# Column B.-25 ftSteel girder(A 32.7 in.2)ATBSteelcolumn14 in.concrete slabC(A-19.2)2 at 12 ft24 ftnSteel floor beamEF

Question

The floor system shown in the attached image consists of a 4in thick reinforced concrete floor slab resting on three steel floor beams (Beams AB, CD, and EF), which in turn are supported by two steel girders (Girders ACE, and BDF). Steel columns support the ends of each girder at point A, B, E, and F. The crosse-sectional area of each steel beam in 19.2in2 and the cross sectional area of each steel girder in 32.7in2. The is also a 6in thick brick wall that is 9ft tall and 25ft long, that bears directly on top of beam CD. (Note: attached image is a plan view image of hte floor plan and the wall come 9ft out of the page). The unit weights for the structural materials 150lb/ft3(normal weight reinforced concrete), 102lb/ft3(brick wall), and 490lb/ft3(steel).

• Provide a FBD of Beam CD, and a FBD of Girder ACE, showing all the dead loads that act on those member along with the reaction forces at the ends of each member.
• Determine the total dead load acting at the top of column B.
• The floor system supports a live load of 100lb/ft2. Show the various live loads(only) on the FBDs of Beam CD and Girder ACE.
Step 1
Step 2

Calculation.

a. Free body diagram of beam CD is follows with all loads.

To provide loads applied we have to calculate all present loads on the beam CD.

The dead load is present in the concrete slab and steel beam is applied on the beam CD as uniformly.

Step 3

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