Determine the design shear strength of the given shape, using Fy=50 ksi. Blank 1W14x53...A=15.80Jin2d=13.900in.tw 3=0.370In.bf =9.030In.n.Jn.0.83010-7:91.2500In.n.k11.00006 1:2In.it2.150n.2.8251.0077.90n.5.090In.ly57.7014.30Syry1.020In.07.10n.Zy22.00In1.94Cw2540Ina=53. 22InWno=23.70n12Sw =35.5018.80Qw =42.80

The yield strength of the steel (Fy) = 50 ksiThe steel section used = W1453The properties of the…

mine the design: shape, A= da tw= bf= T= k it NP l SA IK ly Sy 253 2x Zy it Cw Who Py=50 R$i. Blak W14x53 15.03 13.900 0.370 8.000 0.000 10.7/9 1.2500 1.3300 6.12 2.150 2.32 511.00 77.90 5.090 57.70 11.30 1.920 07.10 22.00 1.94 2540 58.22 29.70 In In In n. Nil 1.²3 M. In. 1.43 1.43 Nu In n.12

mine the design: shape, A= da tw= bf= T= k it NP l SA IK ly Sy 253 2x Zy it Cw Who Py=50 R$i. Blak W14x53 15.03 13.900 0.370 8.000 0.000 10.7/9 1.2500 1.3300 6.12 2.150 2.32 511.00 77.90 5.090 57.70 11.30 1.920 07.10 22.00 1.94 2540 58.22 29.70 In In In n. Nil 1.²3 M. In. 1.43 1.43 Nu In n.12

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

Determine the design shear strength of the given shape, using Fy=50 ksi. Blank 1
W14x53
...
A=
15.80
Jin2
d=
13.900
in.
tw 3=
0.370
In.
bf =
9.030
In.
n.
Jn.
0.830
10-7:9
1.2500
In.
n.
k1
1.0000
6 1:2
In.
it
2.150
n.
2.82
51.00
77.90
n.
5.090
In.
ly
57.70
14.30
Sy
ry
1.020
In.
07.10
n.
Zy
22.00
In
1.94
Cw
2540
In
a=
53. 22
In
Wno=
23.70
n12
Sw =
35.50
18.80
Qw =
42.80

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