] st.max min 0.75 ds 600mm  st.max 600 mm
Palkin leveys on 650mm, jolloin vierekkäisten hakojen väli suojabetoni huomioiden on
650-35-35=580mm. Palkki voidaan raudoittaa kaksileikkeisillä haoilla. OK!
Käytetään jänteen keskialueella
harvempaa leikkausteräsjakoa cch.kesk min smax2 smax1  cch.kesk 616 mm
Minimileikkausterästen
leikkauskestävyys
VRd.min
Asw
cch.kesk
z fywd cot θs  VRd.min 658 kN
CHOSEN SHEAR REBAR FOR BEAM:
USE: nh 4 LEGGED STIRRUPS
ϕh 16 mm DIAMETER OF STIRRUPS
cch 340 mm SPACING OF STIRRUPS

HW02 (e) - SLAB TOP REBAR SHEAR DESIGN ULS
3.8.2 Mitoitus leikkaukselle kriittisessä poikkileikkauksessa
3.8.2.1 Leikkauskestävyys leikkausraudoittamattomana
Kuva 3-16. EC2 kaavan (6.2a-b) mukaisen vetoraudoituksen Asl määrittelykohta.
Momenttivarsi leikkausmitoituksessa zslab 0.9 dslab.top z 1159.2 mm
Kerroin "k" k6.2a min 1
200mm
dslab.top
 2






 k6.2a 1.9
Käytetään yksinkertaistuksen vuoksi leikkauskestävyyden laskennassa vetoraudoituksena
edellä laskettua minimiraudoitusta.
Kriittisen poikkileikkauksen taakse
ankkuroitu vetoraudoitus
leikkauskestävyyden laskennassa
ASL As.slab.top ASL 13404.1 mm
2

Raudoitussuhde ρ1 min
ASL
bslab dslab.top
0.02






 ρ1 0.578 %
Kerroin "k1" k1. 0.15
Kerroin CRd.c CRd.c
0.18
γC
 CRd.c 0.1
Leikkauslujuus (kaava 6.4)
vmin 0.035 k6.2a
3
2

fck
MPa






1
2
 MPa vmin 0.6 MPa
Keskeinen puristusjännitys σcp.slab min
Pd.t
Agr
0.2 fcd






 σcp.slab 0 MPa
Kaava
6.2a VRd.c.a CRd.c k6.2a 100 ρ1
fck
MPa







1
3
 MPa k1. σcp










bw ds VRd.c.a 974 kN
Kaava 6.2b VRd.c.b vmin k1. σcp  bw ds VRd.c.b 857 kN
Leikkauskestävyys
leikkausraudoittamattomana VRd.c max VRd.c.a VRd.c.b  VRd.c 974 kN
Check for
SLAB resistance
VRd.c VEd.slab 1 SLAB SHEAR CAPACITY WITHOUT SHEAR REINF. IS OK!
Note - similar check for slab bot shear resistance should be done at the edges

2.8.4 Laipan leikkautumisen tarkastelu
Tarkistetaan laipan leikkautuminen EN 1992-1-1 kohdan 6.2.4 mukaan:
Kuva 2-18. Laipan leikkautumisen laskentamalli [1, s.90]
Betonin puristussauvojen ja leikkausvoimaa vastaan
kohtisuorassa olevan rakenneosan akselin välinen kulma. θf 26.5deg
Käytetään tarkastelupituutena puolikasta
momentin maksimin ja nollakohdan
välisestä etäisyydestä
Δx 0.5 0.5 Leff Δx 4.3m
Taivutusmomentin muutos
tarkasteluetäisyydellä ΔMEd MEd Δx( ) MEd 0( ) ΔMEd 4586 kN m
Normaalivoiman muutos laipassa
pituudella ∆x ΔFEd
beff.1
beff
ΔMEd
z
 ΔFEd 1655 kN
Pituussuuntainen leikkausjännitys
yhdellä puolen uumaa υEd
ΔFEd
hf Δx
 υEd 1.4 MPa
Vaadittu poikittaisraudoitus
pituusyksikköä kohti (kaava 6.21) Asf.req.f
υEd hf
fyd cot θf 
 Asf.req.f 447
mm
2
m

Tarkistetaan ehto 6.2.4(6) υEd 0.4 fctd 0 Laipan leikkausjännitys on niin suuri, että
taivutuksen edellyttämää raudoitusta on
lisättävä laatassa!
EC2 kohdan 6.2.4(5) mukaan laipan ja uuman välisen leikkautumisen ja poikittaisen
taivutuksen yhteisvaikutuksen tapauksessa teräksen pinta-alaksi valitaan kaavasta (6.21)
saatava arvo tai puolet kaavan (6.21) mukaisesta arvosta lisättynä poikittaisen taivutuksen
edellyttämällä arvolla, sen mukaan, kumpi tuottaa suuremman arvon.
Vinon puristusmurron yläraja
(kaava 6.22)
υRd.max υ fcd sin θf  cos θf  υRd.max 4.1 MPa
Tarkistetaan ehto 6.2.4(4) υEd υRd.max 1 OK!

