Structural
Basics
STRUCTURAL DESIGN
CHEATSHEET
A summary of the most important
formulas, equations & diagrams
𝑞𝑙 2
𝑀=
8
STATICS
Simply supported beam – Line load
q
Reation forces:
𝑙
𝑅𝑎 = 𝑅𝑏 = 𝑞
2
l
Ra
M
Max. shear forces:
𝑙
𝑉𝑎 = 𝑉𝑏 = 𝑞
2
V
Max. bending moment:
𝑙2
𝑀𝑚𝑎𝑥 = 𝑞
8
Rb
Mmax
Va
Vb
Simply supported beam – Point load
Q
Ra
Reation forces:
a
𝑏
𝑙
𝑅𝑎 = 𝑄 , 𝑅𝑏 = 𝑄
b
Rb
l
M
Mmax
Va
linear
𝑎
𝑙
Shear forces:
𝑏
𝑙
𝑉𝑎 = 𝑄 , 𝑉𝑏 = 𝑄
𝑎
𝑙
Max. bending moment:
𝑎∙𝑏
𝑀𝑚𝑎𝑥 = 𝑄
𝑙
V
Vb
For more moment and shear force formulas of the simply supported beam click here.
Cantilever beam – Line load
q
Ma
l
Mmax
Ra
parabolic
Reation forces:
1
2
𝑅𝑎 = 𝑞 ∙ 𝑙, 𝑀𝑎 = − ∙ 𝑞 ∙ 𝑙2
Shear forces:
𝑉𝑎 = −𝑞 ∙ 𝑙
M
V
linear
Structural
Basics
Vb
Max. bending moment:
1
𝑀𝑚𝑎𝑥 = − ∙ 𝑞 ∙ 𝑙 2
2
STATICS
Cantilever beam – Point load
Q
Ma
b
a
l
linear
Reation forces:
𝑅𝑎 = 𝑄
Ra
Max. shear forces:
Mmax
𝑉𝑎 = −𝑄
M
V
Va
Max. bending moment:
𝑀𝑚𝑎𝑥 = 𝑄 ∙ 𝑎
For more moment and shear force formulas of the cantilever beam click here.
2 span continuous beam – Line load on 2 spans
q
Reation forces:
3
8
5
4
𝑅𝑎 = 𝑅𝑐 = 𝑞 ∙ 𝑙, 𝑅𝑏 = 𝑞 ∙ 𝑙
Ra
Rb
l
Mb(-)
+
Rc
l
M
+
Mmax(+)
Va
3
8
5
8
𝑉𝑎 = 𝑉𝑐 = 𝑞 ∙ 𝑙, 𝑉𝑏 = 𝑞 ∙ 𝑙
parabolic
-
Shear forces:
Bending moments:
V
Vb
𝑀𝑚𝑎𝑥 =
9
∙
128
1
8
𝑞 ∙ 𝑙 2 , 𝑀𝑏 = − ∙ 𝑞 ∙ 𝑙 2
Vc
2 span continuous beam – Line load on 1 span
q
Reation forces:
𝑅𝑎 =
Ra
Rb
l
parabolic
+
Va
Mmax(+)
Mb(-)
l
Rc
M
Vc
Vb
V
7
𝑞
16
𝑅𝑐 = −
5
4
∙ 𝑙, 𝑅𝑏 = 𝑞 ∙ 𝑙,
1
𝑞∙𝑙
16
Shear forces:
𝑉𝑎 = 𝑅𝑎 , 𝑉𝑏 = −
9
𝑞
16
∙ 𝑙, 𝑉𝑐 = −𝑅𝑐
Max. bending moment:
49
1
𝑀𝑚𝑎𝑥 =
∙ 𝑞 ∙ 𝑙 2 , 𝑀𝑏 = −
∙ 𝑞 ∙ 𝑙2
512
16
For more moment and shear force formulas of the 2 span continuous beam click here.
