Structural Steel Design (6th Edition)
Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
Question
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Chapter 14, Problem 14.1PFS
To determine

The LRFD design load and the ASD design loads for the given connection members.

Expert Solution & Answer
Check Mark

Answer to Problem 14.1PFS

  97.2k and 64.7k

Explanation of Solution

Given:

  Structural Steel Design (6th Edition), Chapter 14, Problem 14.1PFS

Calculation:

The ultimate load by LRFD is

  Pu=1.6PLPu=1.6(32)Pu=51.2kips

The factored moment is

  Mu=PuL4Mu=51.2( 40)4Mu=512kipsft

The factored shear force is

  Vu=Pu2Vu=51.22Vu=25.6kips

Refer Table 3-2 in the AISC steel construction manual.

Try W30×124

  ϕvVnxVu530kips26.5kipsϕbMpxMu1530kipsft512kipsft

The required moment of inertia is

  L1000=PLL348EI40×121000=32× ( 40×12 )348( 29000)II=5296.55in.4

Refer Table 3-3 in the AISC steel construction manual.

Try W30×124

  Ix=5360in.4>5296.55in.4

Hence, it’s safe.

The ultimate load by ASD is

  Pa=PLPa=32kips

The factored moment is

  Ma=PaL4Ma=32( 40)4Ma=320kipsft

The factored shear force is

  Va=Pa2Va=322Va=16kips

Refer Table 3-2 in the AISC steel construction manual.

Try W30×124

  VuΩVu353kips16kipsM pxΩbMu1020kipsft320kipsft

The required moment of inertia is

  L1000=PLL348EI40×121000=32× ( 40×12 )348( 29000)II=5296.55in.4

Refer Table 3-3 in the AISC steel construction manual.

Try W30×124

  Ix=5360in.4>5296.55in.4

By taking from the minimum values we obtained we get the allowable and safe loads.

Conclusion:

Therefore, the required loads by LRFD and ASD are 97.2k and 64.7k respectively.

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