3.8.1.2.1 Yläpinnan vetoterästen mitoitus laipan leikkautumiselle
Palkin laskennassa määritetty laipan
leikkautuminen edellyttämä raudoitus laatassa Asf.req.f 446.7
mm
2
m

Laatan taivutus edellyttää yläpintaan raudoitusta. As.req.top.b
As.req.slab.top
bslab
 As.req.top.b 1266
mm
2
m

Lopullinen vaadittu
raudoitus (EC2
6.2.4(5)
As.top.req max Asf.req.f
Asf.req.f
2
As.req.top.b






 As.top.req 1490
mm
2
m

Käytetään mitoituksessa hyödyksi myös myös laatan alapinnan teräkset jotka
ankkuroidaan tuen (palkin) yli
As.bot.edge 0.25 As.req.top.b As.bot.edge 316.6
mm
2
m

Käytettävissä oleva ylä
ja alapinnan raudoitus laatassa
laipan leikkautumista vastaan
As.tot As.bot.edge As.req.top.b As.tot 1583
mm
2
m

Johtopäätös: Raudoitusta ei ole tarpeen lisätä laippojen leikkautumisen takia, OK! As.tot As.top.req 1
CONCLUSION;
Shearing of flanges requires that 25% bottom rebar has to fully anchored inside the middle beams!

HW03 (a) - BEAM EFFECTS OF ACTION IN SLS
Bending moments in SLS at beam midspan
Characteristic comb. MC Mc 0.5 Leff  MC 4871.5 kN m
Frequent comb. MF Mf 0.5 Leff  MF 4139.9 kN m
Quasi-permanent comb. MQP Mqp 0.5 Leff  MQP 3847.3 kN m
Dead loads only MG Mg 0.5 Leff  MG 3408.4 kN m
HW03 (b) - BEAM CROSS SEC. PROP. IN SLS
nbot.sls nbot.req
nbot.sls 28 ds 1288 mm
Beam rebar
ϕm 32 mm As.bot nbot.sls ϕm
2
 0.25 π As.bot 22518.9 mm
2

h 1.4m hf 0.3m hw 1120 mm bw 1.2m bf 8.1m beff 7.4m
neff
Es
Ecm
 neff 5.9
Uncracked section properties:
Neutral axis
(from top) yuc
neff 1  ds As.bot  0.5 h bw h hf beff bw  hf 0.5 
neff 1  As.bot  bw h hf beff bw 
 yuc 443.5 mm
Moment of
inertia
Iuc
bwh
3
12
h bw
h
2
yuc





2

bf bw  hf
3

12
 hf bf bw  yuc
hf
2







2

neff 1  As.bot ds yuc 2





Iuc 0.65371 m
4

Rebar
static
moment
Suc.qp As.bot ds yuc  Suc.qp 0.019 m
3

Cracking moment σ
M
I
y= ---> fctm.
Mcr
Iuc.
h yuc. = solve Mcr
Iuc. fctm.
h yuc.
 Mcr
Iuc fctm
h yuc
 Mcr 2193.8 kN m
Cracked section properties:
Assuming ycr < hf
ycr
neff As.bot  As.bot neff 
2
2 beff As.bot neff ds 
beff
 ycr 198 mm
ycr hf  1 OK!
Jäyhyysmomentti
Icr
1
3
bw ycr
3
 neff As.bot ds ycr 
2



 Icr 0.1602 m
4

Rebar static moment
Scr.qp As.bot ds ycr  Scr.qp 0.025 m
3

Nyrkkisääntö :
Halkeilleen poikkileikkauksen
jäyhyysmomentti on 10...60%
halkeilemattoman poikkileikkauksen
jäyhyysmomentista
RATIO OF MOMENT OF INERTIAS
Icr
Iuc
24.5 %

HW03 (c) - STRESS AT TOP OF SECTION CHAR.COMB
Bending moment MC 4871.5 kN m
Assuming uncracked section σc.top.uc
MC
Iuc
yuc  σc.top.uc 3.3 MPa
Mcr MC 0
Mcr
MC
0.5 Cross section cracks!!!!
Check for cracking
For cracked section σc.top.cr
MC
Icr
ycr  σc.top.cr 6 MPa 0.6fck 21 MPa OK
HW03 (d) - STRESS AT BOT REBAR SECTION CHAR.COMB
Bending moment MC 4871.5 kN m
Assuming cracked section σs.bot.cr
MC
Icr
ds yuc 
Es
Ecm
 σs.bot.cr 150.8 MPa 0.8fyk 400 MPa OK
HW03 (e) - CRACK WIDTH BOTTOM REBAR, QP-COMB
Stress in rebar (cracked)
quasi permanent comb.
σs.QP
MQP
Icr
ds yuc 
Es
Ecm
 σs.QP 119.1 MPa σc.top.qp
MQP
Icr
ycr  4.8 MPa
2.9.2.2 Halkeamalevyden laskenta, alapinta 0.45 fck 15.8 MPa OK!
Lasketaan halkeamaleveys EN 1992-1-1 kappaleen 7.3.4 mukaan:
Kerroin tangoilla joilla on hyvä tartunta k1 0.8
Kerroin venymäjakaumalle taivutuksessa k2 0.5
Kansallisessä liitteessä annetut vakiot k3 3.4 k4 0.425
Kerroin pitkäaikaisvaikutuksille kt 0.4
Tankojen jakoväli cc
bw 2 cbot ϕh  ϕm
nbot.sls 1
 cc 39.1 mm
Kimmomodulien suhde neff
Es
Ecm
= neff 5.9
Suojabetoni AP teräkseen cap cbot ϕh cap 56 mm
Vetojännityksen alaisen
betonin tehollinen korkeus hef min
h
2
2.5 h ds 
h ycr 
3







 hef 280 mm
betonin pinta-ala Ac.eff hef bw Ac.eff 336000 mm
2

betonin raudoitussuhde (7.10)
ρp.eff
As.bot
Ac.eff
 ρp.eff 6.7 %
Halkeamaväli (7.11) sr.max k3 cap k1 k2 k4
ϕm
ρp.eff
 sr.max 272 mm
Venymän muutos vetoteräksissä (kaava 7.9)
Δεsm max 0.6
σs.QP
Es

σs.QP kt
fct.eff
ρp.eff
 1 neff ρp.eff 
Es











 Δεsm 0.046 %
Halkeamalaveys (7.8) wk.qp sr.max Δεsm wk.qp 0.13 mm
Sallittu halkeamalaveys alapinnassa
pitkäaikaisyhdistelmälle
wk.all 0.3mm OK!

