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Transcript of M sc steel-08-prof-zahid_siddiqi
Steel Structures
SE-505
Plastic Analysis and Design of Structures
M.Sc. Structural Engineering
Example
Design the following two bay single story frame with the working loads as shown in the figure.
Assume that the positive hinges are formed at the midspans and use the following load combinations:
1. 1.2D + 1.6L
2. 1.2D + 1.6L + 0.8W
3. 1.2D + 0.5L + 1.3W
30m 20m
8m
D = 5 kN/mL = 5 kN/m
25 kN/m
1. Structure And Loading
First And Second CombinationsLet w = 1.2 × 5 + 1.6 × 5 = 14 kN/mHorizontal load = 0.8 × 25 × 8 / 2 = 80 kN
= 5.72 w
Third Combination
Let w = 1.2 × 5 + 0.5 × 5 = 8.5 kN/mHorizontal load = 1.3 × 25 × 8 / 2 = 130 kN
= 15.30 w
5.72 w or2.68 (5.72 w)
Mp
Mp k1 Mp
k1 Mpk2 Mp
A
B
CD
E F
15 w 10 w7.5 w 12.5 w 5 w
2. Relative Mp Values
Let
Plastic moment capacity for beam BC = Mp
Then plastic moment capacity for column AB = Mp
The factor k1 will be calculated according to the ratio of simply supported moments for the spans BC and CD.
This ratio simplifies to the ratio of the square of spans, if the load is the same.
2
2
BC
CD
LL
2
2
3020
k1 = = = 0.44
Mp 0.44 Mp
k2 Mp
Use same Mp value of k1 Mp for the column DE.
Find k2 for the equilibrium of Joint – C, but having certain minimum value.
k2 = 1 − 0.44 = 0.56(not less than 0.5 times the larger Mp value)
3. Analysis i) Beam BC Mechanism 15w
θ θ
2θ
B CWE = 15w × 15θ = 225 w θWI = 4 Mp θw = 0.0178 Mp
10w
θ θ
2θ
C D
ii) Beam CD Mechanism
WE = 10w × 10θ = 100 w θ
WI = 0.44 Mp 4θ
w = 0.0176 Mp
iii) Panel Mechanism
WE = 5.72w × 8θ = 45.76 w θor = 15.3w × 8θ = 122.4 w θWI = 2 Mp θ + 2 × 0.56Mp θ
+ 2 × 0.44Mp θ = 4 Mp θ
w = 0.0874 Mp or 0.0326 Mp
5.72 w or15.3 w
iv) Joint C Mechanism
WE = 0
WI = 2 Mp θ θ
θθ
v) Combined Mechanism [(i) + (iii)]
WE = (225 + 45.76) w θ= 270.76 w θ
or = (225 + 122.4) w θ= 347.4 w θ
WI = 6 Mp θ
w = 0.0222 Mp or 0.0173 Mp
5.72 w or15.3 w
15 w 10 w
4. Reactions For Critical Mechanisms
A. Reactions For Mechanism (v) For 3rd Combination
Mp = 8.5 / 0.0173 = 491.3 kN-m
2 × 491.3 − 7.45 × 8 + 8.5 × 152 / 2 − VA × 15 = 0⇒ VA = 125.28 kN
VE = 8.5 × 20 / 2 = 85 kN
VF = 8.5 × 50 − 85 − 125.28 = 214.2 kN
HA = 130.3 − 68.8 − 54.05 = 7.45 kN
130.3 kN
8.5 kN/m
A
BC D
EF
491.3 275.1 216.2
216.2275.1
491.3 216.2
491.37.4568.8 54.05
54.05
214.2
68.87.45
125.28 85
B. Reactions For Mechanism (v) For 2nd Combination
Mp = 14 / 0.0222 = 630.6 kN-m
630.6 + 77.6 × 8 + 14 × 152 / 2 − VA × 15 = 0⇒ VA = 188.4 kN
VE = 14 × 20 / 2 = 140 kN
VF = 14 × 50 − 188.4 − 140 = 371.6 kN
HA = 80.08 − 88.3 − 69.4
= 77.6 kN (towards the right)
80.08 kN
14 kN/m
A
BC D
EF
630.6 353.2 277.5
277.5353.2
630.6 277.5
630.677.688.3 69.4
69.4
371.6
88.377.6
188.4 140
C. Reactions For Mechanisms (i) And (ii) For 1st Combination
14 kN/m
A
BC D
EF
397.75 222.75 175
350795.5
350
795.5
65.6
350
83.5149.2
210 140
795.5
445.5 350
* Moments at bases of columns may be taken equal to half of the top moments as carry-overs.
* Beams have equal end moments and hence the vertical reactions are equal to simply supported reactions.
5. Bending Moment Diagrams
6. Selection Of Sections
If the sections for dmin of beams are not available or are highly uneconomical, Ireq for partially fixed ends may be checked.
