Cee 312(5)(structural analysis)
18
CEE-312 Structural Analysis and Design Sessional-I (1.0 credit) Lecture: 5 Bijit Kumar Banik Assistant Professor, CEE, SUST Room No.: 115 (“C”building) [email protected] Department of Civil and Environmental Engineering
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Transcript of Cee 312(5)(structural analysis)
CEE-312
Structural Analysis and Design Sessional-I
(1.0 credit)Lecture: 5
Bijit Kumar Banik
Assistant Professor, CEE, SUSTRoom No.: 115 (“C” building)
Department of Civil and Environmental Engineering
Analysis and design of an Industrial roof truss sys tem
6@6 ft = 36 ft
L0L1 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) (3.7
8)
(2.90)
3.24k
L R
0.23k
0.23k
0.12k
0.12k
2.46k
2.46k
1.23k
1.23k
Analysis and design of an Industrial roof truss sys tem
6@6 ft = 36 ft
L0L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
4.54 k 2.60 k
3.24k
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k
0.12k
0.12kR L
Analysis and design of an Industrial roof truss sys tem
1
1
∑V = 0(5/10.3)*L0U1+(9/10.3)*1.23 – 4.54 = 0L0U1= 7.14 k (T) 5
9
10.3
∑H = 0
L0L1 +(9/10.3)*7.14 +3.24 – (5/10.3)*1.23 =0L0L1= 8.88 k (C)
4.54 k
1.23k
L0L1
L0U1
1-1
3.24 k
95
10.3
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
R1
R2
H
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
59
10.3
95
10.3
= 3.24k= 4.54k
= 2.60k
Analysis and design of an Industrial roof truss sys tem
∑V= 0
L1U1 = 0
L1L2 = L0L1 = 8.88k (C)
∑H= 0
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
R1
R2
H
59
10.3
95
10.3
= 3.24k
2 2
L1L2
2-2
L0L1=8.88k
L1U1
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
= 4.54k= 2.60k
Analysis and design of an Industrial roof truss sys tem
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
59
10.3
95
10.3
3
3
9
510.3
95
10.3
95
10.3
∑ML0 = 0– 2.46*6.86+ (5/10.3)*U1L2*6 + (9/10.3)*U1L2*(10/3)= 0
U1L2= 2.90 k (T)∑V = 0
(9/10.3)*1.23+(9/10.3)*2.46 + (5/10.3)*U1U2– (5/10.3)*2.90 – 4.54= 0
U1U2= 5.61 k (T)
+ve 3-3
U1
L0 L1 L2
U22.46 k
1.23 k
4.54k
(10/3)’
8.88k3.24k
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
R1
R2
= 3.24k= 4.54k
= 2.60k
Analysis and design of an Industrial roof truss sys tem
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
59
10.3
95
10.3
4 4
∑H = 0
L2L3= 6.35 k (C)
59
10.3
L2L3
4-4
8.88k
L2U22.90k
∑V = 0L2U2 = (5/10.3)*2.90
L2U2= 1.41 k (C)
L2L3 +8.88 – (9/10.3)*2.90 = 0
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
R1
R2
= 3.24k= 4.54k
= 2.60k
Analysis and design of an Industrial roof truss sys tem
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
59
10.3
95
10.3
5
5
109
13.45
59
10.3
∑ML0 = 0
(9/13.45)*U2L3*(20/3)+ (10/13.45)*U2L3*12 – 2.46*6.86 – 2.46*2*6.86 = 0
U2L3= 3.78 k (T)∑V = 0(5/10.3)*U2U3+ (9/10.3)*(1.23+2.46+2.46)– (10/13.45)*3.78 – 4.54 = 0
U2U3= 4.07 k (T)
+ve
95
10.3
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
R1
R2
= 3.24k= 4.54k
= 2.60k5-5
L0 L1 L2 L3
U1
U2
U3
1.23 k
4.54 k
2.46 k
6.35 k
2*(10/3)
2.46 k
3.24 k
Analysis and design of an Industrial roof truss sys tem
(5/10.3)*U5L6+(9/10.3)*0.12 – 2.60 = 0∑V= 0
