ME31B: CHAPTER FOUR
DESIGN OF STRUCTURAL
MEMBERS
DESIGN OF MEMBERS IN
DIRECT STRESS:

Structural members under
direct stress are mainly ties,
cables, and short columns.
4.1.1 Design of cables and Ties
Example: Two identical ropes support a load P of 5 kN as shown in the figure below.
Calculate the required diameter of the rope, if its ultimate strength is 30 N/mm 2 and a
factor of safety of 4.0 is applied.
F1
60o
30o
F2
P = 5 kN
Fx = 0 ie. F1 cos 60 ° = F2 cos 30 °
F1
= F2 cos 30 ° = 1.732 F2
.........(1)
cos 60 °
y = 0 ie. F1 sin 60 ° + F2 sin 30 ° = 5 kN
0.866 F1 + 0.5 F2 = 5 kN ............(2)
Substitute F1 = 1.732 F2 in (2)
ie. 0.866 x 1.732 F2 + 0.5 F2
From (1), F1
= 1.732 F2
= 5 kN i.e. 1.5 F2 + 0.5 F2 = 5 kN ie. F2 = 2.5 kN
= 1.732 x 2.5 = 4.3 kN
Solution Concluded
The allowable stress in rope = Ultimate stress
= 30
Factor of safety
= 7.5 N/mm2
4
Stress = Force/Area ie. Area = Force/stress = 4.3 x 10 = 573 mm2 (using the
7.5
A   r 2   d 2 / 4, diameter 
4A


4 x 573

bigger force
 27mm (min imum)
Design of Short Columns
4.1.2 Design of Short Columns: A short column has a small height in relation to its
cross sectional area. It is likely to fail due to crushing of the material unlike the slender
columns which fail by buckling. The design of slender columns by buckling is dealt with
later in this chapter.
Slender
Short Columns
Column
Example


A square concrete column(pier) which is 0.5
m high is made of a nominal concrete mix of
1:2:4, with a permissible direct stress of 5.3
N/mm2. What is the required cross-sectional
area if the column is required to carry an
axial load of 300 kN?
Solution:

Area = Force/stress = 300 x 103 N = 56600 mm2
5.3 N/mm2

ie. the column should be minimum 240 mm
square.
4.2 DESIGN OF SIMPLE BEAMS
There are three criteria for the design of simple beams:
a) Bending Moment Criterion: Recall that for a beam: MR = I fc
ymax
where MR is bending resisting moment; I is moment of inertia about
the neutral axis; f c is the maximum compressive stress and
ymax is the distance from neutral axis to end of section.
Since Zc = I/ymax (Z is section modulus)
M = f c . Zc = M or MR = f t . Zt = M
Note


Maximum compressive stress (fc) occurs in
the section where the bending moment is
maximum. In the design of simple beams,
section modulus (Z) should be selected such
that fc
does not exceed the allowable
value. Allowable working stress values can
be found in building codes or Engineering
handbooks.
For safe bending, fw
> f = Mmax/Z
where fw is the allowable bending stress; f
is the actual stress and Mmax is the
maximum bending moment.
b)
Horizontal Shear Force (Q): Acts tangentially to the horizontal
cross sectional plane. The average value of this shear stress is:

= Q/A,
where A is the transverse cross sectional area.
Shear stresses vary from zero at the top to other values elsewhere.
Maximum horizontal shear stress in beams occurs at the
neutral axis of the beam:

For a rectangular section;
max
= 3Q
=
2bd
For a square section, 
For a circular section, 
max
max
= 3Q =
1.5 Q
3 a2
A
=
16

3
For I sections, 
max
=
=
D2
3Q
= 1.5 Q
2A
A
4Q
3A
Q
dxt
(assumes that all shearing resistance is afforded by the web and part
of the flanges that is a continuation of the web).
b)



Deflection of Beams
Excessive deflections cause cracking
of plaster in ceiling and can lead to
jamming of doors and windows.
Most building codes limit the amount of
allowable deflection as a
proportion of the member's length ie.
1/180 , 1/240, or 1/360 of the length.
For standard cases of loading, deflection:

max
= Kc W L3
EI

max
is maximum deflection (mm);
Kc is constant depending on the type of loading and the end
support conditions.
For simple loading, it is 5/384 (See Table 4.4, FAO book for
values of Kc for different loadings).
W is total load (N); L is effective span (mm) = centre to center
distance between supports;
E is the modulus of elasticity (N/mm2 ) and
I is the moment of inertia (mm4).
4.2.1 Steps in the Design of Simple
beams




a) Calculate loading on the beam
b) Calculate the bending moment,
shear forces, etc.
c) For the maximum bending moment
(Mmax), provide a suitable section:
Z = Mmax/fw
, fw is allowable
stress (from tables).
d) Check for shear stress, deflection
and buckling of web if necessary.
Example

