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 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

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.

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

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.

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