Problem #2 A beam having a simple span of 4.2 m. carries a concentrated service live load P at its midspan. The beam is made up of a channel section. Assume that the load is applied through the shear center so that there is no torsional loading. Assume lateral loads are applied at both ends only. Prop. of channel section d = 457.2 mm tr = 15.875 mm Sx = 1141 x 103 mm3 ly = 6.785 x 106 mm4 U = 1.2 DL + 1.6 LL ® Determine the nominal flexural strength of the channel using LRFD. Fy = 248 MPa E = 200000 MPa ® Determine the value of P that can be applied to the beam due to its flexural strength use LRFD. = 26.162 mm Lp = 1.307 L; = 5.341 ® Determine the value of P that can be applied to the beam due to its allowable bending stress using ASD. Cb = 1.32 Zx = 1430 x 103 mm3 Solution: Wt. of channel = 77 kg/m O Nominal flexural strength of channel using LRFD: Lo = 4.2 m 4.2 m
Problem #2 A beam having a simple span of 4.2 m. carries a concentrated service live load P at its midspan. The beam is made up of a channel section. Assume that the load is applied through the shear center so that there is no torsional loading. Assume lateral loads are applied at both ends only. Prop. of channel section d = 457.2 mm tr = 15.875 mm Sx = 1141 x 103 mm3 ly = 6.785 x 106 mm4 U = 1.2 DL + 1.6 LL ® Determine the nominal flexural strength of the channel using LRFD. Fy = 248 MPa E = 200000 MPa ® Determine the value of P that can be applied to the beam due to its flexural strength use LRFD. = 26.162 mm Lp = 1.307 L; = 5.341 ® Determine the value of P that can be applied to the beam due to its allowable bending stress using ASD. Cb = 1.32 Zx = 1430 x 103 mm3 Solution: Wt. of channel = 77 kg/m O Nominal flexural strength of channel using LRFD: Lo = 4.2 m 4.2 m
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Problem #2
A beam having a simple span of 4.2 m. carries
a concentrated service live load P at its
midspan. The beam is made up of a channel
section. Assume that the load is applied
through the shear center so that there is no
torsional loading. Assume lateral loads are
applied at both ends only.
Prop. of channel section
d = 457.2 mm
tr = 15.875 mm
Sx = 1141 x 103 mm³
ly = 6.785 x 106 mm4
U = 1.2 DL + 1.6 LL
® Determine the nominal flexural strength
of the channel using LRFD.
Fy = 248 MPa
E = 200000 MPa
Determine the value of P that can be
ry = 26.162 mm
Lp = 1.307
L; = 5.341
applied to the beam due to its flexural
strength use LRFD.
® Determine the value of P that can be
applied to the beam due to its allowable
bending stress using ASD.
Cb = 1.32
Zx = 1430 x 103 mm³
Solution:
Wt. of channel = 77 kg/m
Lo = 4.2 m
O Nominal flexural strength of channel
using LRFD:
4.2 m
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