Consider a floor where beams are
spaced at 1200 mm and have a span of
4000 mm. The beams are seasoned
cypress with the following properties:
fw = 8 N/mm2 , E = 8400 N/mm 2 ,
density = 500kg/m3 . Loading on
floor and including floor is 2.5 kN/m2 .
Allowable deflection is L/240. Design
the beam. Allowable shear stress is
0.7 N/mm2
b)
Calculate the max. bending moment (Mmax) using the equation
for simple beam, uniformly loaded.
Mmax =
wl2
8
= 3.12 x 42 = 6.24 kN m = 6.24 x 10 6 N mm
8
iv) Find the section modulus(Z)
Z reqd = Mmax
Fw
= 6.24 x 10 6
= 0.78 x 106
mm3
8
iii) Find a suitable beam assuming 100 mm width, from Table 4.3 (FAO book),
section modulus for a rectangular cross section, Z is bd 2 /6
i.e. d =
6 Z = 6 x 0.78 x 106
b
= 216 mm
100
i.e. choose a 100 x 225 mm timber.
Note that since timber required is less than the assumed one, no re-calculations are
necessary.
) Check for shear loading:

= 3 Q = 3 x 6.24 x 10 3
2A
= 0.42 N/mm2
2 x 100 x 225
Since the safe shear stress for the timber is 0.7 N/mm 2
, the section is
adequate in resistance to horizontal shear.
vii) Check deflection to ensure that it is less than 1/240 of the span
 = -5
x W L3 (simple loading)
384
E is 8400 N/mm2
EI
, I = b d3
12
= 100 x 2253
max
=
- 5 x 12.48 x 10 3 x 43 x 109
384
8400 x 95 x 10
3

max
<

mm4
N , L = 4 x 10
= - 13 mm
6
The allowable deflection = 4000/240 = 16.7 mm
Since
6
12
W = 3.12 kN/m x 4 m = 12.48 kN = 12.48 x 10

= 95 x 10
allowable, Beam is satisfactory.
3
mm
4.2.2 Bending Moment Caused By
Askew Loads


When the resulting bending moment on a
beam is not about one of the axis, the
moment need be resolved into components
acting about the main axis.
The stresses are then calculated separately
relative to each axis and the total stress is
found by adding the stresses caused by the
components of the moment.
Roof Truss Showing Purlin
Example

Design a timber purlin, which will span
rafters 2.4 m on centre. The angle of
the roof slope is 30 ° and the purlin will
support a vertical dead load of 250 N/m
and a wind load of 200 N/m acting
normal to the roof. The allowable
bending stress(fw ) for the timber used
is 8 N/mm 2 . The timber density is 600
kg/m3
Forces in the Purlin
Solution







Assume a purlin cross sectional size of 50 x
125 mm
i) Find an estimated self load
w = 0.05 m x 0.125 m x 600 kg/m3 x
9.81 = 37 N/m
Total dead load = 250 + 37 = 287 N/m
ii) Find the components of the loads relative
to the main axes
wx = 200 N/m + 287 N/m cos 30 ° =
448.5 N/m
wy = 287 N/m sin 30 ° = 143.5 N/m
Solution Contd.


iii) Calculate the BM about each axis
for a udl. The purlin is assumed to be a
simple beam
Mmax = w L 2 /8
Mmax x = wx L2 = 448.5 x 2.42
8
8

= 323 x 103 N mm
 Mmax y =
wy L2
= 143.5 x 2.42

8
8
 = 103 x 10 3 N mm

Solution Contd.




