This document contains homework assignments for a reinforced concrete structures course at Aalto University. It includes 5 assignments related to analyzing and designing column-supported slabs and reinforced concrete elements. The first 3 assignments involve forming calculation models, analyzing internal forces, and designing flexural reinforcement for column-supported slabs. The 4th assignment involves checking the capacity of a semi-circular cross-section under biaxial bending and normal force. The 5th assignment provides a problem to design a pile cap foundation using a strut-and-tie model, including forming the model, calculating design forces, sizing reinforcement, and providing a reinforced drawing. Solutions and guidance are provided for each problem.

1. Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 5-Feb-19
Homework assignments and solutions, 2019
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: 2019
Table of contents:
Homework 1. Principles, column supported slab
Homework 2. Design of column-supported flat slab in ULS
Homework 3. Analysis of column-supported flat slab in SLS
Homework 4. Design of semi-circular cross section for biaxial bending and normal force in ULS
Homework 5. Pile-cap design in ULS using Strut and tie models

2. Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 2019 11.1.2019
Homework 1, Principles. Column supported slab 1(1)
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Goal of this assignment is to form simplified calculation models for a column supported two-way slab.
Information:
- Slab geometry: L1=6m ; L2=7m - Slab thickness hL=300mm
- Concrete: C35/45, XC1 - Loads: gk=1 kN/m2 ; qk=5 kN/m2 ; CC2
- Connection between the columns and the slab can be assumed hinged.
- Effect of lateral forces can be neglected
a) Form the calculation model for the slab design strip at MOD C/1-5 *
b) Calculate the effect of actions (Bending moment and Shear force in Ultimate Limit State) at critical
sections for the MOD C/1-5 slab strip*
c) Form the calculation model for the slab design at MOD C/A-F *
d) Calculate the effect of actions (Bending moment and Shear force in Ultimate Limit State) at critical
sections for the MOD 3/A-F slab strip
e) Sketch a principle drawing (plan and sections) showing the placement of tensile reinforcement in
the slab strip MOD C/1-5 (see fig. 2)
f) Sketch a principle drawing (plan and sections) showing the placement of tensile reinforcement in
the slab strip MOD 3/A-F (see fig. 2)
Figure 1. Plan view and section of the column supported slab.
*TIP: How to form simplified model of two-way slab using design strips or the method as known as
“Simplre-frame-method”:
http://www.adaptsoft.com/resources/Structural_Modeling_Slabs_CI_Dec2005.pdf

3. Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 2019 18.1.2019
Homework 2, Design of reinforced column-supported slab in ULS 1(2)
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You are designing a cast-on-situ column-supported-slab (figure 1). Slab thickness is hL=300mm. Slab is
supported by piles that have diameter D400. Connection between the columns and slab is assumed to be
hinged.
- Concrete strength at final condition: C35/45
- Exposure classes XC1. Design working life: 50 years. Consequence class CC2
- Rebar fyk=500MPa, Es=200GPa
- Slab geometry: L1=6m ; L2=7m
- Superimposed dead load: gsDL= 1 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=35mm
- Connection between the columns and the slab can be assumed hinged.
- Effect of lateral forces can be neglected
a) Form the calculation model for the slab strip at MOD C/1-5 and slab strip at MOD C/A-F. Calculate the
effect of actions due to Bending moment in Ultimate Limit State at critical sections. Consider BOTH
NEGATIVE AND POSITIVE BENDING MOMENT.
b) Design the required amount of flexural reinforcement over the column at MOD C/3 in the X-direction
c) Design the required amount of flexural reinforcement over the column at MOD C/3 in the Y-direction
d) Design the required amount of flexural reinforcement between columns (=midspan) at MOD A-B/3 in X-
direction
e) Design the required amount of flexural reinforcement between columns (=midspan) at MOD C/1-2 in
the Y-direction
f) Draw schematic drawings (plan view and cross sections) showing the reinforcement designed in b-e.

4. Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 2019 18.1.2019
Homework 2, Design of reinforced column-supported slab in ULS 2(2)
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Figure 1. Plan view and section of a pile supported slab (paalulaatta). TS=Construction joint. LS=Movement joint.
*TIP: How to form simplified model of two-way slab using design strips or the method as known as
“Simplre-frame-method”:
http://www.adaptsoft.com/resources/Structural_Modeling_Slabs_CI_Dec2005.pdf

5. Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 2019 30.1.2019
Homework 3, Analysis column supported slab in SLS 1(2)
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You are designing a cast-on-situ column-supported-slab (figure 1). Slab thickness is hL=300mm. Slab is
supported by columns that have diameter D400. Connection between the columns and slab is assumed to
be hinged. Goal of this assignment is to calculate the crack width of the slab over the support.
- Concrete strength at final condition: C35/45
- Exposure classes XC1. Design working life: 50 years. Consequence class CC2
- Rebar fyk=500MPa, Es=200GPa
- Slab geometry: L1=6m ; L2=7m
- Superimposed dead load: gsDL= 1 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=35mm
- Connection between the columns and the slab can be assumed hinged.
- Effect of lateral forces can be neglected
a) Form the calculation model for the slab strip at slab strip at MOD 3/A-F. Calculate the effect of actions
due to Bending moment for quasi-permanent combination (Mqp) in SLS over the support C/3.
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) Does the maximum bending moment (for characteristic combination) exceed the cracking moment?
d) Calculate the concrete stress in the compressed part the of section for quasi-permanent combination.
e) Calculate the stress in tensile reinforcement for quasi-permanent combination.
f) Calculate the crack width at tensile reinforcement 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 rebar chosen in HW1-2. Or you can use the following:
Rebar over support: 20T16-c/c100 in one row

6. Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 2019 30.1.2019
Homework 3, Analysis column supported slab in SLS 2(2)
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Figure 1. Plan view and section of a pile supported slab (paalulaatta). TS=Construction joint. LS=Movement joint.
*TIP: How to form simplified model of two-way slab using design strips or the method as known as
“Simplre-frame-method”:
http://www.adaptsoft.com/resources/Structural_Modeling_Slabs_CI_Dec2005.pdf

7. Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 2019 20.1.2019
Homework 5, Design for biaxial bending and normal force in ULS 1(2)
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Goal of the assignment is to check is the resistance of the given cross section (figure 1) and reinforcement adequate
against biaxial bending and normal force. Cross section is affected by eccentric normal force.
- Column concrete class: C50/60, semi-circular column
- Reinforcement fyk=500MPa, 24pcs of 32mm diameter bars placed according to FIG 1.
- Loads affecting the column: Normal force NEd=13100kN
Eccentricities: eX = 100mm, eY=400mm *
*Note: imperfections and 2nd
order effects has already been considered in the given
loads
a) Calculate the effects of action in ultimate limit state around (X-X) and (Y-Y)-axis: MEd.X and MEd.Y
b) Calculate the simplified N-M interaction (capacity) diagram of the cross section about (X-X) axis. Use at least 3
points for the interaction diagrams:
1. Pure tensile failure: Tensile strain of ɛs=1% in top and bottom reinforcement.
2. Balanced failure: Tensile strain of ɛs=fyd/Es in bottom reinforcement and ultimate compressive strain
ɛc=-0,35% in top of concrete section.
4. Pure compression failure: Uniform strain of ɛc=-0,20% in bottom and top of section.
c) Calculate the simplified N-M interaction (capacity) diagram of the cross section about (Y-Y) axis.
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.X and MRd.Y
e) Check the capacity of the cross section against bi-axial bending using to EC2 equation (5.39)
f) Is the capacity of the cross section adequate against biaxial bending and normal force? If not, how could it be
improved?
Figure 1. Cross section and side view of the column

8. Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 2019 20.1.2019
Homework 5, Design for biaxial bending and normal force in ULS 2(2)
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Tip: 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
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.
.
.















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: 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)2
D*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

9. Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 2019 27.1.2019
Homework 5, Strut and tie design of pilecap 1(2)
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You are designing a pilecap foundation. Goal of the task is to form a strut-and-tie model of the structure and
calculate the required rebar.
-Concrete strength at final condition: C30/37 Consequence class CC2
-Rebar fyk=500MPa, Es=200GPa
-Design load in ULS: NEd=3000kN
-Minimum concrete cover to rebar is c=40mm
-Foundation dimensions: L1=L3=500mm ; L2=1400mm ; L4=1400mm ; H=900mm
a=100mm (embedment depth of piles)
D=400mm (diameter of the column on top of the foundation)
-Rectangular piles: Dimensions: 400x400. Length of piles=20m. Concrete used in the piles is C30/37
a) Form the strut-and-tie model of the pile-cap foundation.
b) Calculate the design forces in the struts and ties.
c) Calculate the required amounts of reinforcements in the tensile struts.
d) Check is the allowable compressive stress in concrete exceeded in any struts?
e) Choose the actual amount of reinforcements and place them to the structure. Draw a drawing of the structure
with the reinforcement. Pay attention to detailing and anchoring of tensile struts!
Figure 1. Pland and section of the pile-cap-foundation. 1=Pile. 2=Pilecap foundation. 3=Column above

10. Aalto University Janne Hanka
CIV-E4040 Reinforced Concrete Structures 2019 27.1.2019
Homework 5, Strut and tie design of pilecap 2(2)
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Tips regarding strut-and-ties in EC2:
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.