U5L6 = 5.14k (T)
∑H= 0
L5L6 = 4.43k (C)
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
59
10.3
95
10.3
6
6
59
10.35
910.3
2.60 k
U5L6 0.12k6-6
L5L6(5/10.3)*0.12 – (9/10.3)*5.14 – L5L6= 0
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
R1
R2
= 3.24k= 4.54k
= 2.60k
Analysis and design of an Industrial roof truss sys tem
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
59
10.3
95
10.3
7 7
∑H = 04.43k
7 – 7
L4L5
L5U5
∑V = 0
L5U5= 0
L4L5 = 4.43k (C)
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
R1
R2
= 3.24k= 4.54k
= 2.60k
Analysis and design of an Industrial roof truss sys tem
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
59
10.3
95
10.3
8
8
59
10.3
9510.3
95
10.3
∑ML6 = 00.23*6.86 – (5/10.3)*U5L4*6 – (9/10.3)*U5L4*(10/3)= 0
U5L4= 0.27 k (T)∑V = 0
(9/10.3)*0.23+(9/10.3)*0.12 + (5/10.3)*U4U5– (5/10.3)*0.27 – 2.60= 0
U4U5= 5.0 k (T)
+ve
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
R1
R2
= 3.24k= 4.54k
= 2.60k
L6
U5
L5L4
U4
0.23k
0.12k
2.60k
8-8
Analysis and design of an Industrial roof truss sys tem
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
59
10.3
95
10.3
9 9
∑H = 0
L3L4= 4.19 k (C)
∑V = 0L4U4 = (5/10.3)*0.27
L4U4= 0.13 k (C)
L3L4+ (9/10.3)*0.27 – 4.43 = 0
9510.3
4.43k
9-9L4U4
0.27k
L3U4
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
R1
R2
= 3.24k= 4.54k
= 2.60k
Analysis and design of an Industrial roof truss sys tem
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
59
10.3
95
10.3
10
10
95
10.3
910
13.459
5
10.3
0.23k
0.23k
0.12k
2.60k
L6L5L4L3
U3
U5
U4
4.19k
10 -10 ∑ML6 = 0
–(9/13.45)*U4L3*(20/3) – (10/13.45)*U4L3*12 + 0.23*6.86 +0.23*2*6.86 = 0
U4L3= 0.35 k (T)∑V = 0(5/10.3)*U3U4 + (9/10.3)*(0.12+2*0.23)– (10/13.45)*0.35 – 2.60 = 0
U3U4= 4.85 k (T)
+ve
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
R1
R2
= 3.24k= 4.54k
= 2.60k
Analysis and design of an Industrial roof truss sys tem
6@6 ft = 36 ft
10 ft
L0 L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
59
10.3
95
10.3
11 11
∑V = 0
91013
.45
910
13.45
3.11k
11 – 11
0.95k
L3U3
L3
3.78k 0.35k(10/13.45)*3.78 + (10/13.45)*0.35 – L3U3= 0
L3U3= 3.07 k (C)
2.46k
2.46k
1.23k
1.23k
0.23k
0.23k0.12k
0.12k
R1
R2
= 3.24k= 4.54k
= 2.60k
Analysis and design of an Industrial roof truss sys tem
(- 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
L0L1 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) (0.3
5)
(0.27)
3.24k
R L
Analysis and design of an Industrial roof truss sys tem
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 L5L6
U1
U2
U3
U4
U5
2.60 k4.54 k
(0)
(-0.
13)
(-1.
41)
(0)(-
3.07
)
(3.78)(2.90) (0.3
5)(0.27)
3.24k
L→R
6@6 ft = 36 ft
L0 L1 L2 L3 L4 L5L6
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) (3.7
8)
(2.90)
3.24k
Analysis and design of an Industrial roof truss sys tem
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 L5L6
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 L5L6
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 L5L6
U1
U2
U3
U4
U5
2.94 k2.94 k
(.18
)
(0.6
7)
(0.6
7)
(.18
)
(2.1
4)(- 1.32)(- 1.01) (-
1.32
)
(- 1.01)
Dead Load