iv) The actual stress in the timber must be
less than the allowable stress.
f = Mmax x + Mmax y
< fw
Zx
Zy
v) Try the assumed purlin size of 50 x 125
mm
Zx = b d2
6
= 50 x 1252 = 130 x 10 3 mm3
6
Solution Concluded
Zy =
d b2
= 125 x 50
6
2
= 52 x 103 mm3
6
f = 323 x 103
+
130 x 103
103 x 103
= 2.5 + 2 = 4.5 N/mm2
52 x 103
This size of timber is safe since the actual stress is , < 8 N/mm 2 (the allowable stress)
but a smaller size may be more economical. Try 50 x 100 mm
Zx = bd2 = 50 x 1002 = 83 x 10 3 mm3 ; Zy = d b2 = 100 x 502
6
6
6
6
= 42 x 103 mm3
f=
323 x 103
+ 103 x 103
83 x 103
42 x 103
= 3.9 + 2.5 = 6.4 N/mm 2
This is much closer to the allowable stress so accept ie use 50 x 100 mm size
Try: Use 50 x 75 mm size and see that it is unsafe.
4.2.3 Design Using the Universal Steel
Beams
Steel beams of different cross sectional shapes are commercially
available. The properties of their cross sections can be calculated or
obtained from handbook tables. Using the Steel Designers Manual
(SDM), naming if sections is by using:
t
D
UB
D x B x M e.g.
UB 305 x 102 x 33 ( pp.1014, SDM)
B
UB is Universal beam; D is depth; B is flange width and M is
mass per unit length (kg/m).
Example

A steel beam used as a lintel over a
door opening is required to span 4.5 m
between centres of simple supports.
The beam will carry a 220 mm thick
and 3.2 m high brick wall, weighing 20
kN/m3 . Allowable bending stress is
165 N/mm2. Assume allowable shear
stress of 100 N/mm2, E is 2 x 105
N/mm2. Assume self weight of the
beam as 1.5 kN.
Solution:
udl caused by brick wall = 0.22 m x 3.2 m x 20 kN/m 3
= 14.08 kN/m = 14.08 kN/m x 4.5 m = 63.36 kN.
Self wt. of beam = 1.5 kN
Total udl, W = 63.36 + 1.5 = 64.86 kN
Mmax = W L
= 64.86 x 4.5 = 36.5 kN m = 36.5 x 10 6 N mm
8
8
Zreqd. = Mmax
f
w
=
36.5 x 106
= 0.221 x 10 6 mm3
= 221 cm3
165
The moment of inertia (MI) to limit deflection due to load to L/240 can
be derived from:
L =
240
5 W L3
384 E I
I=
5 x 240 x 64.86 x 1000 N x 4500 2 mm
384 x 2 x 105 N/mm
= 2.07 x 10 7 mm4 = 2.07 x 103 cm 4
= 2070 cm4
With Z = 221 cm3 and I = 2070 cm4 ,
From SDM, choose a UB 254 x 102 x 22 with
Z = 225 .4 cm3 and
I = 2863 cm4
Check for shear stress: Shear stress =
From Table 4.4.
Shear stress =
Q max = W /2
32.43 x 103 N
Qmax
D. t
= 32.43 kN
= 22.01 N/mm2
254 mm x 5.8 mm
Since 22.01 N/mm2 < the allowable shear stress of 100
N/mm2 ,
ie. Beam is very satisfactory.
USE UB 254 x 102 x 22
Composite Beams
4.3 DESIGN OF COLUMNS



Columns are compression members but the
manner in which they tend to fail and the
amount of load which causes failure depend
on:
i) The material of which the column is made
eg. a steel column can carry a greater load
than timber column of similar cross-sectional
size.
ii) The shape of the cross-section of the
column. A column having high c/s area
compared to the height is likely to fail by
crushing rather than by buckling.
Columns Contd.





iii) The end conditions of the column.
To account for buckling of slender columns,
the allowable compressive strength
is
reduced by a factor k , which depends on
the slenderness ratio and the material used.
Pbw = k . cw . A where Pbw
is the
allowable load wrt buckling;
k is the reduction factor which depends on
the slenderness ratio and
A is the cross-sectional area of the column.
4.3.2 Slenderness Ratio






Slenderness ratio can be defined as:
 = kL = l
r
r
Where is slenderness ratio;
k is effective length factor whose value
depends on how the ends of the column are
fixed;
L is the length of the column; r is the radius
of gyration (r = I/A
and l is the effective
length of the columns (k. l)
4.3.3 Types of End Conditions
of a Column:





Columns can either be
(a) fixed in position nor direction (the
weakest condition);
(b) fixed in position but not in direction
(pinned);
(c) fixed in direction but not in position
(d) fixed in position and in direction
4.3.4 Design of Axially Loaded
Timber Columns



Timber columns are designed with the
following formulae:

= kL
and
Pbw = k . cw. A
r
NB: In some building codes, a value of
slenderness ratio in case of sawn timber is
taken as l/b instead of l/r
Example

Design a timber column which is 3 m
long with a compressive load of 15 kN.

