SITUATION. A fully continuous monolithic floor system consists of slab as shown. Use the ACI moment coefficient as provided by the code. -M (interior support)=- 12 +M (midspan) = 16 7.4000 34000 3 4000 30000 0000 3.4000 3.4000 3.0000 3.0000 Q Zoom image Design data: Live load= 4 kPa Floor finish= 1 kPa Ceiling load= 0.25 kPa fc'= 21 MPa fy= 276 MPa Use 12 mm diameter as main reinforcement. Use 10 mm diameter as secondary reinforcement. Using the minimum thickness of the slab, compute the maximum negative factored moment. 5.87 kN-m 7.83 kN-m O 10.06 kN-m 8.19 KN-m
SITUATION. A fully continuous monolithic floor system consists of slab as shown. Use the ACI moment coefficient as provided by the code. -M (interior support)=- 12 +M (midspan) = 16 7.4000 34000 3 4000 30000 0000 3.4000 3.4000 3.0000 3.0000 Q Zoom image Design data: Live load= 4 kPa Floor finish= 1 kPa Ceiling load= 0.25 kPa fc'= 21 MPa fy= 276 MPa Use 12 mm diameter as main reinforcement. Use 10 mm diameter as secondary reinforcement. Using the minimum thickness of the slab, compute the maximum negative factored moment. 5.87 kN-m 7.83 kN-m O 10.06 kN-m 8.19 KN-m
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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