| | L1/2 | L1/2 | | L1/2 L1/2| L1 L1 G L1/8 H L1/8 G L1/4 I- L1/4 Оpen Area L1/4 L1/4 L1/8 L1/8 I- FLOOR BEAMS Problem 1) For a uniform slab loading of 6 kip/ft^2, solve for the load on column 2F and the distributed load on floor beam 3C4C. Use L1 = 24 ft. C IRDE RS -

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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**Text Transcription for Educational Website:**

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I understand that this Zoom session is being recorded, and attest that I have fully complied with the TAMU Code of Honor. You do not need to copy the above statement. Your signature on the front page of your submission will suffice.

---

**Structural Diagram Explanation:**

The diagram represents a structural floor plan with girders and floor beams. The layout includes designated column grids labeled numerically (1 to 5) and alphabetically (A to I), forming rectangular sections.

- **Columns and Beams:**
  - The diagram shows alternating spans of L1 and L/2 along the horizontal direction.
  - Vertical spacings are marked as L/8, L/4, and L/2.

The central portion of the layout is labeled as "Open Area," suggesting an unencumbered space without intermediate supports.

**Problem 1:**

For a uniform slab loading of 6 kip/ft², solve for the load on column 2F and the distributed load on floor beam 3C4C. Use L1 = 24 ft.

--- 

This problem involves calculating the structural loads based on the given parameters and beam arrangements.
Transcribed Image Text:**Text Transcription for Educational Website:** --- I understand that this Zoom session is being recorded, and attest that I have fully complied with the TAMU Code of Honor. You do not need to copy the above statement. Your signature on the front page of your submission will suffice. --- **Structural Diagram Explanation:** The diagram represents a structural floor plan with girders and floor beams. The layout includes designated column grids labeled numerically (1 to 5) and alphabetically (A to I), forming rectangular sections. - **Columns and Beams:** - The diagram shows alternating spans of L1 and L/2 along the horizontal direction. - Vertical spacings are marked as L/8, L/4, and L/2. The central portion of the layout is labeled as "Open Area," suggesting an unencumbered space without intermediate supports. **Problem 1:** For a uniform slab loading of 6 kip/ft², solve for the load on column 2F and the distributed load on floor beam 3C4C. Use L1 = 24 ft. --- This problem involves calculating the structural loads based on the given parameters and beam arrangements.
Expert Solution
Step 1

Load on the column 2F:

Since, the beam self weight and beam size is not given, consider the self weight of beam as 2 kips

The load on the beam 2F3F is w1:

w1=Total Slab LoadBeam Length+Self Weight of beam=L18×L1×6 kip/ft2 L1+2 kips=24 ft8×24 ft×6 kip/ft224 ft+2 kips=20 kips

The load on the beam 2I2F is w3=0

The load on the beam 2F2E is w4:

w4=Total Slab LoadBeam Length+Self Weight of beam=L14×L14×6 kip/ft2 L14+2 kips=24 ft4×24 ft4×6 kip/ft224 ft4+2 kips=38 kips

The load on the beam 1F2F is w2:

w2=Total Slab LoadBeam Length+Self Weight of beam+Load from the secondary beam=L18×L1×6 kip/ft2 L1+2 kips+38 kips2×2=24 ft8×24 ft×6 kip/ft224 ft+2 kips+38 kips=58 kips

The half of the beam loads will be acting on the column 2F.

The total load on the column 2F is W

W=w1+w2+w3+w4=20 kips+58 kips+0+38 kips=116 kips

 

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