Allowable compressive stress ( cw) for
the timber is 5.2 N/mm2, kp is 1.00
Solution: Effective length, l = k L = 1.00 x 3000 = 3000 mm
Try 75 x 125 mm c/section
A = b x d = 75 x 125 = 9375 mm 2
rx =
Ix
A
=
b d3
x 1
12
Likewise, r
= b2 /12 =
=
bd
d2
=
1252
12
= 36.1 mm
12
752 /12 = 21.7 mm.
Use the smaller r, ie ry = 21.7 mm

? = l/r =
3600/21.7 = 138.6 , From Table 4.5 (FAO Book), k
Find the allowable load wrt buckling (P bw)
 . 
cw . A = 0.14 x 5.2 x 9375 = 6825 N = 6.8 kN
?
Pbw = k
Pbw is smaller than the actual load, so the column size is unsafe;
?= 0.14
Try a bigger section e.g. 100 x 125 mm; Area = 100 x 125 = 12500 mm
ry = b2 /12 = 1 1002 /12
= 28.9 mm
?
= l/r = 3000/28.9 = 103.8 ;
k ?= 0.26 (Table 4.5)
Pbw = 0.26 x 5.2 x 12500 = 16900 N = 16.9 kN
Comment: The Pbw on the column is greater than the actual load
of 15 kN, so section is safe. The compressive stress in the column,
 cw = F/A = 15000/12500 = 1.2 N/mm2
This is much less that the allowable compressive stress which made
no allowance for slenderness (I e. 5.2 N/mm 2 )
Note




Actual load/allowable load =
15 kN/16.9 kN = 0.89
This ratio is all right and shows that the
section is economical.
A ratio of 0.85 to 1.00 is acceptable.
4.3.5 Design of Axially Loaded Steel
Columns


The allowable loads for steel with
respect to buckling can be calculated in
the same manner as for timber.
The
relationship
between
the
slenderness ratio and the reduction
factor ( k ) is slightly different (see
Table 4.6).
Example: Calculate the safe load on a hollow square stanchion
whose external dimensions are 120 x 120 mm. The walls of the
columns are 6 mm thick and the allowable compressive stress
 cw is 150 N/mm2 . The column is 4 m high and both ends
are held effectively in position, but one is also restrained in direction.
108 mm
108 mm
120 mm
120 mm
Solution
108 mm
108 mm
120 mm
120 mm
Solution: The effective length of the column, l = 0.85 L =
0.85 x 4000 = 3400 mm
rx = ry =
l
r
Pw  k 
 
I

A

3400
46.6
x  cw
BD 3  bd 3
12 ( BD  bd )
 73, giving
x

1204  1084
12 (120  108 )
2
2
 46.6 mm
k   0.72 (by interpolation from Table 4.6)
A  0.72 x 150 x (1202  1082 )  295 kN
4.3.6 Design of Steel Columns Using BS 5950 and Universal Columns
Example: Design a Universal Column (UC) to carry a load of 1000 kN.
The effective length of the member is 4 m.
Solution: From Table 27a (BS 5950, find enclosed), obtain Pc (allowable
compressive stress). Pc depends on slenderness ratio, l/r
Try
152 x 152 x 37 UC
Area of cross section = 47.4 cm2 (pp 1024)
r min = 3.87 cm (always choose the min of r y and rx as r when
calculating l/r)
Slenderness ratio, l/r = 400 cm/3.87 cm = 103.4
Pc from Table 27 a (BS 5950) = 141- 1.4/2(141 - 137) = 138.2 N/mm2
(Using the design strength of steel, Py as 225 N/mm 2)
Solution Concluded
Check: fc (actual compressive stress) = 1000 x 103 N
= 210.97 N/mm2
47.4 x 102 mm2
As fc > Pc, section is unsafe. Try a larger section e.g. 203 x 203 x 46 UC
Area = 58.8 cm 2
, l/r min = 400 cm/5.11 cm = 78.28
From Table 27a, Pc = 179 - 0.28/2 (179 - 176) = 178.6 N/mm2
fc =
1000 x 103 N
= 170.07 N/mm2
58.8 x 102 mm2
As fc < Pc, the section (203 x 203 x 46 UC) is safe.
Note that the closer the Pc and fc, the more economical the section is.
4.3.7 Axially Loaded Concrete
Columns
Most building codes allow the use of plain
concrete (no reinforcement) only in short
columns ie. where the l/r ratio is < 15 .
For l/r between 10 and 15, the allowable
compressive strength must be reduced.
The tables of figures relating to l/b in place of
a true slenderness ratio are only
approximate, since radii of gyration depend
on both b and d values in the cross section
and must be used with caution.
Example