Structural
Basics
STATICS
3 span continuous beam – Line load on 3 spans
q
Reation forces:
𝑅𝑎 = 𝑅𝑑 = 0.4 ∙ 𝑞 ∙ 𝑙,
𝑅𝑏 = 𝑅𝑐 = 1.1 ∙ 𝑞 ∙ 𝑙
Ra
Rb
l
Mb(-)
+
l
parabolic
-
+
Rc
-
M
+
Mmax(+)
Va
Rd
l
V
Vc
Vb
Vd
Shear forces:
𝑉𝑎 = 𝑉𝑑 = ±0.4 ∙ 𝑞 ∙ 𝑙 ,
𝑉𝑏 − = −𝑉𝑐 + = −0.6 ∙ 𝑞 ∙ 𝑙 ,
𝑉𝑏 + = −𝑉𝑐 − = 0.5 ∙ 𝑞 ∙ 𝑙
Bending moment:
𝑀𝑚𝑎𝑥 = 0.08 ∙ 𝑞 ∙ 𝑙 2 , 𝑀𝑏 = −0.1 ∙ 𝑞 ∙ 𝑙 2
3 span continuous beam – Line load on 2 spans
q
Reation forces:
𝑅𝑎 = 0.383 ∙ 𝑞 ∙ 𝑙, 𝑅𝑏 = 1.2 ∙ 𝑞 ∙ 𝑙,
𝑅𝑐 = 0.45 ∙ 𝑞 ∙ 𝑙, 𝑅𝑑 = −0.033 ∙ 𝑞 ∙ 𝑙
Ra
Rb
l
Mb(-)
-
l
+
+
Mmax(+)
Va
Rc
Rd
l
Mc(-)
-
M
Mbc(+)
Vd
V
Vc
Vb
Shear forces:
𝑉𝑎 = 0.383 ∙ 𝑞 ∙ 𝑙 , 𝑉𝑑 = 0.033 ∙ 𝑞 ∙ 𝑙
𝑉𝑏 − = −0.617𝑞𝑙 , 𝑉𝑏 + = −0.583𝑞𝑙
𝑉𝑐 − = −0.417𝑞𝑙, 𝑉𝑐 + = 0.033𝑞𝑙
Bending moments:
𝑀𝑚𝑎𝑥 = 0.074𝑞𝑙 2 , 𝑀𝑏 = −0.117𝑞𝑙 2 ,
𝑀𝑏𝑐 = 0.053𝑞𝑙 2 , 𝑀𝑐 = −0.033𝑞𝑙 2
3 span continuous beam – Line load on 1 span
q
Reation forces:
Ra
Rb
l
+
Va
l
Rc
Rd
Mb(-)
M
Mc(+)
Mmax(+)
Vb
l
𝑅𝑎 = 0.433 ∙ 𝑞 ∙ 𝑙, 𝑅𝑏 = 0.65 ∙ 𝑞 ∙ 𝑙,
𝑅𝑐 = −0.1 ∙ 𝑞 ∙ 𝑙, 𝑅𝑑 = 0.017 ∙ 𝑞 ∙ 𝑙
Vd
Vc
Bending moments:
𝑀𝑚𝑎𝑥 = 0.094𝑞𝑙 2 , 𝑀𝑏 = −0.067𝑞𝑙 2 ,
𝑀𝑐 = 0.017𝑞𝑙 2
V
For more moment and shear force formulas of the 3 span continuous beam click here.
Structural
Basics
GEOMETRY
Centroid
The centroid is a point of a cross-section that represents the center of mass. It’s the
point at which the entire area of the section can be assumed to be concentrated.
z1
1
z
z2
centroid
σ𝑛𝑖=1 𝐴𝑖 ∙ 𝑧𝑖
𝑧=
𝐴
2
With,
n
Ai
zi
number of “parts” of the cross-section (n=2 in the shape above)
area of the part i
distance of the centroid of part i from the point of origin (top fibre in
the shape above)
area of the section
A
Moment of inertia
The moment of inertia is a measure of an object’s resistance to changes in rotational
motion. It is used to calculate the bending stresses that a structural element will
experience when subjected to a load.
1
𝐴1
z1
𝑛
z2 c
𝐴2
𝐼 = ෍ 𝐼0.𝑖 + 𝐴𝑖 ∙ 𝑧𝑖 2
𝑖=1
2
With,
n
I0.i
Ai
zi
number of “parts” of the cross-section (n=2 in the shape above)
moment of inertia of part i
area of the part i
distance of the centroid of part i from the centroid c
Structural
Basics
GEOMETRY
Section modulus
1
Section modulus of top fibre:
𝐼
𝑊=
𝑧1
z1
centroid
Section modulus of bottom fibre:
𝐼
𝑊=
𝑧2
z2
2
With,
I
z
moment of inertia around strong axis
distance of fibre from centroid
Moment of inertia formulas
Rectangular section
I/H section
𝑤ℎ3
𝐼𝑦 =
12
ℎ
𝐼𝑧 =
𝑡𝑓
𝑡𝑤
12
-
(𝑤−𝑡𝑤 )∙(ℎ−2𝑡𝑤 )3
12
𝐼𝑧 =
ℎ𝑤 3
12
-
(𝑤−𝑡𝑤 )3 ∙(ℎ−2𝑡𝑤 )
12
𝑤
Hollow circular section
(𝐷4 − 𝑑4 ) ∙ 𝜋
𝐼𝑦 =
64
(𝐷4 − 𝑑 4 ) ∙ 𝜋
𝐼𝑧 =
64
𝐷
𝑤ℎ3
12
ℎ
ℎ𝑤 3
𝑤
𝑑
𝐼𝑦 =
Rectangular hollow section
𝐻
ℎ
𝑏
𝐵𝐻 3 − 𝑏ℎ3
𝐼𝑦 =
12
𝐼𝑧 =
𝐵
For more moment of inertia formulas of other sections click here.