HW03 (f) - BEAM DEFLECTION FOR QP-COMB.
Creep factor φ 2
Kuva 2-23. Taipuman laskentamalli kuormien pitkäaikaisyhdistelmällä
Considrefation of creep: Ec.eff
Ecm
1 φ
 Ec.eff 11359 MPa
Uncracked defledtion
Δqp.uc
5 pqp Leff
4

384 Ec.eff Iuc
 Δqp.uc 15.6 mm
Lasketaan taipuman suuruus kuormien tavallisella yhdistelmällä huomioiden halkeilun
vaikutus jäykkyyteen.
Halkeilun vaikutus taipumaan voidaan arvioida
EC2 kaavalla (7.18).
α ζ αII 1 ζ( ) αI=
Jossa α on tarkasteltava siirtymä- tai muodonmuutosparametri. Tässä esimerkissä muodonmuutosparametrina käytetään
taipumaa (∆). αI on paramatrin (taipuman) arvo halkeilemattoman tilan mukaan laskettuna ja αII on parametrin (taipuman) arvo
laskettuna täysin halkeilleen tilan mukaan laskettuna.
Kerroin pitkäaikaisille kuormille jolla huomioidaan kuormituksen
keston vaikutus keskimääräiseen venymään
β 0.5
Vetojäykistyksen huomioiva
jakaumakerroin ζ 1 β
Mcr
MF






2
 ζ 0.9
Uncracked defledtion
Δqp.cr
5 pqp Leff
4

384 Ec.eff Icr
 Δqp.uc 15.6 mm
Taipuman suuruus kuormien
tavalliselle yhdistelmälle
Δqp ζ Δqp.cr 1 ζ( ) Δqp.uc Δqp 56.9 mm
Leff
250
68 mm
DEFLECTION TOO BIG! Rebar amount has to be increased -> by iterration -> to value 26T32
2.10.3.2 Lisätaipuma kutistumasta / Extra deflection due to deflection
εcs 0.03 %
Raudoituksen staattinen momentti
neutraaliakselin suhteen halkeilemattomana
Suc As.bot ds yuc  Suc 19016992.7 mm
3

Raudoituksen staattinen momentti
neutraaliakselin suhteen halkeilleena
Scr As.bot ds ycr  Scr 24548637.4 mm
3

Kutistumisen aiheuttama kaarevuus EC2 kaava (7.20) huomioiden halkeilun vaikutus
κcs ζ εcs neff
Scr
Icr
 1 ζ( ) εcs neff
Suc
Iuc
 κcs 0.00000024
1
mm

Kutistuman aiheuttama lisätaipuma
yksiaukkoisessa palkissa
Δcs
1
8
Leff
2
 κcs Δcs 8.6 mm
Kokonaistaipuma pitkäaikaisyhdistelmällä
huomioiden palautuva rajatila
Δqp.tot Δqp Δcs Δqp.tot 65.6 mm
Sallittu taipuma jänteen keskellä Δqp.all
Leff
250
 Δqp.all 68 mm
Laskettu taipuma on pienempi kuin sallittu, OK!

HW02 (f) - REBAR SKETCH

CIV-E4040 Reinforced Concrete Structures 2018 25.1.2018
Homework 4, Design of rectangular column in ULS 1(2)
Return to MyCourses in PDF-format.
Goal of this assignment is to design a typical rectangular column (AT MODULE 5/F) in a warehouse hall
against biaxial bending and normal force. Column is supporting roof made of steel trusses. Structure is
surfaced with claddings around the perimeter. Structure is surfaced with claddings around the warehouse.
Cladding walls are supported laterally by the columns.
- Column concrete strength at final condition: C40/50 Rebar: fyk=500MPa
- Consequence class CC2.
- Structure main geometry: see the attachments. Length of the column is 8,4m
- Concrete selfweight ρc=25kN/m3. Concrete cover to rebar c=40 mm
- Vertical loads acting on the roof that is supported by the columns:
o Total weifght of one steel-truss Gtruss= 70kN
o Deadload: gsDL= 0,8 kN/m2. Liveload qLL=2 kN/m2. Comb. factors: ψ0=0,7; ψ1=0,5; ψ2=0,3
- Horizontal loads acting on the cladding walls:
o Wind load: qwind=0,5 kN/m2
Comb. factors: ψ0=0,6 ; ψ1=0,2 ; ψ2=0,2
a) Choose the dimensions of the columns cross section (h x h). Form the calculation model of the column.
Calculate the loads acting on the column.
b) Calculate the design axial force NEd and design bending moment MEd for the column.
c) Choose the amount+diameter of main rebars. Calculate the simplified N-M interaction (capacity)
diagram of the cross section.
d) Place the calculated effects of action from (a) to the N-M interaction diagram calculated in (b).
Determine the bending moment capacity of the cross section MRd.
e) Is the capacity of the cross section adequate against biaxial bending and normal force? If not, how
could it be improved?
Figure 1. Structure and cross section of typical column.

