Compute the maximum acceptable tensile SERVICE LOAD that may act on asingle tee section that is connected to a gusset plate using welds 12 inches long, asshown in the figure. The service live load is three times the dead load. Use A992steel. USE LRFD ONLY, no block shear will occur.WT12 x 38Longitudinalwelds11.2 in? y = 3.0 in.Given: Properties of WT12 × 38: Ag =Use A992 Steel: F, = 50 ksi Fu = 65 ksibf = 8.99in.%3D%3DLL = 3 DL%3D%3Dtwy = centroidal distancebfC. What is the Governing Ultimate Tensile Capacity based on Net FractureRound your answer to 3 decimal places.

Answer We know that LRFD capacity of the connection is given by minimum of the below three capacity…

Compute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate using welds 12 inches long, as shown in the figure. The service live load is three times the dead load. Use A992 steel. USE LRFD ONLY, no block shear will occur.

Compute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate using welds 12 inches long, as shown in the figure. The service live load is three times the dead load. Use A992 steel. USE LRFD ONLY, no block shear will occur.

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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Concept explainers

Connections

In steel structures, connections are the members that join two or more steel members. The connections are the most important members of any steel structure, and hence, they are designed more carefully. The majority of the construction cost of steel structures is incurred due to the connections. The design of connections is done as per the recommendations and specifications of the American Institute of Steel Construction (AISC).

Question

Compute the maximum acceptable tensile SERVICE LOAD that may act on a
single tee section that is connected to a gusset plate using welds 12 inches long, as
shown in the figure. The service live load is three times the dead load. Use A992
steel. USE LRFD ONLY, no block shear will occur.
WT12 x 38
Longitudinal
welds
11.2 in? y = 3.0 in.
Given: Properties of WT12 × 38: Ag =
Use A992 Steel: F, = 50 ksi Fu = 65 ksi
bf = 8.99in.
%3D
%3D
LL = 3 DL
%3D
%3D
tw
y = centroidal distance
bf
C. What is the Governing Ultimate Tensile Capacity based on Net Fracture
Round your answer to 3 decimal places.

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