Structural
Basics
𝐻𝐵3 − ℎ𝑏 3
12
LOADS
Overview of loads used in structural design
The “common” characteristic loads that are used in the structural design
of buildings are:
• Dead load
• Live load
• Horizontal wind load on walls
• Wind loads on roofs
• Snow load
• Soil pressure
• Seismic load
Dead load
The dead load represents permanent loads, such as the self-weight of
structural and non-structural building materials. The self-weight of a
concrete slab, a timber truss roof and windows are examples of the dead
load. The weight is calculated and then applied to the structural member
that carries it.
Area dead load:
𝑔𝑘 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 ⋅ 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 [
𝑘𝑁
]
𝑚
Click here for the in-depth article.
Live load
The live load represents variable loads such as weight of people, furniture,
cars, office equipment, etc. that can change over time. It’s an
approximation for structural engineers to estimate the additional weight
(excluding self-weight) that can act on structures due to different room
categories.
You’ll find the values of the live load for different room classes in EN 19911-1 Table 6.2 and the National Annex of your country.
Click here for the in-depth article.
Structural
Basics
LOADS
Horizontal wind loads on walls
The horizontal wind load on buildings are split up in 4 or 5 areas (A, B, C, D,
E) with different load values.
C
E
B
A
C
B
A
D
Wind loads on walls:
𝑤𝑘 = 𝑞𝑝 ⋅ 𝑐𝑝𝑒
With,
qp
cpe
peak velocity pressure
pressure coefficient for each area according to EN 19911-4 Table 7.1
Click here for step-by-step guide for the peak velocity pressure.
And here for the full calculation guide for horizontal wind loads.
Structural
Basics
LOADS
Wind loads on pitched roofs
The wind loads on pitched roofs are split up in 4 or 5 areas (F, G, H, I, (J))
with different load values.
J
F
I
H
G
F
Wind loads on pitched roofs:
𝑤𝑘 = 𝑞𝑝 ⋅ 𝑐𝑝𝑒
With,
qp
cpe
peak velocity pressure
pressure coefficient for each area according to EN 19911-4 Table 7.4a.
Click here for the calculation guide for wind loads on pitched roofs.
Structural
Basics
LOADS
Wind loads on flat roofs
The wind loads on flat roofs are split up in 4 areas (F, G, H, I) with different
load values.
F
H
G
I
F
Wind loads on flat roofs:
𝑤𝑘 = 𝑞𝑝 ⋅ 𝑐𝑝𝑒
With,
qp
cpe
peak velocity pressure
pressure coefficient for each area according to EN 19911-4 Table 7.2.
Click here for the calculation guide for wind loads on flat roofs.
Structural
Basics
LOADS
Snow loads on flat and pitched roofs
The snow load on pitched and flat roofs is calculated as (EN 1991-1-3 (5.1)):
𝑠 = 𝜇𝑖 ⋅ 𝐶𝑒 ⋅ 𝐶𝑡 ⋅ 𝑠𝑘
With,
snow load shape coefficient
exposure coefficient
thermal coefficient
characteristic snow load value on ground
μi
Ce
Ct
sk
Click here for a calculation guide for snow loads on pitched roofs.
And here for flat roofs.