Homework 4, Design of rectangular column in ULS 2(2)
Tip b: Simplified N-M interaction diagram of the cross section can be calculated using the following strain
distributions of the cross section according (refer to EC2 figure 6.1):
1. Pure tensile failure: Tensile strain of ɛs=1% in top and bottom reinforcement.
2. Balanced failure: Tensile strain of ɛs=fyk/Es in bottom reinforcement
and ultimate compressive strain ɛc=-0,35% at the top of concrete section.
3. Ultimate compressive strain ɛc=-0,35% at the top of concrete section and compressive strain of ɛc=-0,20% at
the centroid of the cross section.
4. Pure compression failure: Uniform strain of ɛc=-0,20% in bottom and top of cross section.
Tip c: How to evaluate bending moment capacity MRd for the given normal force NEd from the N-M diagram:
Tip (d): Resistance of the cross section against biaxial bending and normal force can be checked using the
following criterion: [EN 1992-1-1 5.8.9(4) equation (5.39)]
1
.
.
.
.















a
zRd
zEd
a
yRd
yEd
M
M
M
M
MEd z/y = design moment around the respective axis
MRd z/y = moment resistance in the respective direction
a = exponent for rectangular cross sections with linear interpolation for intermediate values:
NEd/NRd = 0,1 0,7 1,0
a = 1,0 1,5 2,0
NEd = design value of axial force
NRd = Acfcd + Asfyd, is the design axial resistance of section.
Ac = area of the concrete section As = area of longitudinal reinforcement
Tip b: Design bending moment and moments due to imperfection and second order effects can be estimated
with the following equations (According to RakMK B4 §2.2.5.4)* :
Design bending moment: MEd = MEd.0 + Mi + M2
Moment due to actions (hor. force etc) MEd.0
Moment due to imperfections Mi = D/20 + L0/500
Moment due to 2nd order effects M2 = (λ/145)2D*NEd
NEd = Design normal force
D = Diameter of circular column or height of rectangular column
L0 = L*μ = Buckling length of column
λ = 4L0/D = Slenderness ratio for circular columns
λ = 3,464*L0/D = Slenderness ratio for rectangular columns
μ = Buckling factor. μ=2 for mast columns. μ=1 for braced columns
* RakMK method can be used in exercise because, EC2 calculation method for 2nd
order effects is rather cumbersome. RakMK is yields generally
more conservative results, thus the design on the safe side. Detailed design method acc. to EC2 has been shown in:
http://www.elementtisuunnittelu.fi/fi/runkorakenteet/pilarit/nurjahduspituus
http://eurocodes.fi/1992/paasivu1992/sahkoinen1992/Leaflet_5_Pilarit.pdf

LASTAUSLAITURI
JA -KATOS
VARASTO
A A
BB
~1585,2m2
MYYMÄLÄ-
TILA
111,6m2
SÄVYTYS
81,9m2
TEKN.
TILAT
19,1m2
PALAVAN
AINEEN
VARASTO
24,4m2
hissi-
varaus
+47.8
+46.6
+47.8
NOSTO-OVI +
PIKARULLAOVI
6m x ?m
NOSTO-OVI +
PIKARULLAOVI
4m x ?m
sähköinen
kuormaussilta
sähköinen
kuorm
aussilta
NOSTO-OVI
4m x ?m
ovilämpöpuhallin
SOSIAALI-
TILAT
51,9m2
VSS/
SOS.TILAT
32,2m2
NÄYTTELY-
TILA
57,1m2
KOULUTUS-
TILA
46,1m2
KONE-
HUOLTO
22,7m2
TK
TSTOH
2 hlöä
SIIV. WC
TSTOH
2 hlöä
TSTOH
2 hlöä
TSTOH
2 hlöä
PRH
20,9m2
KÄY
34,3m2
23,5m2 23,5m2
23,5m2
23,5m2
KÄY
9,6m2
KÄY
9,6m2
LE-WC
5,7m2
3,3m2
3,5m2
5,9m2
kt
kt
kt
TRUKKIEN LATAUSPISTEET 6 KPL
0m 2 5 10 20
VARASTO-/TUOTANTOTILAA
POHJA-LAYOUT, 1.KRS 1:250
HAKKILA, VANTAA
08.12.2017
RAK5
ALUSTAVA, VE2
1. KRS:
HUONEISTO-ALA: n. 2237 m²
KERROSALA: n. 2282 m²
Column to be designed
A
A
BB
PLAN VIEW OF A
WAREHOUSE

NÄYTTELYTILA
AULA
VARASTO
KONE-
HUOLTO
IV-KONEH.
KOULUTUS-
TILA
NEUV.
0m 2 5 10 20
LEIKKAUS A-A 1:250
HAKKILA, VANTAA
08.12.2017
RAK 5
ALUSTAVA, VE2
A-A
Roof and columns are braced
with truss members in
longer (x-) direction

VARASTO
0m 2 5 10
LEIKKAUS B-B, 1:100
HAKKILA, VANTAA
08.12.2017
RAK 5
ALUSTAVA, VE2
8,400mm
8,400mm
B-B
Steel trusses.
Total weight of ONE truss is
Gtruss=70kN
Pad (ground bearing) foundations.
Size of foundations:
Lenght x Width = 3000 x 3000mm
Thickness = 1000mm
Coefficient of subgrade reaction for the soil under the
foundation is: cu=100MN/m^3
Columns are masts in lateral (short) direction.
Lenght of column = 8400mm
Fixed connection be-
tween foundation and
column
Pin-connection be-
tween roof truss and
column
Dead load of the roof
DL=0,8 kN/m2
Live load on the roof
LL = 2 kN/m2
Wind load on the
walls
qwind=0,5kN/m2
Wind load on the
walls
qwind=0,5kN/m2