7. Check Moment MagnificationRevise the column and beam sections accordingly.
8. Check Compactness Of Sections9. Satisfy Interaction Equations For Columns10. Check Interaction Equations For Other
Combinations11. Check Shear12. Decide Lateral BracingConsider both the critical lengths Lp and Lpd. Also decide the unbraced lengths of columns to prevent buckling about minor axis.
13. Design Three Connections
14. Design Three Base Plates
Example
Design the given gable frame with haunched connections for the following load cases:
1.2D + 1.6L
1.2D + 0.5L + 1.3W
9m 9m
6m
D = 5 kN/m, L = 10 kN/m
10 kN/m
2.5m
1. LoadingCase I
Let w = 1.2 × 5 + 1.6 × 10 = 22 kN/m
Case II
Let w = 1.2 × 5 + 0.5 × 10 = 11 kN/m
Horizontal load = 1.3 × 10 × 8.52 / 2 = 78.27 kN= 7.12 w
2. Relative Plastic Moment ValuesMp – values of the girder at the end of the haunch (point – A) and at the maximum moment section (point – B) close to the crown should be the same, although the moments will be opposite in direction.
Mp – values of the column may be different and can be found from the moment diagram.
Let
Mp = value of Mp required at the haunch point
22 95.2 +
3341.99
×
MA and MB = value of Mp required for the girder
Length of haunch along the girder = L / 10 to L / 6
= 3 m
Length of the haunch along the column
= 2/3rd of length along the beam = 2 m
Total inclined length of one girder =
= 9.341 m
Horizontal length of haunch = = 2.89 m
3. AnalysisCase I:
x
2.89m
A
A
B
B
6H 6H8.5H
891 kN-m
289.222 2×
95.2 H
Let
Plastic moment required at the haunch point
= Mp = 6H
Due to the presence of the haunch, let the hinges be formed at section A and B.
MA = 22 × 9 × 2.89 − − 6H −
= 480.3 − 1.134 Mp
× 2.89
222 2x×
95.2 HMB = 22 × 9 × x − − 6H −
= 198 x − 11 x2 − Mp − 0.046 Mp × x
× x
Now, MB = − MA
(as both must be equal and opposite)198 x − 11 x2 − Mp − 0.046 Mp × x
= 480.3 − 1.134 Mp
Mp = x
xx046.0134.2
3.48019811 2
+++−
To get the minimum collapse load or the maximum required Mp value,
dxdM p
= 0
(−11x2 + 198 x + 480.3) (0.046) − (2.134 + 0.046 x) (−22 x + 198) = 0
0.506 x2 + 46.948 x − 400.436 = 0
x = 7.863
Mp = 543.8 kN-m
MA = − 136.3 kN-m
MB = 136.3 kN-m
Column moment = 362.53 kN-m
22 kN/m
120.65543.8
362.53
543.8
2 m
543.8 / 6 = 90.63
198
Case II:11 kN/m
2.5 m
HH − T
T = 78.27 kN
6 m
A
B
18 m
x
A
A
B
B
6H 6H8.5H
445.5 kN-m
469.62 kN-m
The negative moment on the right side (section A) is more as the simply supported positive moment here is less.
However the positive moment (section B) is greater on the left side.
289.211 2×
1889.262.469 ×
95.2 H
Let plastic moment required at the haunch point
= Mp = 6H
Due to the presence of the haunch, let the hinges be formed at section A and B.
MA = 99 × 2.89 − +
− 6H − × 2.89
= 315.57 − 1.134 Mp
211 2x×
1862.469 x×
95.2 H
MB = 99 × x − + 469.62 −
− 6H − × x
MB = 72.91 x − 5.5 x2 − Mp − 0.0463 Mp × x+ 469.62
Now, MB = − MA
(as both must be equal and opposite)
72.91 x − 5.5 x2 − Mp − 0.0463 Mp × x + 469.62
= 315.57 − 1.134 Mp
Mp = xxx
0463.0134.219.78591.725.5 2
+++−
To get the minimum collapse load or the maximum required Mp value,
dxdM p = 0
(−5.5x2 + 72.91 x + 785.19) (0.0463) − (2.134 + 0.0463 x) (−11 x + 72.91) = 0
0.25465 x2 + 23.474 x − 119.236 = 0
x = 4.827 m
Mp = 428.0 kN-m
MA = − 169.8 kN-m
MB = 169.8 kN-m
Column moment = 285.3 kN-m
11 kN/m
428.0
285.3
78.27
2 m
428 / 6 = 71.33
125.1
6.94
72.91
41.645
4. Initial Selection Of SectionsA. Girder:
Mp = 169.8 kN-m
Z = 2509.0108.169 6
××
W 360 × 44 is selected with the following properties:
A = 5710 mm2; d = 352 mm; bf = 171 mm; tf = 9.8 mm; tw = 6.9 mm
φb Mp = 174.40 kN-m; Lp = 1.88 m; Lr = 5.53 m; rx = 146mm; ry = 37.8 mm
= 755 × 103 mm3
Concluded