Structural
Basics
LOAD COMBINATIONS
Ultimate limit state (ULS)
෍ 𝛾𝐺.𝑗 ⋅ 𝐺𝑘.𝑗 + 𝛾𝑄.1 ⋅ 𝑄𝑘.1 + ෍ 𝛾𝑄.𝑖 ⋅ 𝜓0.𝑖 ⋅ 𝑄𝑘.𝑖
𝑗≥1
𝑖>1
With,
partial factor for permanent actions
characteristic value of permanent action (e.g. dead load)
partial factor for variable actions
characteristic value of variable action (e.g. live, snow and
wind load)
factor for combination value of a variable action
γG
Gk
γQ
Qk
Ψ0
Serviceability limit state (SLS)
Characteristic combination:
෍ 𝐺𝑘.𝑗 + 𝑄𝑘.1 + ෍ 𝜓0.𝑖 ⋅ 𝑄𝑘.𝑖
𝑗≥1
𝑖>1
Frequent combination:
෍ 𝐺𝑘.𝑗 + 𝜓1.1 ⋅ 𝑄𝑘.1 + ෍ 𝜓2.𝑖 ⋅ 𝑄𝑘.𝑖
𝑗≥1
𝑖>1
Quasi-permanent combination:
෍ 𝐺𝑘.𝑗 + ෍ 𝜓2.𝑖 ⋅ 𝑄𝑘.𝑖
𝑗≥1
𝑖≥1
With,
Ψ1
Ψ2
factor for frequent value of a variable action
factor for quasi-permanent value of a variable action
Structural
Basics
LOAD COMBINATIONS
Accidental limit state (ALS)
෍ 𝐺𝑘.𝑗 + 𝐴𝑑 + ( 𝜓1.1 𝑜𝑟 𝜓2.1 ) ⋅ 𝑄𝑘.1 + ෍ 𝜓2.𝑖 ⋅ 𝑄𝑘.𝑖
𝑗≥1
𝑖>1
Seismic combination:
෍ 𝐺𝑘.𝑗 + 𝐴𝐸𝑑 + ෍ 𝜓2.𝑖 ⋅ 𝑄𝑘.𝑖
𝑗≥1
𝑖≥1
With,
design value of an accidental action
design value of seismic action
Ad
AEd
Full in-depth article about load combinations: here
Structural
Basics
TIMBER DESIGN
Bending verification
Design bending stresses:
𝜎𝑚.𝑦.𝑑 =
𝑀𝑦.𝑑 ℎ
⋅
𝐼𝑦 2
𝜎𝑚.𝑧.𝑑 =
𝑀𝑧.𝑑 𝑤
⋅
𝐼𝑧 2
With,
design bending moment around y/z-axis
moment of inertia around y/z-axis
cross-sectional height
cross-sectional width
Md
I
h
w
Design bending strengths (EN 1995-1-1 (2.17)):
𝑓𝑚.𝑦.𝑑 = 𝑘𝑚𝑜𝑑
𝑓𝑚.𝑦.𝑘
𝛾𝑚
𝑓𝑚.𝑧.𝑑 = 𝑘𝑚𝑜𝑑
𝑓𝑚.𝑧.𝑘
𝛾𝑚
With,
modification factor (EN 1995-1-1 Table 3.1)
characteristic bending strength of timber material
partial factor (EN 1995-1-1 Table 2.3)
kmod
fm.k
γm
Bending verification (EN 1995-1-1 (6.11) + (6.12))
𝑘𝑚 = 0.7/1.0
𝜎𝑚.𝑦.𝑑
𝜎𝑚.𝑧.𝑑
+ 𝑘𝑚 ⋅
≤1
𝑓𝑚.𝑦.𝑑
𝑓𝑚.𝑧.𝑑
𝜎𝑚.𝑦.𝑑 𝜎𝑚.𝑧.𝑑
𝑘𝑚 ⋅
+
≤1
𝑓𝑚.𝑦.𝑑 𝑓𝑚.𝑧.𝑑
Structural
Basics
EN 1995-1-1 6.1.6 (2)
TIMBER DESIGN
Shear verification
Design shear stress:
𝜏𝑣.𝑑 =
3 𝑉𝑑
⋅
2 𝑤⋅ℎ
With,
design shear force
cross-sectional height
cross-sectional width
Vd
h
w
Design shear strengths (EN 1995-1-1 (2.17)):
𝑓𝑣.𝑑 = 𝑘𝑚𝑜𝑑
𝑓𝑣.𝑘
𝛾𝑚
With,
modification factor (EN 1995-1-1 Table 3.1)
characteristic shear strength of timber material
partial factor (EN 1995-1-1 Table 2.3)
kmod
fv.k
γm
Shear verification (EN 1995-1-1 (6.13))
𝜏𝑣.𝑑
≤1
𝑓𝑣.𝑑
Structural
Basics
TIMBER DESIGN
Compression verification (columns)
Design compression stress:
𝜎𝑐.0.𝑑 =
𝑁𝑑
𝑤⋅ℎ
With,
design normal force
cross-sectional height
cross-sectional width
Nd
h
w
Design compression strength – parallel to grain (EN 1995-1-1 (2.17)):
𝑓𝑐.0.𝑑 = 𝑘𝑚𝑜𝑑
𝑓𝑐.0.𝑘
𝛾𝑚
With,