Cross section Geometry SIZE of COLUMN SIZE OF FOUNDATION
Height of member Hcol 380mm Hfound 3m
Width of member Bcol 380mm Bfound 2m
Column lenght Leff 8.4m
MATERIALS
EC2 §3.1 Concrete
Strength class of concrete:
Material properties of Concrete according to EC2 §3.1 table 3.1
fck 40 MPa
Characteristic compressive strength
Selfweight of concrete ρc 25
kN
m
3

Equation (3.19, 3.20) λ 0.8
Equation (3.21, 3.22) η 1
EC2 §3.2 Reinforcing steel
Yield strength and class of reinforcement:
Material properties of reinforcement according to EC2 §3.2 & Annex C
Characteristic yield strength fyk 500 MPa
Design value of modulus of elasticity §3.2.7(4) Es 200000 MPa
EC2 §3.2 Prestressing steel §3.3
Strenght value of prestressing steel
Material properties of prestressing steel according to EC2 §3.2 & Annex C
2 Material design values
2.1 Design values for concrete §3.1.6
Coefficients taking account of long term
effects and unfavourable effects resulting from
the way the load is applied.
Partial safety factor for concrete
§2.4.2.4 Table 2.1N
γc 1.35 αcc 0.85
Design compressive strength (3.15) fcd
αcc fck
γc
 fcd 25.185 MPa
2.2 Design values for reinforcement steel
Partial safety factor for reinforcement
§2.4.2.4 Table 2.1N
γs 1.15
fyd
fyk
γs
 fyd 434.783 MPa
Design yield strength (3.15)

a) Calculation model of the column
Iyy
Hcol
3
Bcol
12
 Iyy 1.738 10
3
 m
4

Slenderness ratio around YY-axis
Slenderness ratio around YY-axis
Izz
Hcol Bcol
3

12
 Izz 1.738 10
3
 m
4

Modulus of subgrade reaction cu 100
MN
m
3

Rotational stiffness around YY kf.yy
1
12
cu Hfound
3
 Bfound kf.yy 450
MN m
rad

Rotational stiffness around ZZ kf.zz
1
12
cu Hfound Bfound
3
 kf.zz 200
MN m
rad

Calculation for buckling lenghts around YY & ZZ axis
NOTE! Flexibility of the foundation shall be taken into account acc.to EC2 5.8.3.2
km.yy_AP
1
kf.yy
Ecm Iyy
Leff
 km.yy_AP 0.016 rad
km.yy_YP 99999999999 rad
μyy max 1 10
km.yy_AP km.yy_YP
km.yy_AP km.yy_YP
 1
km.yy_AP
1 km.yy_AP







1
km.yy_YP
1 km.yy_YP














 μyy 2.03
km.zz_AP
1
kf.zz
Ecm Izz
Leff
 km.zz_AP 0.04
km.zz_YP 999999999999999 rad
μzz 0.5 1
km.zz_AP
0.45 km.zz_AP







1
km.zz_YP
0.45 km.zz_YP







 μzz 0.73
Buckling lengths: L0.yy μyy Leff L0.yy 17.068m L0.zz μzz Leff L0.zz 6.158m
λyy
L0.yy
Iyy
Hcol Bcol
 λyy 155.59 λzz
L0.zz
Izz
Hcol Bcol
 λzz 56.138
Slenderness ratio

b) Calculate the design axial force NEd and design bending moment MEd for the column.
Calculation of loads effecting the column
Spacing of columns cccol 6m
Span of roof truss Ltruss 30m
Weight of the truss NEk.truss 70kN
ψ0 0.7
Live load on roof qroof 2
kN
m
2

Dead load roof groof 0.8
kN
m
2

Wind load effecting envelope qwind 0.5
kN
m
2
 ψ0w 0.6 ψ0w 1.5 0.9
LOAD COMBINATIONS TO BE CONSIDIRED IN ULS
CO1: 1.35SW+1.35DL
CO2: 1.15SW+1.15DL+1,5LL+1,5*0,6WL (Max M, max N)
CO3; 0.90SW+0.90DL+1,5WL (Max M, min N)
CO4: 1.15SW+1.15DL+1,5*0,7LL+1,5WL (Max M, max N)
Effects of actions without partial factors
Dead load of truss and roof
Selfweight of column
NEk.DL
NEk.truss
2
cccol
Ltruss
2
 groof
25
kN
m
3
 Hcol Bcol Leff
 NEk.DL 137.324 kN
Live load on roof NEk.LL cccol
Ltruss
2
 qroof NEk.LL 180 kN
Wind load (load to column) qEk.WL cccol qwind qEk.WL 3
kN
m

QEk.WL qEk.WL Ltruss QEk.WL 90 kN
Shear force at bottom of column
Bending moment at bottom of col. MEk.WL.yy 0.5 qEk.WL Leff
2
 MEk.WL.yy 105.84 kN m

Effects of imperfections and 2nd order effects around YY axis
Bending moment due to
imperfections DL Mi.DL.yy
Hcol
20
L0.yy
500







NEk.DL  Mi.DL.yy 7.297 kN m
imperfections LL Mi.LL.yy
Hcol
20
L0.yy
500







NEk.LL  Mi.LL.yy 9.564 kN m
2order actions DL M2.DL.yy
λyy
145






2
Hcol NEk.DL  M2.DL.yy 60.084 kN m
2order actions LL M2.LL.yy
λyy
145






2
Hcol NEk.LL  M2.LL.yy 78.756 kN m
Effects of imperfections and 2nd order effects around ZZ axis
imperfections DL Mi.DL.zz
Bcol
20
L0.zz
500