A concrete column, with an effective length
of 4 m has a cross section of 300 x 400 mm.
Calculate the allowable axial load, if a
nominal concrete mix is 1:2:4 is to be used.
Solution: Slenderness ratio, l/b = 4000/300
= 13.3
Hence Table 4.7, gives Pcc = 3.47 N/mm2
by interpolation.
Pw = Pcc . A = 3.47 x 300 x 400
= 416.4 kN
4.3.8 Plain & Centrally Reinforced
Concrete Walls



Walls are designed in the same
manner as columns, but there are a
few differences.
A wall is different from a column by
having a length which is more than five
times its thickness.
Plain concrete walls should have a
minimum thickness of 100 mm.
Plain and Reinforced Concrete Walls
Contd.


Where the load is eccentric, the wall
must
have
centrally
placed
reinforcement of at least 0.2 % of the
cross section if the eccentricity > 0.20.
This reinforcement may not be
included in the load carrying capacity of
the wall.
Example

Determine the maximum allowable load
per metre of a 120 mm thick wall, with
an effective height of 2.8 m and made
from concrete grade C 15: (a) when
the load is central and (b) when the
load is eccentric by 20 mm.
Solution: Slenderness ratio, l/b = 2800/120 = 23.3
a. Interpolation gives: Pcw = 2.8 - 3.3/5(2.8 - 2.0) = 2.27 N/mm2
Allowable load, Pw = A x Pcw = 1.00 x 0.12 x 2.27 x 106 = 272.4 kN/m wall.
1000
b) Ratio of eccentricity e/b = 20/120 = 0.167
A double interpolation is needed to get Pcw
For C15, and e = 0.1 & l/r = 23.33, Pcw = 1.9 - 3.3/5(1.9 - 1.3) = 1.50 N/mm2
For C15 and e = 0.2 & l/r = 23.33, Pcw = 1.1 - 3.3/5(1.1 - 0.7) = 0.836 N/mm2
Now for e = 0.1, Pcw = 1.50 and for e = 0.2 , Pcw = 0.836
ie. for e = 0.167,
Pcw = 1.50 - 0.067/0.1(1.50 - 0.836) = 1.06 N/mm2
Allowable load, Pw = 1.0 x 0.12 x 1.06 x 106
= 127.2 kN/m wall
1000
* Central reinforcement is not required since e/b < 0.20.m
Example: A steel column in a farm workshop building is to carry a total axial load of
2400 kN. Design a double I- section for the column if its length is 5.25 m (kp = 0.65).
Solution
D = 161.8 mm
154.4 mm
12.7 cm
x = 100 mm
Conditions: 1.
2.
0.6 < x < 0.7 D
Iyy > Ixx
Try 2 Column Sections, 152 x 152 x 37 UC
Distance x has to be between 0.6 and 0.7 D i.e. between 113.3 and 97.1 mm.
Choose x = 100 mm
Ixx = 2 I x = 2 x 2218 = 4436 cm 4
Iyy = 2 [Iy + Ay2] = 2 [709 + 47.4 x 12.7 2] = 8354.1 cm4
Condition 2 is satisfied
Rxx = 6.84 cm (SDM); r yy =
Iyy/2A
=
8341.1/(2 x 47.4)
= 9.4 cm
Choose rxx since it is smaller than ryy
Slenderness ratio, l/rmin
= 5.25 x 0.65 = 50
6.84 cm
From Table 27a, Pc = 208 N/mm2
Safe axial load = 208 N/mm2 x 2 x 47 x 10 2
= 1971.6 kN
1000
Comment: Since actual load (2400 kN) is greater than the safe load, section is unsafe.
Choose a larger section.
Try 203 x 203 x 46 UC
Depth of section = 203 .2 mm; Width = 203.2 mm
Max. distance (x) = 203.2 x 0.7 = 142.2 mm; Min x = 121.9 mm
Choose x = 130 mm
Ixx = 2 Ix = 2 x 4564 = 9128 cm 4
Ixx = 2 I x = 2 x 2218 = 4436 cm 4
Iyy = 2 [Iy + Ay2] = 2 [1539 + 58.8 x 16.7 2] = 17,937.4 cm4
Condition 2 is satisfied
rxx = 8.81 cm (SDM);
Slenderness ratio, l/rmin
= 5.25 x 0.65 = 38.7
8.81 cm
From Table 27a, (BS5950), Pc = 214.8 N/mm 2 (by interpolation)
Safe axial load = 214.8 N/mm2 x 2 x 58.8 x 10 2
= 2526 kN
1000
Since actual load is less than the safe load, Section is safe
Economy ratio: 2400 kN/2526 kN = 0.95
Section (203 x 203 x 46 UC) is safe and economical.
4.4
DESIGN
OF
ROOF
TRUSSES
4.4.1 Truss Components
Ridge
Internal bracing
Rise
Rafter
Purlin
Main tie
Span
Roof trusses consist of sloping rafters which meet at a ridge,
a main tie connecting the foot of the rafters and the internal bracing
members. Roof trusses are used in conjuction with purlins to
support a roof covering. Purlins are horizontal beams which span the
distance between trusses and carry the roof structure.
Roof Trusses Contd.



