Phase 5
Design Calculations
Class of Concrete
3 months
going to use the C40/50 concrete class during
ruction. The first number, 40 in this case, is the
class of concrete and it can be denoted as fck which
represents the compressive strength of concrete. It is
a very strong grade of concrete, making it especially
effective for big construction projects such as a
student accommodation that is 5 Storeys high. This
class of concrete can be commonly found in
foundations and beams for structural support in
construction sites.
Beam Calculations
The first part of my design is going to the beam
which is a horizontal element, but on the 2D sketch
of a floor plan, it is the vertical component. The
horizontal component will therefore be the slab.
4.33
433
BED ROOM
BED ROOM
3.5m
A
The image above shows that the beam I have chosen
to evaluate has a span of 3.5m and it is in the middle
of 2 bedrooms. This beam is also going to be on the
ground floor, so when I need to consider the loading
conditions, the imposed loading will be that of the
floor only which is 3KN//m². As the beam is going to
take half the load from both the rooms, the
characteristic imposed load would be the imposed
loading multiplied by the total length across.
Characteristic imposed load=3KN//m² x (2.15+2.15)
m=12.9KN/m
beff
Open Word
bw
b This type of section is known as the Reinforced
Concrete T section, as it is shaped like T. T sections
play pivotal roles in construction as the web can
resist shear stresses effectively and the flanges can
resist bending stresses. The beams are vertical, and
the slabs are horizontal.
All my calculations have assumed that the slab is
one-way
• The value b(width) is 4.3m as it shows the length
halfway between the 2 rooms.
Characteristic Imposed
4.3 12.9KN/m
load=3KN/m² X
•The value hf is the thickness of the slab which I
have chosen to be 150mm=0.15m
•The value bw is the thickness of the beam which I
have chosen to be 250mm-0.25m
I have assumed the total vertical distance (depth)
is 400mm-0.4m
The span/depth is 3.5/0.4-8.75>2.5 so it is a
shallow beam
• Dimensions of beam are 0.4mx0.25m=0.1m²
Reinforced Concrete beams contain steel bars, and
this is because of the nature of both materials.
Concrete works well in compression as it has a high
compressive strength and steel works well under
tension as it has a high tensile strength. The purpose
of the steel bars in the concrete beams are to provide
additional strength to the beam so it can withstand
larger loads.
• I have assumed the diameter of the bar in the
reinforced concrete beam to be 20mm
• The cross-sectional area of the bar is therefore
πD²/4=π x 202/4=314mm²
The distance between the bottom of the steel bar and
the bottom of the beam is called the nominal cover
which I have assumed to be 40mm.
• Weight of concrete=25KN/m³
⚫fyk-500N/mm2=500x10°N/m²
Beam self-weight Weight of concrete x Area of T
section
Open Word
Design au
1
Design load=135gb+1 5gb – (1 35-17 6875)+
d • Dimensions of beam are 0.4mx0.25m=0.1m²
Reinforced Concrete beams contain steel bars, and
this is because of the nature of both materials.
Concrete works well in compression as it has a high
compressive strength and steel works well under
tension as it has a high tensile strength. The purpose
of the steel bars in the concrete beams are to provide
additional strength to the beam so it can withstand
larger loads.
• I have assumed the diameter of the bar in the
reinforced concrete beam to be 20mm
• The cross-sectional area of the bar is therefore
лD²/4=π x 20²/4=314mm²
The distance between the bottom of the steel bar and
the bottom of the beam is called the nominal cover.
which I have assumed to be 40mm.
• Weight of concrete=25KN/m³
• fyk=500N/mm²-500x10°N/m²
Beam self-weight Weight of concrete x Area of T
section
Design load
=25[(4.3x0.15) + (0.25x0.25)]
=17.6875KN/m=17.69KN/m(2dp)
Design load-1.35gk+1.5qk = (1.35x17.6875) +
(1.5x12.9) 43.228125KN/m = 43.23KN/m(2dp)
The Design load is the uniformly distributed load
across the beam's length, 3.5m.
Maximum Bending moment
The equation for the maximum bending moment
is wL2/8=43.23x 3.52/8=66.20KNm(2dp)
Below is the Bending Moment diagram for the
simply supported beam with a Uniformly Distributed
Load.
66.20 kNm
Reaction forces on a simply supported
beam
Open Word
1
maximum shear force is the design load x half the
length=43.23, 1.75 75.65KN(24p). Design load
=25[(4.3x0.15) + (0.25x0.25)]
=17.6875KN/m=17.69KN/m(2dp)
Design load 1.35gk+1.5qk = (1.35x17.6875) +
(1.5x12.9) 43.228125KN/m = 43.23KN/m(2dp)
The Design load is the uniformly distributed load
across the beam's length, 3.5m.
Maximum Bending moment
The equation for the maximum bending moment
is wL2/8=43.23x 3.52/8=66.20KNm(2dp)
Below is the Bending Moment diagram for the
simply supported beam with a Uniformly Distributed
Load.
66.20 KNM
Reaction forces on a simply supported
beam
There will be 2 reaction forces on the beam (1 at
each end) as it is a simply supported beam, so the
maximum shear force is the design load x half the
length=43.23x1.75=75.65KN(2dp).
Below is the Shear Force diagram for the simply
supported beam with a Uniformly Distributed Load.
7565 к
-756541
Effective Depth
The effective depth is the depth from the top of
the beam to the halfway point of the steel bar. In
reinforced concrete beams, there are vertical links
that provide shear reinforce to help maintain the
beam's stability under loading. We can assume
the vertical link to be 10mm.
Effective depth= Total depth of beam - Nominal
Open Word
beffective (Effective flange width)
b1=b2=(h-bw)/2= (4 3-0 25№/2=2.025m.
tic link 400-40-10-10-340mm=0.34m
beffective (Effective flange width)
bl=b2 (b-bw)/2= (4.3-0.25)/2=2.025m
lo=3.5m (span)
beffective, 1= beffective,2=0.2b1 +0.110= (0.2x2.025)
+ (0.1x3.5)=0.755m
beffective beffective, 1+ beffective, 2 + bw=0.755 +
0.755+0.25=1.76m
Area of Concrete
==
Ac (Area of concrete) = (1.76x0.15) + (0.4x(0.4-
0.15))=0.364m²
Calculating K
Assume the neutral axis lies in the flange
The value of b we use is the effective width and
for d we use the effective depth
fck 40MPa 40x10°N/m²=
K-MED/bd2fck=
(66.2x103)/1.76x0.342x40x106-8.13x103<0.207
As our value of K is less than 0.207, there is no
compression reinforcement required
Calculating Z
Z=
Z=
=0.338
0.95d 0.95x0.34=0.323
0.338 0.95d Therefore z=0.95d=0.323m=323mm
Neutral Axis Calculation
x=(d-z)/0.4 (0.34-0.323)/0.4-0.0425m
hf/0.8 0.15/0.8=0.1875m
X