NEk.DL  Mi.DL.zz 4.3 kN m
imperfections LL Mi.LL.zz
Bcol
20
L0.zz
500







NEk.LL  Mi.LL.zz 5.637 kN m
2order actions DL M2.DL.zz
λzz
145






2
Bcol NEk.DL  M2.DL.zz 7.822 kN m
2order actions LL
M2.LL.zz
λzz
145






2
Bcol NEk.LL 
M2.LL.zz 10.252 kN m
NOTE. Calculations on safe side because imperfections has been considired in both directions.
Calculation for combination of actions
CO1 NEd.CO1 1.35 NEk.DL  MEd.yy.CO1 1.35 Mi.DL.yy
M2.DL.yy






 MEd.zz.CO1 1.35 Mi.DL.zz
M2.DL.zz







NEd.CO1 185.387 kN MEd.yy.CO1 90.964 kN m MEd.zz.CO1 16.365 kN m
CO2 NEd.CO2 1.15 NEk.DL 
1.5 NEk.LL 
 MEd.yy.CO2 1.15 Mi.DL.yy
M2.DL.yy







1.5 Mi.LL.yy 

1.5 ψ0w MEk.WL.yy 

 MEd.zz.CO2 1.15 Mi.DL.zz
M2.DL.zz







1.5 Mi.LL.yy 

CO3 NEd.CO3 0.9 NEk.DL MEd.yy.CO3 0.9 Mi.DL.yy
M2.DL.yy







1.5 MEk.WL.yy 
 MEd.zz.CO3 0.9 Mi.DL.zz
M2.DL.zz







CO4 NEd.CO4 1.15 NEk.DL 
1.5 ψ0 NEk.LL 
 MEd.yy.CO4 1.15 Mi.DL.yy
M2.DL.yy







1.5 ψ0 Mi.LL.yy 

1.5 MEk.WL.yy 

 MEd.zz.CO4 1.15 Mi.DL.zz
M2.DL.zz







1.5 Mi.LL.yy 


14.2.2018 1(4) CIV-E4040 Reinforced Concrete Structures 2018 Janne Hanka, HW04
CALCULATIONS
h= 380 mm
c) Rebars and N-M diagram b= 380 mm
e= 64.5 mm
c= 40 mm a'= 83.7 mm
Østir = 12 mm Ømain = 25 mm
number of bars side h = 4
number of bars side b = 4
Stress-Strain model of concrete and steel:
Simplified stress block for concrete:
Figure. Principle for calculating N-M diagram for the given stress-strain curve.
Y Y
Z
Z

c) N-M Capacity diagrams of the cross section
Design values for materials in ULS
fck= 30 MPa fyk= 500 MPa
fcd=0,85*fck/1,5= 17.00 MPa fyd=fyk/1,15 434.78 MPa
εcu2= 3.5 ‰ η = 1 Es= 200000 MPa
εc2= 2 ‰ λ = 0.8
N-M diagram about Y-Y axis
Height Strain at the section at differenght heights
comp.zone
Steel line d'n [mm] x = -∞ 184.0 400 ∞ =Height of x
0 0 = εYP -0.01 0.0035 0.003373 0.002 =Strain at top of section
1 64.5 strain εs1 -0.01 0.002273 0.00283 0.002 =Strain at steel line 1…
2 148.2 strain εs2 -0.01 0.000682 0.002124 0.002
3 231.8 strain εs3 -0.01 -0.00091 0.001418 0.002
4 315.5 strain εs4 -0.01 -0.0025 0.000713 0.002
380 00 = εAP -0.01 -0.00373 0.000169 0.002 =Strain at bot of section
Steel line 0 conc. σc = 0.0 17.0 17.0 17.0 =Stress in concrete
1 0 stress σs1 -434.8 417.8 417.8 383.0 =Stress at steel line 1…
2 0 stress σs2 -434.8 136.5 407.8 383.0
3 0 stress σs3 -434.8 -181.8 266.7 383.0
4 0 stress σs4 -434.8 -434.8 125.5 383.0
Steel line As,n [mm2
] Fc = 0 951127.3 2067200 2454800 =Force resultant in conc.
1 1963.5 force Fs1 -853694 820314.2 820314.2 752018.7 =Force resultant in steel
2 981.7 force Fs2 -426847 133959.6 400336.2 376009.4
3 981.7 force Fs3 -426847 -178457 261787.6 376009.4
4 1963.5 force Fs4 -853694 -853694 246477.8 752018.7
Nrd,yy [kN] = Fc + ∑Fs = -2561.08 873 3796 4711 Normal force capacity
Mrd,yy [kNm] = about top = 0 334 140 0 Moment capacity
Point 1 Point 2 Point 3 Point 4
Pure tension
failure
Balanced
failure
Pure compr.
Fail
NOTE! N-M diagram around Z-Z axis is same as around Y-Y because
cross section is symmetrically reinforced
FORCE/VOIMA
[F=N]
STRESS/JÄNNITYS
[σ=N/mm2
]
STRAIN/
MUODONMUUTOS
[ε]