modification factor (EN 1995-1-1 Table 3.1)
characteristic compression strength parallel to grain
partial factor (EN 1995-1-1 Table 2.3)
kmod
fc.0.k
γm
Compression verification (EN 1995-1-1 (6.2))
𝜎𝑐.0.𝑑
≤1
𝑓𝑐.0.𝑑
Structural
Basics
TIMBER DESIGN
Buckling verification (columns)
Buckling length:
𝑙
Radius of inertia:
𝑖=
Slenderness ratio:
𝜆=
𝐼
𝑤⋅ℎ
𝑙
𝑖
𝜆
Relative slenderness ratio (EN 𝜆𝑟𝑒𝑙.𝑦 = ⋅
𝜋
1995-1-1 (6.21)):
𝑓𝑐.0.𝑘
𝐸0.𝑔.05
βc factor (EN 1995-1-1 (6.29))
𝛽𝑐
Instability factor (EN 1995-1-1
(6.27))
𝑘 = 0.5 ⋅ (1 + 𝛽𝑐 ⋅ 𝜆𝑟𝑒𝑙.𝑦 − 0.3 + 𝜆2𝑟𝑒𝑙.𝑦 )
Buckling reduction factor
𝑘𝑐 =
coefficient (EN 1995-1-1 (6.25))
𝑘+
Utilization check one axial
bending (EN 1995-1-1 (6.23))
1
𝑘 2 − 𝜆2𝑟𝑒𝑙.𝑦
𝜎𝑐.0.𝑑
𝜎𝑚.𝑑
+
≤1
𝑘𝑐 ⋅ 𝑓𝑐.0.𝑑 𝑓𝑚.𝑑
With,
fc.0.k
E0.g.05
I
h
w
Structural
Basics
characteristic compression strength parallel to grain
E-modulus parallel to grain, 5% fractile
moment of inertia
cross-sectional height
cross-sectional width
STEEL DESIGN
Material properties
E-modulus:
𝐸 = 0.21 ∙ 106 𝑀𝑃𝑎
Density:
𝜌 = 7850
Poissons ratio:
𝜈 = 0.3
Shear modulus:
𝐺=
Yield and ultimate strength:
𝑓𝑦 , 𝑓𝑢
𝑘𝑔
𝑚3
𝐸
2 ⋅ (1 + 𝜈)
Check out the tables on Eurocodeapplied.com (link)
Strength properties of bolts (EN 1993-1-8 Table 3.1)
Bolt class
Property
4.6
4.8
5.6
5.8
6.8
8.8
10.9
Yield strength fyb (MPa)
240
320
300
400
480
640
900
Ultimate tensile
strength fub (MPa)
400
400
500
500
600
800
1000
Ultimate tensile strength of
welds:
fu = fu of the weaker steel plate
Correlation factor
(dependent on steel strength
(EN 1993-1-8 Table 4.1)
𝛽0
Structural
Basics
STEEL DESIGN
Geometric properties
Cross-sectional height:
Cross-sectional width:
ℎ
𝑤
Web thickness:
𝑡𝑤
Flange thickness:
𝑡𝑓
Root radius:
𝑟
You can find these and many more parameters on Eurocodeapplied.com
(link)
Bolt nominal diameter:
𝑑
Bolt Hole diameter:
𝑑0
Stress area (threaded part of
bolt):
𝐴𝑠
You can find these and many more parameters on Eurocodeapplied.com
(link)
Weld throat thickness:
Structural
Basics
𝑎
STEEL DESIGN
ULS compression verification
𝐴 ⋅ 𝑓𝑦
𝑁𝑐.𝑅𝑑 =
Compressive design
𝛾𝑀0
resistance (EN 1993-1-1 (6.10)):
With,
full/effective cross-sectional area depending on the crosssection class
steel yield strength
partial factor
A
fy
γM0
Compression verification
(EN 1993-1-1 (6.9):
𝑁𝐸𝑑
≤ 1.0
𝑁𝑐.𝑅𝑑
With,
design compression force (normal force)
NEd
ULS bending verification
Bending design resistance
(EN 1993-1-1 (6.13)):
𝑀𝑐.𝑅𝑑 =
𝑊𝑝𝑙 ⋅ 𝑓𝑦
𝛾𝑀0
With,
plastic section modulus (cross-section classes 1+2). You
can find the value here.