Steel and timber trusses are usually
designed assuming pin-jointed members. In
practice, timber trusses are assembled with
bolts, nails
or special connectors and steel trusses are
bolted, riveted or welded.
These rigid joint impose secondary stresses
which are negligible and therefore not used
in design.
4.4.2 Steps in Designing a Truss



a)
Select general layout of truss
members and truss spacing
b)
Estimate external loads to be
applied including self weight of truss,
purlins and roof covering, together with
wind loads.
c)
Determine the critical (worst
combinations) loading. It is usual to
consider dead loads alone, and then
dead and imposed load combined.
Steps in Designing a Truss Contd.

d) Analyse framework to find forces in
members.

e)
Select material and section to
provide in each member a stress value
which does not exceed the permissible
value.
4.4.2.1 Spacing and Layout

Roof trusses should as far as possible
be spaced to achieve a minimum of
weight and economy of materials used
in the total roof structure.

For spans up to 20 m, the spacing of
steel trusses is likely to be about 4 m,
and in case of timber, 2m.
Spacing and Layout Contd.


Short spans up to 8 m should have
pitched timber rafters or light weight
trusses either pitched or flat. Medium
spans of 7 to 15 m or 16 m require
truss frames designed of timber or
steel.
Long spans of over 16 m should if
possible be broken into small units or
the roofs should be designed by
specialists.
Pitch or Slope of a Roof
The pitch or slope of roof depends on locality, imposed loading and type
of covering. Heavy rainfall may require steep slopes for rapid drainage,
a slope of 22 ° is common for corrugated steel or asbestos roofing sheets.
Slope = H/Run
H
Pitch = H/Span
½ Span
Run
Span
+ Truss spacing could be ¼ to 1/5 th of the span
Estimation of Roof Loads
4.4.2.2 Estimation of Loads
C Trusses 4 m c/c spacing
B
F
3.2 m

AD
Purlin
2.2 m
E
G
4m
12 m
+Belgian Truss
  tan 1
2.20
 20o
6
Example

Determine the critical forces for design
for each of the members in the left
hand of the truss shown above.
Assume the following: Trusses are
spaced 4 m on centres. The roof deck
is of galvanised sheet weighing 6 kg/m2
. All purlins weigh 22kg/m. Openings
constitute 20% of the wall surface.
Calculate the panel loads.
Solution






(a) Dead load
i) Determine the panel area supported by
one purlin
= Distance between the purlins x spacing
of trusses = 3.2 m x 4 m = 12.8 m2
ii) Calculate the roof deck load supported by
one purlin
= 12.8 m 2 x 6 kg/m 2 x 9.81 = 753.4
N = 0.75 kN.
iii) Weight of each purlin = 22 kg/m x 4 m
x 9.81 = 863.28 N = 0.86 kN
Solution Contd.