N-M diagrams that describe the maximum capacity of the cross section
16.0; 185.0
-3000
-2000
-1000
0
1000
2000
3000
4000
5000
6000
0 50 100 150 200 250 300 350 400
N[kN]
Mzz [kNm]
N-M diagram around Z-Z-axis
Exact capacity diagram z-z
Load ZZ
Simplified
91.0; 185.0
-3000
-2000
-1000
0
1000
2000
3000
4000
5000
6000
0 50 100 150 200 250 300 350 400
N[kN]
Myy [kNm]
N-M diagram around Y-Y-axis
Exact capacity diagram y-y
Load YY
Simplified
Y Y
Z
Z
NEdNEd
NEd
NEd

d) Moment capacity for the given normal force NEd
NEd= 427 kN Normal force
Moment capacity around Z-Z and Y-Y axis MRd.yy= 295.0 kNm
is (due to symmetry) MRd.zz= 295.0 kNm
Moment capacity for different values of NEd
CO1 NEd= 185 kN MRd.yy=MRd.zz= 270 kNm
Place the normal force NEd to the capacity diagram
and solve for bending moment capacity MRd
185.0
427.0
123.0
346.00; 427 295.0; 427295.0; 0
-3000
-2000
-1000
0
1000
2000
3000
4000
5000
6000
0 50 100 150 200 250 300 350 400
N[kN]
Myy [kNm]
Solution for bending moment capacity for
the given NEd (about Y-Y axis)

e) CHECK FOR BIAXIAL CAPACITY
NRd Hcol Hcol fcd fyd 12
π 25mm( )
2

4
 NRd 6.198 10
3
 kN
aco1 1 MRd.yy 270kN m
CO1
NEd.CO1
NRd
0.03
MRd.zz MRd.yy
MEd.yy.CO1
MRd.yy






aco1
MEd.zz.CO1
MRd.zz






aco1
 0.4 <1 OK!
MRd.yy2 295kN m
CO2
NEd.CO2
NRd
0.069 aco2 1
MRd.zz2 MRd.yy
MEd.yy.CO2
MRd.yy2






aco2
MEd.zz.CO2
MRd.zz2






aco2
 0.7 OK!
NEd.CO3
NRd
0.02 aco3 1 MRd.yy3 265kN m
CO3
MEd.yy.CO3
MRd.yy3






aco3
MEd.zz.CO3
MRd.zz3






aco3
 0.9 OK!
aco4 1 MRd.yy4 260kN mNEd.CO4
NRd
0.056
CO4
MEd.yy.CO4
MRd.yy4






aco4
MEd.zz.CO4
MRd.zz4






aco4
 1.1 OK!
OK! Resistance against Biaxial bending is adequate!
Sketch of cross section:

Homework 5, Strut and tie design 1(2)
You are designing a corbel connection within a cast-in-situ slab structure. Casting area #1 RC-BEAMS are
supporting the casting area #2 RC-BEAMS with beam corbels. Bearings are placed to the connection (figure2)
that transfers only the vertical force.
-Beam concrete strength at final condition: C35/45 Consequence class CC2
-Rebar fyk=500MPa, Es=200GPa
-Beam span length: L1=17m. Spacing of beams (slab span lengths) L2=8,1m. Cantilever L3=3,4m
-Superimposed dead load: gsDL= 0,5 kN/m2. Concrete selfweight ρc=25kN/m3.
-Liveload qLL=5 kN/m2. Combination factors: ψ0=0,7; ψ1=0,5; ψ2=0,3 (EN 1990 Class G, garages)
-Concrete cover to rebar is c=40mm
-Slab thickness hL=300mm. Beam height H=1500mm, width Bw=1600mm.
a) Form calculation model of the RC-BEAM-3 and RC-BEAM-1. Calculate the ULS design force to be used in the
Corbel design.
b) Formulate and draw the calculation model of the corbel connection between RC-BEAM-1 and RC-BEAM-2
using strut-and-tie theory.
b) Calculate the design forces in the struts and ties.
d) Calculate the required amounts of reinforcements in the tensile struts.
e) Check is the allowable compressive stress in concrete exceeded in any struts?
f) Choose the actual amount of reinforcements and place them to the structure. Draw a sketch of the structure
with the reinforcement. Pay attention to detailing and anchoring of tensile struts!
Figure 1. Plan view of the structure.
Figure 2. Detail and dimenstions of the beam corbel.

Homework 5, Strut and tie design 2(2)
Tip (d-e):
Allowable stress for a concrete strut may be calculated using equation:
σRd.max = fcd (with compressive transverse stress or without transverse stress, EC2 fig.6.23)
σRd.max= 0,6 (1 – fck/250) fcd (with tensile transverse stress, EC2 fig.6.23)
Allowable stress for a node may be calculated using equation:
σRd.max = k v’ fcd where:
k= 1 (compression node without tensile ties, EC2 fig.6.26)
k=0,85 (compression-tension node with reinforcement from one direction, EC2 fig.6.27)
k=0,75 (compression-tension node with reinforcement provided in two directions, EC2 fig.6.28)
Figure 6.26: Compression node
without ties.
Figure 6.27: Compression tension
node with reinforcement provided
in one direction.
Figure 6.28: Compression
tension node with
reinforcement provided in
two directions.

Cross section Geometry
Beam lengths L1 17m
Cantilever L3 3.4m
Beam spacing bf 8.1m
Beam width and thickness bw 1600mm hw 1500mm
Slab thickness hf 300mm
MATERIALS
EC2 §3.1 Concrete
Strength class of concrete:
Material properties of Concrete according to EC2 §3.1 table 3.1
fck 35 MPa
Characteristic compressive strength
Selfweight of concrete ρc 25
kN
m
3

Equation (3.19, 3.20) λ 0.8
Equation (3.21, 3.22) η 1
EC2 §3.2 Reinforcing steel
Yield strength and class of reinforcement:
Material properties of reinforcement according to EC2 §3.2 & Annex C
Characteristic yield strength fyk 500 MPa
Design value of modulus of elasticity §3.2.7(4) Es 200000 MPa
Material properties of prestressing steel according to EC2 §3.2 & Annex C
2 Material design values
2.1 Design values for concrete §3.1.6
Coefficients taking account of long term
effects and unfavourable effects resulting from
the way the load is applied.
Partial safety factor for concrete
§2.4.2.4 Table 2.1N
γc 1.35 αcc 0.85
Design compressive strength (3.15) fcd
αcc fck
γc
 fcd 22.037 MPa
2.2 Design values for reinforcement steel
Partial safety factor for reinforcement
§2.4.2.4 Table 2.1N
γs 1.15
fyd
fyk
γs
 fyd 434.783 MPa
Design yield strength (3.15)
LIVE LOADS ON FLOOR
Dead loads gk 0.5
kN
m
2

