Shown below is a floor plan of an office, balcony and storage at the second floor lev of a commercial building. Refer to NSCP 2015 for the values of minimum live loads. Determine the live load intensity and load
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- The building section associated with the floor plan in Figure P2.4 is shown in Figure P2.8. Assume a live load of 60 lb/ft2 on all three floors. Calculate the axial forces produced by the live load in column C2 in the third and first stories. Consider any live load reduction if permitted by the ASCE standard.Calculate uniform load frame reactions, and normal force, shear force and bending moment and draw diagrams. L= 6m H= 4m F1= -10kN F2= -6 Kn P= 3kN/mA two story office building has interior column spaced at 6.7m apart in two perpendicular directions. If the flat of the roof live loading is allowed to be reduced live load of an interior column located at the ground floor level. Using NSCP 2010, determine the following: a.) basic roof live load in kPa b.) roof live load in kN forthe column c.) basic floor live load due to occupancy in kPa d.) reduced floor live load in kN for the column e.) total live load supported by the ground floor column
- A floor slab 100 mm thick is cast monolithically with beams 300 mm wide 500 mm deep spaced 2.0 mon centers, on simple supports over a span of 6.0 m. The floor supports a superimposed servicedead load of 1.77 kPa and service live load of 4.8 kPa. Using f’c = 21 MPa, long bar fy = 415 MPa,calculate: 7. Factored uniform load on a typical interior beam in KN/m using load factors of 1.2 for dead load and 1.6 forlive load.The reinforced concrete column isreinforced with 6 – 28 mm Ø bars as shown. Steel yield strength fy = 414 MPa and concrete strength fc’ = 21 MPa. Determine the nominal balanced load.a. 1721.252 kN c. 1821.225 kNb. 1921.253 kN d. 1521.522 KNDetermine the nominal balanced eccentricity.a. 362.544 mm c. 352.544 mmb. 372.544 mm d. 382.544 mmDetermine the nominal balanced moment.a. 677.326 kN.m. c. 777.321 kN.m.b. 312.852 kN.m d. 563.285 kN.m. Please answer asap. Thanks•Timber beamSafe working stress = 25 MPaCross-section dimensions:height = 119 mm,width = 66 mmProving ring number 293. Steel beam Safe working stress = 200 MPa Cross-section dimensions:external height = 149 mm,external width = 50 mm,wall thickness = 3.0 mm,Proving ring number 1522.1)Calculate safe max bending moment in the central part of the beam using Mmax = (I)(max stress)/(D/2)2) calculate Safe applied force in the proving ring or tie rod, assuming a symmetrical setup as in the photowith a ‘lever arm’ of 600 mm
- Pls solve with FBD’s it’s imperative. Handwritten answers only. The building section associated with the floor plan in Figure P2.3 is shown in Figure P2.7. Assume a live load of 60 lb/ft on all three floors. Calculate the axial forces produced by the live load in column C2 in the third and first stories. Consider any live load reduction if permitted by the ASCE standard.A column in the ground floor of a 5-story RC commercial building has an unfactored dead load of 401 kN and an unfactored live load of 334 kN. The loads act at an eccentricity of 75 mm at the top and 50 mm at the bottom. Design a pin-ended tied column, nonsway, if its clear height is 6 meters. Use f’c = 28 MPa, fy = 420 MPa. Assume an equal distribution of longitudinal bars in two opposite faces of the column. NOTE:!! Attach the reference interaction diagram used.A wide flange steel beam shown in Figure P2.3 supports a permanent concrete masonry wall, floor slab, architectural finishes, mechanical and electrical systems. Determine the uniform dead load in kips per linear foot acting on the beam. The wall is 9.5-ft high, non-load bearing and later- ally braced at the top to upper floor framing (not shown). The wall consists of 8-in. lightweight reinforced concrete masonry units with an average weight of 90 psf. The com- posite concrete floor slab construction spans over simply supported steel beams, with a tributary width of 10 ft, and weighs 50 psf. The estimated uniform dead load for structural steel framing, fireproofing, architectural features, floor finish, and ceiling tiles equals 24 psf, and for mechanical duct- ing, piping, and electrical systems equals 6 psf.
- Hello, can someone help me with this pls. Based on the following figure, determine the vertical reaction at the hinge-type support for the loads shown on the beam. Assume that the dimensions are in meters. For the exercise use a value of x equal to 2.7 m, a value for the distributed load w1 equal to 17 kN/m and a value for the distributed load w2 equal to 28 kN/m.Draw the FBD All distributed loads should be simplified to their equivalent loads Draw the FBD where the system is the child and the chairA 5.0 simply supported beam is supposed to carry a 2m width concrete slab on its center(unit weight of concreteis 24 kN/m3). The slab thickness is 150mm. On top of slab is a super imposed live load pressure of 14000Pa and supper imposed dead load pressure of 24000Pa. (Use NSCP 2015 Provisions) A.Determine the critical factored moment that the beam will carry. B.If critical factored moment Mu = 380 kN.m, determine the additional unfactored concentrated dead load at midspan of beam. C.If critical factored moment Mu = 400 kN.m, determine the additional service live load pressure on the slab.