structural analysis

1. CEE-312
Structural Analysis and Design Sessional-I
(1.0 credit)
Lecture: 6
Bijit Kumar Banik
Assistant Professor, CEE, SUST
Room No.: 115 (“C” building)
bijit_sustbd@yahoo.com
Department of Civil and Environmental Engineering

2. Analysis and design of an Industrial roof truss system
R L
(- 8.88) (-8.88) (-6.35) (-4.19) (-4.43) (-4.43)
(7.14) (5.61) (4.07) (4.85) (5.00) (5.14)
6@6 ft = 36 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
2.60 k4.54 k
(0)
(-0.13)
(-1.41)
(0)
(-3.07)
(3.78)
(2.90)
(0.35)
(0.27)
3.24k
L→R
6@6 ft = 36 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
4.54 k2.60 k
(- 1.19) (-1.19) (-0.95) (-3.11) (-5.64) (-5.64)
(5.14) (5.00) (4.85) (4.07) (5.61) (7.14)
(0)
(-0.13)
(-1.41)
(0)
(-3.07)
(0.35)
(0.27)
(3.78)
(2.90)
3.24k
(4.41) (4.41) (3.53) (3.53) (4.41) (4.41)
(- 5.05) (- 4.04) (- 3.03) (- 3.03) (- 4.04) (- 5.05)
6@6 ft = 36 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
2.94 k2.94 k
(.18)
(0.67)
(0.67)
(.18)
(2.14)
(-1.32)
(- 1.01)
(-1.32)
(- 1.01)
Dead Load

3. Analysis and design of an Industrial roof truss system
Load combinations
Condition 1: F1 = DL [No wind]
Condition 2: F2 = DL + WL (L→R)
Dead Load →DL
Wind Load →WL
Condition 3: F3 = DL + WL (R→L)
For design compressive force → Minimum of F1, F2 & F3 (If
positive leave it !)
For design tensile force → Maximum of F1, F2 & F3 (If
negative leave it !)

4. Analysis and design of an Industrial roof truss system
Compr.TensionWL(R L)WL(L R)DL
-1.23+4.41-4.43-5.64+4.41L5L6
-1.23+4.41-4.43-5.64+4.41L4L5
Member
-0.66+3.53-4.19-3.11+3.53L3L4
Chord
-2.82+3.53-6.35-0.95+3.536.00L2L3
Bottom
-4.47+4.41-8.88-1.19+4.41L1L2
-4.47+4.41-8.88-1.19+4.41L0L1
-5.05+2.09+5.14+7.14-5.05L6U5
-4.04+1.57+5.00+5.61-4.04U4U5
Member
-3.03+1.82+4.85+4.07-3.03U3U4
Chord
-3.03+1.82+4.07+4.85-3.036.86U2U3
Top
-4.04+1.57+5.61+5.00-4.04U1U2
-5.05+2.09+7.14+5.14-5.05L0U1
Design Member Force (k)Member Force (k)Length
(ft)
MemberRemarks

5. Analysis and design of an Industrial roof truss system
-1.32+2.46+0.35+3.78-1.328.97L3U4
-0.93+2.14-3.07-3.07+2.1410.0L3U3
Member
-1.32+2.46+3.78+0.35-1.328.97L3U2
Web
-0.74+0.67-1.41-0.13+0.676.67L2U2
-1.01+1.89+2.90+0.27-1.016.86L2U1
-+0.1800+0.183.33L1U1
Compr.TensionWL(R L)WL(L R)DL
Length
(ft)
MemberRemarks
-+0.1800+0.183.33L5U5
-1.01+1.89+0.27+2.90-1.016.86L4U5
-0.74+0.67-0.13-1.41+0.676.67L4U4

6. Analysis and design of an Industrial roof truss system
During analysis we assumed as if truss members are pin ended
Actually they are continuous or welded/riveted to other part of
the truss
So, ends are somewhat rigid
We assume the effective length factor k = 0.6
Whereas k = 0.5 for fixed end
k = 1.0 for pin end
Le = 1.0 L Le = 0.5 L
L/4
L/4Pin
Fixed

7. Analysis and design of an Industrial roof truss system
2L0.25One end fixed
other free
L1Both end hinged
0.7L2One end fixed
other hinged
0.5L4Fixed
Le = Effective
length
N = Number of
times strength
of hinged
columns
End connection

8. Design of chord members
y
c
F
E
C
2
π=
Cc = slenderness ratio
E = 29,000 ksi
Fy = 36 ksi
Fa = allowable stress
P = maximum bar force
Choose angle (based on Areq)
Take A, rmin
rmin = min. radius of gyration out
of rx, ry & rz
3
2
/
8
1/
8
3
3
5
/
5.01






−





+














−
=
cc
c
y
a
C
rKL
C
rKL
C
rKL
F
F
( )2
/
000,149
rKL
Fa =
If KL/r < CcIf KL/r < Cc
Compare Fa with f Take another member with greater ‘L’
(if any); check for that L & P
Choose another angle of
greater secn; Take A, rmin
If Fa < f If Fa > f
Final secn
of anglestop If all members have been checked
a
req
ya
F
P
A
FFAssume
=
= 5.0
A
P
f
stressecompressivActual
=
Compare KL/r with Cc