Total dead load to be carried by truss
= 0.75 + 0.86 kN = 1.61 kN
iv)Estimate total truss weight per panel
as 10 % of the total load to be carried =
0.16 kN
v) Total dead load per panel point , P
= 1.61 + 0.16 = 1.77 kN
With this, calculate all the loads in all the members noting those under
tension and compression. See table below.
1.77 kN
C
1.77 kN
1.77 kN
B
A
F
D
E
G
2.66 kN
2.66 kN
vi) Calculate wind load: q = 0.0127 V 2 K = basic velocity pressure, Pa
Assuming eave height of building = 2 m
Wind velocity at 6.1 m = 14 m/s; K = (h/6.1) 2/7
=
q = 0.0127 x 14 2 x 0.727 = 1.81 kN/m2 (Pa).
(2/6.1)2/7
= 0.727
Solution Contd.
From Table 5.2 (FAO book), coefficients for buildings open at both sides can be
assumed as + 0.6 for windward slope and - 0.6 on leeward slope.
Full panel load, windward side = 1.81 kN/m 2 x 0.6 x 12.8 m2
= 13.9 kN
Full panel load, leeward side = 1.81 x - 0.6 x 12.8 = - 13.90 kN
With this again, calculate the forces in all the members (See table below)
C
13.9 kN
13.9 kN
B
F
A
G
D
20.85 kN
E
20.85 kN
Tabulate Results as Follows:
....................................................................................................…………………………….
.
Member
Dead
Coeffs Imposed
load(D) (Load/
Dead + Imposed
Load (I)
Wind load
load (D + I)
(W)
Design
Load
....................(kN)..........1.77).....................................................................
AB, FG
7.8 (C)
4.41
61.3 (C)
AD, EG
7.8 (T)
4.41
BC,CF
6.0 (C)
3.39
47.1 (C)
47.1 (C)
BD, EF
5.0 (C)
2.84
39.5 (C)
39.5 (C)
CD, CE
6.4 (T)
3.62
50.3 (T)50.3 (T)
DE
1.8 (T)
1.02
14.2 (T)14.2 (T)
61.3 (T)
61.3 (C)
61.3 (T)
The design load should be the largest of the combinations.
A simplified
approach can be used if the intention is to use a common section throughout. In this
case, the designer concentrates on the top chord or rafter members which normally have
maximum compression loads. A force diagram or method of sections can then be used
to determine the load on these members and the necessary size.
Example: A farm building composed of block walls carries steel roof trusses over a span
of 8 m. Roofing sheets determine the purlin spacing. Assume a force analysis shows
maximum rafter forces of approximately 50 kN in compression (D + I) and 30 kN in
tension (D + W), outer main tie member 50 kN tension (D + I0 and 30 kN compression
(D + W). A reversal of forces due to the uplift of wind cause the outer tie member to
have 50 kN of tension and 30 kN of compression. Design the truss members.
2m
2m
8m
Solution
(1) Rafter Design
Maximum force on rafter = 50 kN
In roof design, it is assumed that the joints are pinned i.e. effective length, l = 1 x 1.38
m = 1.38 m
Try angle section 64 x 38 x 4.6
rmin = 1.10 cm (SDM); Area = 3.56 cm2
Slenderness ratio, l/rmin
= 1.38 x 100 cm = 125.5
1.10 cm
From Table 27a, (BS5950), Pc = 104.75 N/mm 2 (by interpolation)
Safe axial load = 104.75 N/mm2 x 4.56 x 102
1000
= 47.7 kN
Try 64 x 38 x 6.3
rmin = 1.09 cm (SDM); Area = 6.08 cm2
Slenderness ratio, l/rmin
= 1.38 x 100 cm = 126.6
1.09 cm
From Table 27a, (BS5950), Pc = 103.1 N/mm 2 (by interpolation)
Safe axial load = 103.1 N/mm2 x 6.08 x 102
= 62.68 kN
1000
Since the safe load is greater than the actual load, section is safe. Check economy
Actual load/safe load = 50/62.68 = 80.0% ……..Manageable
ii) Main Tie Design
Maximum tensile force by main tie = 50 kN
For tension member, stress = F/A
Area = Force
Allowable stress, Pt
Assuming Pt is 155 N/mm2, Area = (50 x 103 N)/155 N/mm2
= 322.6 mm2
= 3.26 cm2
To allow for uniform size of angles, use the same section as that for the
rafters i.e. 64 x 38 x 6.3 angle section. This will also allow for
enough area for putting rivets.
Check for Compressive Forces in the Main Tie
Effective length of main tie = 1 x 2 = 2 m
For section 64 x 38 x 6.3 , area = 6.08 cm 2, rmin = 1.09 cm
rmin = 1.09 cm (SDM); Area = 6.08 cm2
Slenderness ratio, l/rmin
= 2 x 100 cm = 183.5
1.09 cm
From Table 27a, (BS5950), Pc = 53.9 N/mm 2
Safe axial load = 53.9 N/mm2 x 6.82 x 102
= 32.77 kN
1000
Section is safe since it can carry the 30 kN maximum compressive forces.
Other Designs
7
Purlins:
See section 4.2.2 for problems on asymmetric bending.
Purlins are designed as simple beams on asymmetrical bending.
8
Joints: Joints are either riveted, bolted or welded (refer to Strength of
Materials 2).
Final Design Diagram
4.5 FOUNDATIONS AND
FOOTINGS DESIGN



