| | 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 -
| | 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 -
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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**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.
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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:
The load on the beam 2I2F is w3=0
The load on the beam 2F2E is w4:
The load on the beam 1F2F is w2:
The half of the beam loads will be acting on the column 2F.
The total load on the column 2F is W
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