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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.
Column B.
-25 ft
Steel girder
(A 32.7 in.2)
AT
B
Steel
column
1
4 in.
concrete slab
C
(A-19.2)
2 at 12 ft
24 ft
n
Steel floor beam
E
F
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Column B. -25 ft Steel girder (A 32.7 in.2) AT B Steel column 1 4 in. concrete slab C (A-19.2) 2 at 12 ft 24 ft n Steel floor beam E F

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Step 1
Given:
Area of the steel girder is A=32.7in2
Area of the steel floor is A=19.2in2
Thickness of reinforced concrete slab is 4in
And the given floor system is given below
25ft
B
A=32.7in2
Steel girder
2 at 12 ft-24ft
D
A=19.2in2
Steel floor beam
E
LL
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Given: Area of the steel girder is A=32.7in2 Area of the steel floor is A=19.2in2 Thickness of reinforced concrete slab is 4in And the given floor system is given below 25ft B A=32.7in2 Steel girder 2 at 12 ft-24ft D A=19.2in2 Steel floor beam E LL

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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.

 Calculating dead load for concrete slab is

Dead load unit weight x area x thick ness
150x12x 4in
Dead load
4
Dead load 150x12x-
12
Dead load 600lb/ft
Now, calculating the dead load for steel beam is
Dead load unit weight x area
Dead load 490 1b19.2
ft 122
Dead load 65.33lb/ft
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Dead load unit weight x area x thick ness 150x12x 4in Dead load 4 Dead load 150x12x- 12 Dead load 600lb/ft Now, calculating the dead load for steel beam is Dead load unit weight x area Dead load 490 1b19.2 ft 122 Dead load 65.33lb/ft

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Step 3
Now total load for the slab is
Total load of the slab Dead of the concrete slab Dead of the steel beam
Total load Dead load of the steel beam + Dwad load of the concrete slab
Total load 65.33lb/ft+ 600lb/ft
Total load 665.33lb/ft
Therefore, the beam with dead load is shown below.
665.33lb/ft
25ft
Cy
Dy
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Now total load for the slab is Total load of the slab Dead of the concrete slab Dead of the steel beam Total load Dead load of the steel beam + Dwad load of the concrete slab Total load 65.33lb/ft+ 600lb/ft Total load 665.33lb/ft Therefore, the beam with dead load is shown below. 665.33lb/ft 25ft Cy Dy

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