4.5.1 Introduction: Foundation is the part
of the structure through which the load of the
structure is transmitted to the ground.
A combination of footing and foundation
distributes the load on the bearing surface
(the soil) and keeps the building level and
plumb and reduces settling to a minimum.
Footing and foundation are normally made of
concrete, no matter the construction
material.
Footing and Foundation Contd.

Before design of footing and foundation
can be made, the total load from the
building as well as soil bearing
characteristics should be determined.
Soil Bearing Capacity


Strength of the soil required to resist
the loads resting in it.
This is after the top soil has been
removed (See Table 5.6, FAO book for
values of soil bearing capacities).
Detailed investigations of the soil is not
usually required for small scale
buildings
Soil Bearing Capacities
4.5.3 Foundation Footings
4.5.3.1 Description: A footing is an
enlarged base for a foundation
designed to distribute the building load
over a larger area of soil and to provide
a firm, level surface for constructing the
foundation wall.
Foundation wall
Footing
Footings and Foundation Contd.


The footing is wider than the foundation
wall because the soil's bearing stress is
less than that of the material (concrete)
of the wall e.g. concrete has a strength
of about 1000 kN/m2 .
A 1 : 3 : 5 ratio of cement, dry sand,
and gravel is suggested for footings
with 31 litres of water per 50 kg sack of
cement.
4.5.3.2 Dimensions of Footings
Area of footing =
Total load
Bearing capacity of the soil
The total load includes an estimated mass of the footing itself. The width
is determined by dividing the area by the length. It is at times common
to pour a concrete footing that is as deep as the wall is thick and twice
as wide. The foundation for large buildings may require reinforcing.
2a
a
a (12.5 cm)
a
2a (25 cm)
1.5 a
4a (50 cm)
Walls and Piers
Columns
Footing Proportions
For farm structures, typical value of a is 10 to 12.5 cm.
Note




Unlike the continuous wall footings, the pier
and column footings are heavily loaded.
It is very important to correctly estimate the
proportion of building load to be carried by
each pier or column.
If wall footings are very lightly loaded, it is
advisable to design any pier or column
footings required for the building with
approximately the same load per unit area.
If any settling occurs, it should be uniform
throughout.
Example

For the building below, the loads are as
follows:

a) The roof framing plus the expected wind load =
130 kN
b) The wall above the foundation is 0.9 kN/m
c) The floor will be used for grain storage and will
support as much as 7.3 kN/m2 . The floor structure
is an additional 0.5 kN/m2
The foundation walls and piers are each 1 m high
above the footing. The wall is 200 mm thick and the
piers 300 mm square. The soil on the site is
compacted clay in a well drained area. Find the size
of the foundation and pier footing that will safely
support the loads. Assume that the weight of the
mass 1 kg approximately equals 10 N. The mass of
concrete is 24 kN/m3.



Solution







Note that each pier(column) carries 4/32 ie.
1/8 of total floor load. Each wall carries 7/32
of floor load ie. {(2 x 3) + 1/2 + 1/2}
1) The division of load on each foundation
wall is as follows:
a) Roof load = 50 % on each wall i.e. 50 %
of 130 kN = 65 kN
b) Wall load = for each side: 16 m x 0.9
kN/m = 14.4 kN
c) Floor load = each side of wall carries 7/32
Total floor load = (7.3 + 0.5) x 16 x 8 = 998.4 kN
i.e. floor load carried by each wall = 7/32 x 998 kN
= 218.4 kN
Solution Contd.