Wpl
Bending verification
(EN 1993-1-1 (6.12):
𝑀𝐸𝑑
≤ 1.0
𝑀𝑐.𝑅𝑑
With,
design bending moment
MEd
Structural
Basics
STEEL DESIGN
ULS shear verification
Shear design resistance
(EN 1993-1-1 (6.18)):
𝑉𝑝𝑙.𝑅𝑑
𝑓𝑦
3
= 𝐴𝑣
𝛾𝑀0
With,
shear area. You can find the values here.
Av
Check if shear buckling
verification is required
(EN 1993-1-1 (6.22)):
𝜀≤
235𝑀𝑃𝑎
𝑓𝑦
𝜂 ≤ 1.0
ℎ𝑤
𝜀
≤ 72 ⋅
𝑡𝑤
𝜂
Shear verification
(EN 1993-1-1 (6.17):
𝑉𝐸𝑑
≤ 1.0
𝑉𝑐.𝑅𝑑
With,
design shear force
VEd
Structural
Basics
STEEL DESIGN
ULS flexural buckling verification (columns)
Buckling length:
𝑙
Radius of inertia:
𝑖=
Slenderness (EN 1993-1-1
6.3.1.3 (1)):
𝜆1 = 𝜋
𝐼
𝑤⋅ℎ
𝐸
𝑓𝑦
𝑙 1
𝑖 𝜆1
Non-dimensional slenderness: 𝜆 = ⋅
Buckling curve (EN 1993-1-1
Table 6.1):
𝛼
Reduction factors
(EN 1993-1-1 (6.49)):
𝜙 = 0.5 ⋅ (1 + 𝛼 ⋅ 𝜆 − 0.2 + 𝜆2 ))
𝜒=
Design buckling resistance
(EN 1993-1-1 (6.47))
1
𝜙 + 𝜙 2 − 𝜆2
𝑁𝑏.𝑅𝑑 =
𝜒 ⋅ 𝐴 ⋅ 𝑓𝑦
𝛾𝑀1
With,
full/effective cross-sectional area depending on the crosssection class
A
Utilization check
(EN 1993-1-1 (6.46))
Structural
Basics
𝑁𝐸𝑑
≤1
𝑁𝑏.𝑅𝑑
STEEL DESIGN
Fillet weld design – directional method
Normal stress perpendicular
to throat:
𝜎90 =
Shear stress parallel to the
axis of weld:
𝜏0 =
𝑁
2 ⋅ 𝑙𝑒𝑓𝑓 ⋅ 𝑎 ⋅ 2
+
𝑀
2
2 ⋅ 𝑙𝑒𝑓𝑓
⋅
𝑉
2 ⋅ 𝑙𝑒𝑓𝑓 ⋅ 𝑎
Shear stress perpendicular to 𝜏90 = 𝜎90
the axis of weld:
𝑀
𝑁
𝑉
𝜏90
𝜏0
𝜎90
Verification criteria 1
(EN 1993-1-8 (4.1)):
2
2
𝜎90
+ 3(𝜏02 + 𝜏90
)≤
Verification criteria 2
(EN 1993-1-8 (4.1)):
𝜎90 ≤ 0.9
Structural
Basics
𝑓𝑢
𝛾𝑀2
𝑓𝑢
𝛽𝑤 ⋅ 𝛾𝑀2
𝑎
⋅ 2
6
STEEL DESIGN
Fillet weld design – simplified method
Normal stress in weld:
𝜎𝑁 =
𝑁
𝑀
+
𝑙𝑤 ⋅ 𝑎 𝑊
With,
length of weld
section modulus of weld
lw
W
Shear stress in weld:
𝑉𝑑
Shear stress perpendicular to 𝜏𝑣 =
𝑙𝑤 ⋅ 𝑎
the axis of weld:
𝜎𝑁2 + 𝜏𝑣2
Resulting stress:
𝐹𝑤.𝐸𝑑 =
Resistance stress:
𝐹𝑤.𝑅𝑑 =
Utilization check
(EN 1993-1-8 (4.2)):
𝐹𝑤.𝐸𝑑 ≤ 𝐹𝑤.𝑅𝑑
𝑓𝑢
3 ⋅ 𝛽𝑤 ⋅ 𝛾𝑀2
Full in-depth article about fillet weld design here.