9. Design of Top chord members
Maximum compressive force found on L0U1 , having L = 6.86 ft ; P = 5.05 k
y
c
F
E
C
2
π=
36
000,29*2
π= 1.126=
Fa = 0.5Fy = 0.5*36 = 18 ksi
a
req
F
P
A =
18
05.5
= = 0.281 in2
From AISC chart, select
16
3
4
1
1
4
1
1 XXL

10. Design of Top chord members
0.244
0.434
A = 0.434 in2
rmin = 0.244

11. Design of Top chord members
ksif 64.11
434.0
05.5
==
4.202
244.0
)12*86.6(*6.0
==
r
KL
> Cc
From AISC chart, now select
16
3
22 XXL
( )
fksiFa <== 64.3
4.202
000,149
2 Not ok

12. Design of Top chord members
0.244
0.434
A = 0.715 in2
rmin = 0.394

13. Design of Top chord members
ksif 06.7
715.0
05.5
==
36.125
394.0
)12*86.6(*6.0
==
r
KL
< Cc
So, Design Top chord is
16
3
22 XXL
ok
3
2
/
8
1/
8
3
3
5
/
5.01






−





+














−
=
cc
c
y
a
C
rKL
C
rKL
C
rKL
F
F
3
2
1.126
36.125
8
1
1.126
36.125
8
3
3
5
1.126
36.125
5.0136






−





+














−
=
= 9.50 ksi > f

14. Design of Bottom chord members
Maximum compressive force found on L0L1 , having L = 6 ft ; P = 4.47 k
y
c
F
E
C
2
π=
36
000,29*2
π= 1.126=
Fa = 0.5Fy = 0.5*36 = 18 ksi
a
req
F
P
A =
18
47.4
= = 0.248 in2
From AISC chart, select
16
3
4
1
1
4
1
1 XXL

15. Design of Bottom chord members
0.244
0.434
A = 0.434 in2
rmin = 0.244

16. Design of Bottom chord members
ksif 3.10
434.0
47.4
==
05.177
244.0
)12*6(*6.0
==
r
KL
> Cc
From AISC chart, now select
8
1
22 XXL
( )
fksiFa <== 75.4
05.177
000,149
2 Not ok

17. Design of Bottom chord members
A = 0.484 in2
rmin = 0.398
1/8

18. Design of Bottom chord members
ksif 24.9
484.0
47.4
==
54.108
398.0
)12*6(*6.0
==
r
KL
< Cc
So, Design Bottom chord is
8
1
22 XXL
ok
3
2
/
8
1/
8
3
3
5
/
5.01






−





+














−
=
cc
c
y
a
C
rKL
C
rKL
C
rKL
F
F
3
2
1.126
54.108
8
1
1.126
54.108
8
3
3
5
1.126
54.108
5.0136






−





+














−
=
= 11.88 ksi > f

19. Design of Web members
Maximum tensile force found on U2L3 , having L = 8.97 ft ; P = 2.46 k
y
c
F
E
C
2
π=
36
000,29*2
π= 1.126=
Fa = 0.5Fy = 0.5*36 = 18 ksi
a
req
F
P
A =
18
46.2
= = 0.137 in2
From AISC chart, select
16
3
4
1
1
4
1
1 XXL

20. Design of Web chord members
0.244
0.434
A = 0.434 in2
rmin = 0.244

21. Design of Web chord members
ksif 67.5
434.0
46.2
==
69.264
244.0
)12*97.8(*6.0
==
r
KL
> Cc
From AISC chart, now select
16
5
2
1
1
2
1
1 XXL
( )
fksiFa <== 13.2
69.264
000,149
2 Not ok

22. Design of Web chord members
A = 0.84 in2
rmin = 0.291
1/8
0.291
0.84 in2

23. Design of Web chord members
ksif 93.2
84.0
46.2
==
94.221
291.0
)12*97.8(*6.0
==
r
KL
> Cc
ok
( )
fksiFa >== 02.3
94.221
000,149
2
But we have a member U3L3 of length 10 ft with P = 2.14 k
Now check the section for this member
ksif 55.2
84.0
14.2
==
42.247
291.0
)12*10(*6.0
==
r
KL
> Cc
( )
fksiFa <== 43.2
42.247
000,149
2 Not ok

24. Design of Web chord members
A = 0.984 in2
rmin = 0.289
1/8
0.291
0.984 in2

25. Design of Web chord members
ksif 17.2
984.0
14.2
==
13.249
289.0
)12*10(*6.0
==
r
KL
> Cc
( )
fksiFa >== 40.2
13.249
000,149
2 ok
So, Design web member is
8
3
2
1
1
2
1
1 XXL

26. Design summery of members
Design web member
8
3
2
1
1
2
1
1 XXL
Design Bottom chord member
8
1
22 XXL
Design Top chord member
16
3
22 XXL