Cee 312(2)

CEE-312
Structural Analysis and Design Sessional-I
(1.0 credit)
Lecture: 2
Bijit Kumar Banik
Assistant Professor, CEE, SUST
Room No.: 115 (“C” building)
bijit_sustbd@yahoo.com
Department of Civil and Environmental Engineering

Design philosophy
A general statement assuming safety in engineering design
Resistance (of material & x-section) >= Effect of applied load
It is essential in the above equation that both sides are evaluated
in same conditions; e.g. if of load to produce compressive stress
on soil, then it should be compared with the bearing capacity of soil

Design methods
1. ASD (Allowable Stress Design)
2. LRFD (Load & Resistance Factor Design)

ASD
Safety in design is obtained by specifying that the effects
of the loads should produce stresses that were the fraction
of yield stress, Fy
This is equivalent to:
FOS = Resistance, R / Effect of load, Q

ASD
Since the specifications sets limit on stresses, it became
Allowable Stress Design (ASD)
Mostly reasonable where stresses are distributed uniformly
(such as determinate trusses, arches, cables etc)

Drawback of ASD
1. ASD does not give reasonable measures of strength
2. Factor Of Safety (FOS) is applied only to stress not in load

LRFD
Considers variability not only in resistance but also in the
effects of load
Quality of material should maintain highest level
Now different factored load can be assign to different
types of loads (dead, live, wind etc)

Trusses
In architecture and structural engineering, a truss is a structure comprising
one or more triangular units constructed with straight members whose ends
are connected at joints referred to as nodes (hinge or pin).
Assumptions
1. Members are connected at their end by frictionless pins
2. Loads and reactions applied only at joints
3. Two force members

Types of Trusses
Two broad categories:
Roof truss
Bridge truss

Roof Trusses
Flat: The most economical flat truss for a roof
is provided when the depth of the truss in inches
is approximately equal to the span in inches.
Mono: Used where the roof is required to slope in only
one direction. Also in pairs with their high ends abutting
on extremely long span with a support underneath the
high end.
Common: Truss configurations for the most
widely designed roof shapes.
Scissors: Provides a cathedral or vaulted ceiling.
Most economical when the difference in slope
between the top and bottom chords is at least 3/12
or the bottom chord pitch is no more than half the
top chord pitch.

Stability and Determinacy of Truss
Minimum members required to make a rigid framework isThree
3 members
3 joints
5 members
4 joints
7 members
5 joints
Thus to form a rigid and stable truss of ‘n’ joints number of
members (j) required is:
j = 3+2*(n-3)
For first 3 joints For remaining joints
j= 2n-3

Stability and Determinacy of Truss
If j < 2n-3 Unstable truss
If j > 2n-3 Indeterminate truss
Bottom chord
Diagonal
Top chord
Vertical

Industrial roof truss system
Span
Bay
Parlin
Top chord bracing
Sagrod
Column
Beam

Vertical bracing
Bottom chord strut
Rise

When Span > 40 ft Truss system
Trusses are normally spaced 15-25 ft apart
For longer span trusses must be simply supported
Hinge support
Roller support
Why this is called simply supported ??

Loads on structures
Load
DynamicStatic
e.g. Earthquake
Live loadDead load
e.g. Self wt. or fixed object e.g. Wind, snow, wheel of vehicle
or any moving objects

Wind load analysis for roof truss
Basic wind pressure at 30 ft, q (psf) = 0.00256 V2
where, V is wind speed (mph)
Pressure on a pitched roof
α
P
Windward side
Wind
Leeward side
P

For windward surface







≤
≤≤−
≤≤−
≤≤−
=
α
αα
αα
α
0
00
00
0
609.0
6030)9.003.0(
3020)1.207.0(
2007.0
q
q
q
q
P
For leeward surface
qP 7.0−=
where, ‘P’ is the pressure normal to the roof surface
If P is +ve
If P is -ve
Pushing the roof surface
Pulling or suction or uplift
Leeward side always experience uplift
Maximum P will be taken

Analysis and design of an Industrial roof truss system
Steps
1. Selection of truss type
2. Estimation of loads
3. Analysis and design of purlins
4. Analysis and design of sagrods
5. Dead load and wind load analysis
6. Combination of D.L and W.L to determine the
design bar forces
7. Design of members
8. Design of bracing system
9. Design of connections (welded)
10. Detailing

Your Truss !!!
6@ A ft
B ft
L0
L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
Bay = 20 ft; Fy=36 ksi; E = 29 ksi
11.511109.5B
76.55.55A
4321Group

1. Selection of truss type
6@6 ft = 36 ft
10 ft
3@6.86 ft
L0
L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
Bay = 20 ft; Fy=36 ksi; E = 29 ksi
29.050

2. Estimation of loads
Dead loads
1. CGI (Corrugated Galvanized Iron) 2.0 psf
2. Purlins 1.5 psf
3. Sagrods, bracings 1.0 psf
Sub total 4.5 psf
4. Self weight 60 lb/ft of span
Live loads
Design wind speed 100 mph
Given loads should be transformed into Equivalent joint loads

Roofing
Purlin
Sagrod
Top chord
Purlins are nothing but beams. They span between the adjacent
trusses, i.e the spacing of the trusses (Bay) is the span of purlins.
Normally channel section is used to design purlins.
Design criteria for purlin is “bending stress”
First we will check for dead load and then for the wind load

Since the principal axes of the purlins section are inclined, the
dead load causes biaxial bending in the purlins.
Y
X
W
Check for Dead Load

