17.1.1 Assignment No. 1

lifegunbarrelcityΠολεοδομικά Έργα

26 Νοε 2013 (πριν από 3 χρόνια και 9 μήνες)

81 εμφανίσεις

17.1.1
Assignment No. 1

TASK 1

a)
Explain briefly, the effects of gravitational force on the structural elements (walls,

floors, roof, beams and columns) of a building.


Answer:

Every object is attracted towards the Earth due to the gravitational pull. The mass of an object
can be determined from its density and volume, but for structural calculations we need to find
the weight of the object, as part of the design load. As explain
ed in chapter 3 of the book, the
weight of an object is the force with which it is attracted towards the Earth. A structural
element exerts dead load and imposed load on the building and both of them are due to the
gravitational pull (or force). There are
some exceptions like wind load, the effect of which is
different.

Any object which is situated above the ground surface has to be supported otherwise it will
fall due to the gravitational pull. For example, the roof of a building is supported on load
-
bear
ing walls. Similarly, the floors are supported on the walls, and the walls are supported on
the foundation. Finally the foundation is supported on subsoil which is also known as the
natural foundation.

The floor of a building would deflect under the action

of dead load and imposed load, and as
a consequence of the deflection the joists are subjected to compression and tension. The joists
are also subjected to shear force due to the combined load (dead + imposed load). The roof
trusses, which are the structu
ral part of the roof structure, are subjected to tension and
compression due to the combined load. The dead load of the walls creates compression in
bricks, limiting the height of the walls; similarly in concrete columns compressive stress in
concrete is c
reated.

b) Explain the effect of gravitational forces on the shape of structural elements like

walls,
columns and beams.

Answer:

Beams deflect under the combined effect of dead load and imposed load. Simply supported
beams sag whereas cantilevers hog und
er the effect of forces. For more details refer to
Chapter 10 of the book
-

Figures 10.3 and 10.5.

The forces acting on walls and columns cause compression in these elements. Compression
may cause the shortening of walls and columns.

c)
Explain how the
structural elements of a building may fail as a result of

overloading.

Answer:

Overloading in timber joists and reinforced concrete beams may cause failure due to
excessive deflection and excessive shear force. This has been explained in Chapter 10,
sectio
n 10.3. Overloading will create stress in a material which is more than its elastic limit,
therefore, there is a permanent change in the shape. This may cause distortion in the shape of
the building and could result in failure.

Overloading in walls may cau
se crushing of the bricks or blocks. Aerated concrete blocks are
more vulnerable to failure as their strength is far less than that of bricks.

d) A 6.0 m long concrete beam which is 200 mm × 400 mm in cross
-
section, is resting

on
two walls. Calculate the

dead load of the beam if the density of concrete is 2400 kg/m
3
.

Answer:


Cross
-
sectional area of the beam =
0.2 m
×

0.4 m

= 0.08 m
2


Volume = 6.0 m × 0.08 = 0.48 m
2



Mass = Density
×

Volume = 2400
×

0.48 = 1152 kg


Weight or dead load of the beam = M
ass
×

g = 1152
×

9.81 =
11 301.12 N


e) Calculate reactions R
1

and R
2

for the beam shown in Figure 17.1



C
A
B
R
1
R
2
3 kN
5 kN
D
1.5 m
1.5 m
3 m



Figure 17.1

Answer:


Calculation of reaction R
1


R
1

will cause clockwise moment, and bot
h 3 kN and 5

kN

forces will cause anti
-
clockwise moments.


Clockwise moment = R
1

×
6

m



Anti
-
clockwise moment due to the 3 kN force




= 3 kN ×
4.5 m = 13.5

kNm


Anti
-
clockwise moment due to the 5

kN force




= 5 kN × 3 m = 15

kNm


Total anti
-
clockwise moment = 13.5 +

15

=
28.5

kNm

Moment due to R
2

is zero as it acts at point B (see example 11.2 for explanation).


Clockwise moments must be equal to the anti
-
clockwise moments to maintain the
stability of the beam at point B:


R
1

× 6 m = 28.5

kNm


or R
1

=
m
6
kNm
28.5

=
4.75 kN


R
2

= (3 + 5)


4.75 =
3.25 kN