Structural
Basics
RC DESIGN
Material properties
E-modulus concrete
(dependent on concrete
strength class):
𝐸
Partial factor – concrete:
𝛾𝑐
Partial factor – reinforcement: 𝛾𝑠
Characteristic cylinder
compressive strength:
𝑓𝑐𝑘
Design compressive strength: 𝑓𝑐𝑑 =
𝑓𝑐𝑘
𝛾𝑐
Mean tensile concrete
strength:
𝑓𝑐𝑡𝑚
Design tensile strength:
𝑓𝑐𝑡𝑑 =
Yield strength of
reinforcement:
𝑓𝑦𝑘
Design yield strength:
𝑓𝑦𝑑 =
𝑓𝑐𝑡𝑚
𝛾𝑐
𝑓𝑦𝑘
𝛾𝑠
You can find the values of these properties on Eurocodeapplied.com
(link)
Structural
Basics
RC DESIGN
ULS Bending verification
𝑑
Lever arm of longitudinal
reinforcement to compression
fibre
𝜇=
𝑀𝑑
𝑤 ⋅ 𝑑 2 ⋅ 𝜂 ⋅ 𝑓𝑐𝑑
𝜔 =1− 1−2⋅𝜇
Required longitudinal
reinforcement:
𝐴𝑠.𝑟𝑒𝑞 = 𝜔 ⋅
𝑤 ⋅ 𝑑 ⋅ 𝜂 ⋅ 𝑓𝑐𝑑
𝑓𝑦𝑑
With,
bending moment
width of RC beam
factor dependent on concrete class
Md
w
𝜂
Degree of reinforcement
checks:
𝜔𝑚𝑖𝑛 = max(0.26
𝜔𝑏𝑎𝑙 = 𝜆
𝜀𝑐𝑢3
𝜀𝑐𝑢3 ⋅ 𝜀𝑦𝑑
𝜔𝑚𝑎𝑥 = 0.044
Check 1:
𝜔 > 𝜔𝑚𝑖𝑛
Check 2:
𝜔 < 𝜔𝑏𝑎𝑙
Structural
Basics
𝑓𝑦𝑑
𝑓𝑐𝑡𝑚 𝑓𝑦𝑑
⋅
; 0.0013
)
𝑓𝑦𝑘 𝑓𝑐𝑑
𝑓𝑐𝑑
𝑓𝑦𝑑
𝜂 ⋅ 𝑓𝑐𝑑
RC DESIGN
𝜔 < 𝜔𝑚𝑎𝑥
Check 3:
Minimum reinforcement (EN
1992-1-1 9.2.1.1 (9.1N)):
𝐴𝑠.𝑚𝑖𝑛
𝑓𝑐𝑡𝑚
0.26
⋅𝑤⋅𝑑
= 𝑚𝑎𝑥 ቐ
𝑓𝑦𝑘
0.0013 ⋅ 𝑤 ⋅ 𝑑
ULS Shear verification
First, check if shear reinforcement is required.
Members not requiring design shear reinforcement
EN 1992-1-1 6.2.2 (1):
Design value of shear
resistance (EN 1992-1-1
(6.2.a)):
200
𝑑
𝑚𝑚
𝐴𝑠
𝜌1 = min(
; 0.02)
𝑤⋅𝑑
1
0.18
⋅ 𝑘 ⋅ (100𝜌1 ⋅ 𝑓𝑐𝑘 )3
𝛾𝑐
𝜐𝑅𝑑.𝑐 = 𝑚𝑎𝑥
0.051 3
⋅ 𝑘 2 ⋅ 𝑓𝑐𝑘
𝛾𝑐
𝑘 =1+
𝑉𝑅𝑑.𝑐 = 𝜐𝑅𝑑.𝑐 ⋅ 𝑤 ⋅ ℎ
Shear reinforcement required
if:
𝑉𝐸𝑑 > 𝑉𝑅𝑑.𝑐
With,
height of RC beam
width of RC beam
h
VEd
Structural
Basics
RC DESIGN
Members requiring design shear reinforcement
EN 1992-1-1 Figure 6.5
𝑧 = 0.9 ⋅ 𝑑
Coefficient taking into
account state of stress in
compression chord:
𝛼𝑐𝑤
Strength reduction factor for 𝜐1
concrete cracked in shear (EN
1992-1-1 (6.9)):
Angle between concrete
compression strut and beam
