structural analysis

1. CEE-312
Structural Analysis and Design
Sessional-I
(1.0 credit)
Lecture: 3
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
Progress so far
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

3. Analysis and design of an Industrial roof truss system
L0
L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
Is this a stable truss?
Is this statically determinate?
j= 2n-3 j= 21; n = 12
Yes
Yes

4. Dead load and wind load analysis
Dead load calculations
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
Point load = 4.5 psf * 6.86 ft * 20 ft = 617.4 lbs
Purlin spacing Bay
Self wt. of truss will be equally divided among the top & bottom chord

5. 6@6 ft = 36 ft
10 ft
3@6.86 ft
L0
L1 L2 L3 L4 L5
L6
Point load at top chord joints (U1,U2,U3,U4 & U5) due to
self wt. = ½ * 60 * 6 = 180 lbs
6 ft
U1
U2
U3
U4
U5
Total D.L. at top chord joints = 617.4 + 180 = 797.4 lb ≈ 0.8 kips
Point load at bottom chord joints (L1,L2,L3,L4 & L5) due to
self wt. = ½ * 60 * 6 = 180 lbs ≈ 0.18 kips
Loads at joints L0 & L6 = (0.80+0.18)/2 = 0.49 kips
Dead load and wind load analysis

6. 7 ft 7 ft 14 ft 7 ft 7 ft
4@5.815ft = 23.36 ft
5.25ft 5.25ft 5.25ft 5.25ft
5 ft
5 ft
Self weight 60 lb/ft of span
CGI+Purlin+Bracing+Sagrod 4.5 psf
Dead load
Bay = 20 ft
L0
Please calculate the joint load at L0
Dead load and wind load analysis

7. 6@6 ft = 36 ft
10 ft
L0
L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
0.49k 0.49k
0.80k
0.80k
0.80k
0.80k
0.80k
0.18k 0.18k 0.18k 0.18k 0.18k
Taking moment about L0
0.98*6+0.98*12+0.98*18+0.98*24+0.98*30+0.49*36-R1*36 = 0
R1
R2
R1= 2.94 k
R2= 2.94 k (Symmetric loading)
= 2.94 k= 2.94 k
0
Dead load and wind load analysis

8. 6@6 ft = 36 ft
10 ft
L0
L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
0.49k 0.49k
0.80k
0.80k
0.80k
0.80k
0.80k
0.18k 0.18k 0.18k 0.18k 0.18k
∑V = 0
R1
R2
2.94 – 0.49 – (5/10.3)*L0U1 = 0
L0U1= 5.05 k (C)
= 2.94 k= 2.94 k
0
1
1
2.94 k
0.49k
L0L1
L0U1
1-1
5
9
10.3
∑H = 0
(9/10.3)*L0U1 = L0L1
L0L1= 4.41 k (T)
Dead load and wind load analysis

9. 6@6 ft = 36 ft
10 ft
L0
L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
0.49k 0.49k
0.80k
0.80k
0.80k
0.80k
0.80k
0.18k 0.18k 0.18k 0.18k 0.18k
R1
R2
= 2.94 k
= 2.94 k
0
2 2
0.18 k
L1L2
2-2
L0L1=4.41k
L1U1
∑V = 0
L1U1= 0.18 k (T)
∑H = 0
L1L2= 4.41 k (T)
Dead load and wind load analysis

10. 3
3
9
510.3
9
5
10.3
3-3∑ML2 = 0
2.94*12 – 0.49*12–0.18*6+
(5/10.3)*U1U2*6 +(9/10.3)*U1U2*(10/3)= 0
U1U2= 4.04 k (C)
∑H = 0
4.41-4.04*(9/10.3)+U1L2*(9/10.3) = 0
U1L2= 1.01 k (C)
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
0.49k 0.49k
0.80k
0.80k
0.80k
0.80k
0.80k
0.18k 0.18k 0.18k 0.18k 0.18k
R1
R2 = 2.94 k= 2.94 k
0
+ve
U1
L0 L1 L2
U2
0.80 k
0.49 k
2.94 k 0.18 k
(10/3)’
4.41k
Vertical component of U1U2
Horizontal component of U1U2
Dead load and wind load analysis

11. ∑H = 0
L2L3= 3.53 k (T)
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
0.49k 0.49k
0.80k
0.80k
0.80k
0.80k
0.80k
0.18k 0.18k 0.18k 0.18k 0.18k
R1
R2 = 2.94 k= 2.94 k
0 4
4
5
9
10.3
0.18 k
L2L3
4-4
4.41k
L2U2
1.01k
∑V = 0
L2U2 - 0.18 – 1.01*(5/10.3) = 0
L2U2= 0.67 k (T)
L2L3 + 1.01*(9/10.3)-4.41 = 0
Dead load and wind load analysis

12. 6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
0.49k 0.49k
0.80k
0.80k
0.80k
0.80k
0.80k
0.18k 0.18k 0.18k 0.18k 0.18k
R1
R2 = 2.94 k= 2.94 k
0
5
5
10
9
13.45
5-5
5
9
10.3
∑ML3 = 0
2.94*18 – 0.49*18 – 0.18*12 – 0.18*12 – 0.80*6
– 0.18*6 + (9/10.3)*U2U3*(2*10/3) +(5/10.3)*U2U3*6= 0
U2U3= 3.03 k (C)
∑H = 0
U2L3*(9/13.45) – 3.03*(9/10.3) + 3.53 = 0
U2L3= 1.32 k (C)
+ve Horizontal component of U2U3
Vertical component of U1U2
L0 L1 L2 L3
U1
U2
U3
0.49 k
2.94 k
0.80 k
0.18 k 0.18 k
0.80 k
3.53 k
2*(10/3)
Dead load and wind load analysis

13. 6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
0.49k 0.49k
0.80k
0.80k
0.80k
0.80k
0.80k
0.18k 0.18k 0.18k 0.18k 0.18k
R1
R2 = 2.94 k= 2.94 k
0
10
9
13.45
6 6
6-6
0.18 k
L3L4
3.53k
L3U3
1.32k 1.32k
10
9
13.45
∑V = 0
L3U3= 2.14 k (T)
L3U3 + 2*1.32*(10/13.45) – 0.18 = 0
Dead load and wind load analysis

14. (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 and wind load analysis