A simply supported beam having a simple span o 8 m must support a uniformly distributed service load of 10 kN/m including its own weight. A36 steel is used. Properties of the steel beam: 2 A = 10452 mm d=598.68 mm b₁ = 177.93 mm t=12.33 mm a. b. C. t = 10.03 mm F, = 248 MPa 1,= 562 x 105 mm² ly = 12 x 105 mm² Z, 2196 x 10³ mm³ r₂ = 231.39 mm Determine the allowable bending stress using elastic design (ASD). Determine the safe uniform live load it could support using ASD method due to its flexural strength. Determine the safe uniform live load it could support using plastic design method due to its flexural sobrid strength. U = 1.2 DL +1.6 LL 3
A simply supported beam having a simple span o 8 m must support a uniformly distributed service load of 10 kN/m including its own weight. A36 steel is used. Properties of the steel beam: 2 A = 10452 mm d=598.68 mm b₁ = 177.93 mm t=12.33 mm a. b. C. t = 10.03 mm F, = 248 MPa 1,= 562 x 105 mm² ly = 12 x 105 mm² Z, 2196 x 10³ mm³ r₂ = 231.39 mm Determine the allowable bending stress using elastic design (ASD). Determine the safe uniform live load it could support using ASD method due to its flexural strength. Determine the safe uniform live load it could support using plastic design method due to its flexural sobrid strength. U = 1.2 DL +1.6 LL 3
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Steel Design
I need a complete solution from A,B and C in Steel Design with complete answer thank you..
Transcribed Image Text:A simply supported beam having a simple span o 8 m must support a uniformly distributed service load of 10
kN/m including its own weight. A36 steel is used.
Properties of the steel beam:
a.
b.
C.
2
A= 10452 mm
d=598.68 mm
b = 177.93 mm
t= 12.33 mm
t = 10.03 mm
Fy = 248 MPa
1, 562 x 106 mmª
ly = 12 x 105 mm*
Z, 2196 x 10³ mm³
r₂ = 231.39 mm
Determine the allowable bending stress using elastic design (ASD).
Determine the safe uniform live load it could support using ASD method due to its flexural strength.
Determine the safe uniform live load it could support using plastic design method due to its flexural
strength. U = 1.2 DL +1.6 LL
4011
D
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