axis perp. to shear force:
𝜃
𝑓𝑐𝑑
Shear resistance (EN 1992-1-1 𝑉𝑅𝑑.𝑚𝑎𝑥 = 𝛼𝑐𝑤 ⋅ 𝑤 ⋅ 𝑧 ⋅ 𝜐1 ⋅
cot 𝜃 + tan 𝜃
(6.9)):
𝑉𝐸𝑑
Verification
𝜂=
Reduction of design yield
strength of reinforcement (EN
1992-1-1 (6.8)):
𝑓𝑦𝑤𝑑 = 0.8 ⋅ 𝑓𝑦𝑘
Shear links (EN 1992-1-1 (6.8)):
𝐴𝑠𝑤 =
Inclination of shear
reinforcement:
𝛼
Max. spacing (EN 1992-1-1
(9.6N)):
𝑠𝑙.𝑚𝑎𝑥 = 0.75 ⋅ 𝑑 ⋅ (1 + cot 𝛼 )
Structural
Basics
𝑉𝑅𝑑.𝑚𝑎𝑥
𝑉𝐸𝑑
𝑧 ⋅ 𝑓𝑦𝑤𝑑 ⋅ cot(𝜃 )
RC DESIGN
SLS Crack width verification
Crack width limit (EN 1992-1-1 𝑤𝑚𝑎𝑥
Table 7.1N)
Final creep coefficient
𝜙
E-modulus of concrete longterm (quasi-permanent):
𝐸𝑐.𝑒𝑓𝑓 =
Long-term steel-concrete
ratio:
𝛼𝑠 =
𝐸𝑐𝑚
1+𝜙
𝐸𝑠
𝐸𝑐.𝑒𝑓𝑓
Fc
x
d
d – x/3
Fs
Equilibrium of 1. moment of
area:
𝑤⋅𝑥⋅
Neutral axis (solving for x):
𝑥=
Structural
Basics
𝑥
= 𝛼𝑠 ⋅ 𝐴𝑠 ⋅ (𝑑 − 𝑥)
2
𝛼𝑠 ⋅ 𝐴𝑠
2⋅𝑤⋅𝑑
⋅ (−1 + 1 +
)
𝑤
𝛼𝑠 ⋅ 𝐴𝑠
RC DESIGN
Stress in reinforcement:
𝜎𝑠 =
𝑀𝑞𝑝
𝑥
𝑑 − ⋅ 𝐴𝑠
3
Factor dependent on duration 𝑘𝑡
of load (EN 1992-1-1)
𝑓𝑐𝑡.𝑒𝑓𝑓 = 𝑓𝑐𝑡𝑚
Effective height (EN 1992-1-1
(7.3.2 (3)):
ℎ𝑐.𝑒𝑓𝑓 = min(2.5 ⋅ ℎ − 𝑑 ;
.
𝜌𝑝.𝑒𝑓𝑓 =
𝐴𝑠
ℎ𝑐.𝑒𝑓𝑓 ⋅ 𝑤
EN 1992-1-1 (7.9):
𝜎𝑠 − 𝑘𝑡 ⋅
𝜀𝑠𝑚 − 𝜀𝑐𝑚 = max
ℎ−𝑥 ℎ
; )
3 2
𝑓𝑐𝑡.𝑒𝑓𝑓
⋅ (1 + 𝛼𝑠 ⋅ 𝜌𝑝.𝑒𝑓𝑓 )
𝜌𝑝.𝑒𝑓𝑓
𝐸𝑠
𝜎𝑠
0.6 ⋅
𝐸𝑠
Coefficient (EN 1992-1-1 (7.11)) 𝑘1
Coefficient (EN 1992-1-1 (7.11)) 𝑘2
Max. cracking spacing:
𝑠𝑟.𝑚𝑎𝑥 = 3.4 ⋅ 𝑐 + 0.425 ⋅ 𝑘1 ⋅ 𝑘2 ⋅
Verification:
𝑠𝑟.𝑚𝑎𝑥 < 5 ⋅ (𝑐 +
Structural
Basics
𝑑𝑠
)
2
𝑑𝑠
𝜌𝑝.𝑒𝑓𝑓
RC DESIGN
Crack width:
𝑤𝑘 = 𝑠𝑟.𝑚𝑎𝑥 ⋅ (𝜀𝑠𝑚 − 𝜀𝑐𝑚 )
Utilization:
𝜂=
𝑤𝑘
𝑤𝑚𝑎𝑥
SLS Deflection verification
The deflection calculation of reinforced concrete elements is not as
straightforward as for timber or steel structures. However, according
to Eurocode EN 1992-1-1 the deflection requirement is likely to be
satisfied if the span – effective depth ratio is less than the values given
in EN 1992-1-1 Table 7.4N.
Click here for a reinforced concrete beam design guide.
Structural
Basics