So, different support condition of purlin for X and Y direction
For Y-direction Y
X
W
20 ft
For X-direction
10 ft 10 ft
Additional support for sagrod

Dead loads coming on purlins
Roofing
Purlin
Sagrod
Roofing 2.0 psf
Self wt. of purlin 1.5 psf
Total 3.5 psf (neglecting sagrod)

3. Analysis and design of purlins Parlin
6.86 ft
3.5 psf
UDL on purlin, wDL= 3.5 psf X 6.86 ft = 24.01 lb/ft

wDLx= wDLsinθ = 24.01 X sin29.010 = 11.66 lb/ft
X
Y
wDL= 24.01 lb/ft
29.050
29.050
wDLy= wDLcosθ = 24.01 X cos29.010 = 20.99 lb/ft

For the detail computation of BMD please go through:
Strength of materials-by Singer; pp-285, Prob.-828
For Y-direction
L = 20 ft
For X-direction
L/2 =10 ft L/2 = 10 ft
BMD
wDLy
wDLx
8
2
Lw
Mxx
DLy
=
32
2
Lw
Myy DLx
=

Y
X
wDLy
wDLx
L
Mxx = (wDLy *L) * L/2
Myy = (wDLx *L) * L/2

8
2
Lw
Mxx
DLy
=
32
2
Lw
Myy DLx
=
ftkip −== 05.1
8
20*99.20 2
ftkip −== 15.0
32
20*66.11 2

Select American standard channel: C 3 X 4.1
Sxx = 1.10 in3 Syy=0.202 in3

Maximum bending stress:
= 0.66*36 = 23.76 ksi
yy
yy
xx
xx
b
S
M
S
M
f +=
ksi37.20
202.0
12*15.0
10.1
12*05.1
=+=
For bending; Allowable stress,
> fb(20.37 ksi)
Fb= 0.66Fy
So, section C 3 X 4.1 is ok for dead load

Check for Wind Load
3@6 ft = 18 ft
10 ft
L0
L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
29.050
Pitch angle, 01
05.29)
18
10
(tan == −
α V = 100 mph
psfVq 6.25100*00256.000256.0 22
===Basic wind pressure,

For windward surface







≤
≤≤−
≤≤−
≤≤−
=
α
αα
αα
α
0
00
00
0
609.0
6030)9.003.0(
3020)1.207.0(
2007.0
q
q
q
q
P
For leeward surface
qP 7.0−=
P = (0.07*29.05-2.1)*25.6 = -1.7 psf
UDL on the windward side = -1.7*6.86 = -11.66 lb/ft
Purlin spacing
UDL on the Leeward side = - 17.92 *6.86 = -122.93 lb/ft
P = -0.7*25.6 = -17.92 psf
Here Leeward side load will govern
What does the negative sign mean? Suction

Maximum bending stress:
yy
yy
xx
xx
b
S
M
S
M
f +=
202.0
12*15.0
10.1
12*10.5
+=
Why pound is expressed as ‘lb’ ?
libra→ lbpound weight which was libra pondo in Latin.
Resultant load in y-direction = -122.93+20.99 = -101.94 lb/ft
Dead load in y-direction = 20.99 lb/ft (from previous DL calculation)
→101.94 lb/ft
8
2
Lw
Mxx
y
= ftkip −== 10.5
8
20*94.101 2
ftkipsMyy −= 15.0
> Fb(23.76 ksi)ksi57.63= Not ok

Next trail channel: C 4 X 7.25
Sxx = 2.29 in3 Syy=0.343 in3

yy
yy
xx
xx
b
S
M
S
M
f +=
343.0
12*15.0
29.2
12*10.5
+=
ksi97.31= > Fb(23.76 ksi)
Not ok

Next trail channel: C 5 X 6.7
Sxx = 3.00 in3 Syy=0.378 in3

yy
yy
xx
xx
b
S
M
S
M
f +=
378.0
12*15.0
00.3
12*10.5
+=
ksi16.25= > Fb(23.76 ksi)
Not ok
e oe oe oe o !!!

Next trail channel: C 5 X 9
Sxx = 3.56 in3 Syy=0.450 in3

i.e. 9/6.86 = 1.31 psf
yy
yy
xx
xx
b
S
M
S
M
f +=
450.0
12*15.0
56.3
12*10.5
+=
So, C 5 X 9 is the final Purlin section
ksi19.21= < Fb(23.76 ksi)
যযযয !!!!!!!! ok
Check
Self weight of purlin = 9 lb/ft
< 1.5 psf (assumed self wt.)

For the detail computation of reaction please go through:
Strength of materials-by Singer; pp-285, Prob.-828
For X-direction
L/2 =10 ft L/2 = 10 ft
wDLx = 11.66 lb/ft
LwDLx
8
3 LwDLx
8
3
LwDLx
8
5
Tensile Force on sagrod = Midspan reaction
LwF DLx
8
5
=
A round bar of dia (3/8) inch will be adequate.
≈== lbF 75.14520*66.11*
8
5
kips15.0

Assuming that the bolt threads will reduce the effective
diameter by (1/16) inch.
Net X-sectional area =
2
2
077.0
16
1
8
3
*
4
in=





−
π
Allowable stress in tension, Ft = 0.6 Fy = 21.6 ksi
The rod is able to carry a load of 21.6*0.077 = 1.66 kips>> F
So, #3 or (3/8) inch round rod will be used as sagrod

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Cee 312(2)

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