4. Internal Plaster area of a room of Size 4m*3m and height of 3m with a door of 1mx2m and two windows of size 2mx1.5m in square meter is (A) 46 (С) 38 (B) 50 (D) 54

人工知能を使用せず、 すべてを段階的にデジタル形式で解決してください。 ありがとう SOLVE STEP BY STEP IN DIGITAL FORMAT DON'T USE CHATGPT Design the longitudinal steel and the cross section of the beam shown, consider concrete that indicates fy = 4200 kg/(cm²), 1/2 stirrups and 2.5 cm coatings. 1m 0.5T 2m 0.2T 6.5m Calculate top rods per steel (min) asmin If f'c = 200kg/cm² 1.5T/m

Refer to Figure P2.4 for the floor plan. Calculatethe tributary areas for (a) floor beam B3, (b) floor beamB4, (c) girder G3, (d) girder G4, (e) edge column C3, and(f) corner column C4.

estimation

Determine the losses in the post-tensioned concrete beam as shown below. 560 mm Given: POST-TENSIONED BEAM with A PARABOLIC TENDON PROFILE Span Length = 20 m fi = 1200 N/mm² A, = 2850 mm² Ac 4.23x105 mm² 1=9.36x10¹0 mm e-560 mm (@midspan, from N.A.) fc= 40 N/mm² Es = 210,000 N/mm² μ = 0.19 K = 0.0005 (per meter) Anchorage slip = 5 mm (jacked at left-end only). (make sure to draw the force variation diagram for anchorage slip) V/S= 6 RH = 70 Dead Load 9.97 kN/m fpy = 1714 N/mm² Bridge Width III

The following diagram shows a partial floor framing plan of an elevated walkway building connecting an airport terminal building to the adjacent garage. The concrete floor is not post- tensioned. Material properties are: - Concrete weight: wt = 150 pcf - Concrete compressive strength: f'c = 4000 psi - Steel tensile strength: fy = 60 ksi - Stirrup consists of #4 bar - Framing members are not exposed to weather or in contact with soil. Per the ACI, what is the minimum slab depth? Consider both the interior slab and the cantilevered slab at the edge of the building. 30-0¹ B4 B4 B4 83 Column 52 SI 30-0 -Sab Mark Dean- B4 B4 B4 83 B 30-0¹ BI B2 B2 BI -Girder 12 A 1 3-4 Ei 13:4 05 40-0 to 2

Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when 0=35° and P = 200 N. 800 N Hs = 0.20 Hk= 0.15 25° Previous