Resumen De Articulos
Páginas: 2 (262 palabras)
Publicado: 24 de septiembre de 2012
EXAMPLE 1
Design a welded plate girder to support a uniform load of 3 kips per ft., and two concentrated loads of 70 kips located 17 ft.,from each end. The compression flange of the girder will be laterally supported only at points of concentrated load.
Givens: Maximumbending moment: 2054 kip-ft
Maximum vertical shear: 142 kips
Span: 48 ft.
Maximun depth: 72in.
Steel: Fy = 36 ksi.
SOLUTION:
A. Preliminary web design:
1. Assume web depth, h = 70 in.
For no reduction in flangestress, h/t ≤760/(22)1/2 = 162
Corresponding thickness of web = 70/162 = 0.43 in.
2. Minimum thickness of web = 70/322 = 0.22 in.
Try webplate 5/16 X 70: Aw = 21.9 in;
h/t = 70/0.313 = 224
B. Preliminary flange design:
1.Required flange area:
An approximate formula for area of one flange is:
Af=M/Fbh – Aw/6 = 2054X12/(22X70) – 21.9/6 = 12.4 in2.
Try ¾ X 18plate: Af = 13.5 in2
2. Check for adequancy against local buckling:
bf/2tf =18/2X0.75 =12.0 < 15.8 o.k.
C. Trial girder section:
Web5/16 X 70; 2 flange plates ¾ X18
1. Check by “moment of inertia” method:
section A y ∑Ay2 Io Igr
1 Web 5/16X70 21.9 0 0 8932 8932.29
1flange ¾ X18 13.5 37.375 18858.02 0.63 18858.65
1 flange ¾ X18 13.5 37.375 18858.02 0.63 18858.65
Moment of Inertia 37716.04 1.26 46649.6
Design a welded plate girder to support a uniform load of 3 kips per ft., and two concentrated loads of 70 kips located 17 ft.,from each end. The compression flange of the girder will be laterally supported only at points of concentrated load.
Givens: Maximumbending moment: 2054 kip-ft
Maximum vertical shear: 142 kips
Span: 48 ft.
Maximun depth: 72in.
Steel: Fy = 36 ksi.
SOLUTION:
A. Preliminary web design:
1. Assume web depth, h = 70 in.
For no reduction in flangestress, h/t ≤760/(22)1/2 = 162
Corresponding thickness of web = 70/162 = 0.43 in.
2. Minimum thickness of web = 70/322 = 0.22 in.
Try webplate 5/16 X 70: Aw = 21.9 in;
h/t = 70/0.313 = 224
B. Preliminary flange design:
1.Required flange area:
An approximate formula for area of one flange is:
Af=M/Fbh – Aw/6 = 2054X12/(22X70) – 21.9/6 = 12.4 in2.
Try ¾ X 18plate: Af = 13.5 in2
2. Check for adequancy against local buckling:
bf/2tf =18/2X0.75 =12.0 < 15.8 o.k.
C. Trial girder section:
Web5/16 X 70; 2 flange plates ¾ X18
1. Check by “moment of inertia” method:
section A y ∑Ay2 Io Igr
1 Web 5/16X70 21.9 0 0 8932 8932.29
1flange ¾ X18 13.5 37.375 18858.02 0.63 18858.65
1 flange ¾ X18 13.5 37.375 18858.02 0.63 18858.65
Moment of Inertia 37716.04 1.26 46649.6
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