Live loads qk 5
kN
m
2


a) Calculation for loads on the corbel connection
Total dead load
(RC-BEAM3 support
reaction)
Gk
L1 L3( )
2
bf gk 25
kN
m
3
bf hf hw hf  bw 





 Gk 767.04kN
Total live load (RC-BEAM3
support reaction) Qk
L1 L3( )
2
bf qk  Qk 275.4kN
Design load in ULS VEd max 1.35 Gk 1.15 Gk 1.5 Qk  VEd 1295.2kN
b) STRUT and TIE model of the Corbel
c) Calculation for forces in the struts
Forces shall be calculated using FEM model in which all members are trusses (carry only axial forces)

Software licensed to Sweco
Job Title
Client
Job No Sheet No Rev
Part
Ref
By Date Chd
File Date/Time
1
14-Feb-18
14-Feb-2018 23:56RCS2018_HW5.std
Print Time/Date: 15/02/2018 00:05 Print Run 1 of 1STAAD.Pro V8i (SELECTseries 6) 20.07.11.33
7
896.7 kN896.7 kN
8
-227.8 kN
-227.8 kN
1
1e+003 kN
1e+003 kN
5
-1e+003 kN
-1e+003 kN
2
-730.0 kN -730.0 kN
3
-1e+003 kN
-1e+003 kN
4
1e+003 kN
1e+003 kN
6
-896.7 kN-896.7 kN
X = -896.7 kN
Y = 0.0 kN
Z = 0.0 kN
X = 896.7 kN
Y = 1295.2 kN
Z = 0.0 kN
-1e+003 kN
Axial ForceLoad 1 :
Force - kN
X
Y
Z
Whole Structure Loads 6642.05kN:1m Fx 20000kN:1m 1 LOAD CASE 1
Beam End Forces
Sign convention is as the action of the joint on the beam.
Axial Shear Torsion Bending
Beam Node L/C Fx
(kN)
Fy
(kN)
Fz
(kN)
Mx
(kNm)
My
(kNm)
Mz
(kNm)
1 4 1:LOAD CASE 1486.8 0.0 0.0 0.0 0.0 0.0
2 1:LOAD CASE -1486.8 0.0 0.0 0.0 0.0 0.0
2 4 1:LOAD CASE -730.0 0.0 0.0 0.0 0.0 0.0
5 1:LOAD CASE 730.0 0.0 0.0 0.0 0.0 0.0
3 2 1:LOAD CASE -1139.8 0.0 0.0 0.0 0.0 0.0
3 1:LOAD CASE 1139.8 0.0 0.0 0.0 0.0 0.0
4 3 1:LOAD CASE 1450.3 0.0 0.0 0.0 0.0 0.0
5 1:LOAD CASE -1450.3 0.0 0.0 0.0 0.0 0.0
5 5 1:LOAD CASE -1295.2 0.0 0.0 0.0 0.0 0.0
7 1:LOAD CASE 1295.2 0.0 0.0 0.0 0.0 0.0
6 7 1:LOAD CASE -896.7 0.0 0.0 0.0 0.0 0.0
3 1:LOAD CASE 896.7 0.0 0.0 0.0 0.0 0.0
7 6 1:LOAD CASE 896.7 0.0 0.0 0.0 0.0 0.0
2 1:LOAD CASE -896.7 0.0 0.0 0.0 0.0 0.0
8 2 1:LOAD CASE -227.8 0.0 0.0 0.0 0.0 0.0
5 1:LOAD CASE 227.8 0.0 0.0 0.0 0.0 0.0

d) Required amount of rebar in tensile ties
TIE 2 (horizontal) NEd.2 730kN
TIE 3 (vertical) NEd.3 1139.8kN
TIE 5 (vertical) NEd.5 1295.2kN
Note. It is recommendable to limit the stress in order to
avoid problems with anchorage of tiesDesign stress allowed in ties: σULS 300MPa
Diameter of rebars: ϕm1 16mm Asm1 ϕm1
2
0.25 π 201.062mm
2

ϕm2 20mm Asm2 ϕm2
2
0.25 π 314.159mm
2

Rebar amounts Number of bars required
TIE 2 (horizontal) As2
NEd.2
σULS
 As2 2433mm
2
 Ceil
As2
Asm2
1






8
As3
NEd.3
σULS
 As3 3799mm
2

TIE 3 (vertical) Ceil
As3
Asm2
1






13
TIE 5 (vertical)
As5
NEd.5
σULS
 As5 4317mm
2
 Ceil
As5
Asm2
1






14
e) Check for comressive stress
Allowable stress: fcd.3 0.75 1
fck
250MPa







 fcd fcd.3 14.214MPa
Note. assuming tensile struts only to node
Maximum comp. stress in strus NEd.1 1468kN
Assumed height of strus h1 100mm
Compressive stress σc.1
NEd.1
h1 bw
 σc.1 9.175MPa σc.1 fcd 1 OK!

f) Rebar sketch:

Reinforced concrete Course assignments, 2018

Recommended

Recommended

More Related Content

Similar to Reinforced concrete Course assignments, 2018

Similar to Reinforced concrete Course assignments, 2018 (20)

Recently uploaded

Recently uploaded (20)

Reinforced concrete Course assignments, 2018