For the Integrated Project (office building), design the Floor Girders in the gravity-only system, assuming that the top flange is fully restrained with the floor diaphragm. 1. Use Table 3-2 from the Steel Manual to select the most economical compact W-shape section. Demands: Stories 1 and 2 Edge: 114.4 k.ft From Table 3-2, use W12x50 D/C should be less than 0.95: D/C = Stories 1 and 2 Interior: 226.4 k.ft Use W12x50: D/C = Roof Edge: 33.3 k.ft Use W12x40: D/C = Roof Interior: 58.4 k.ft Use W12x40: D/C =

Which of the following best gives the purlin load wuy? See loaded purlin. The figure shown has trusses 6m apart with midpoint sag rods and Purlins C150x12.5 spaced 2.5m on center. The roof truss is inclined 1 vertical to 2 horizontal. The weight of roofing materials is 0.9 kPa (includes self wt.), wind load is 0.8 kPa perpendicular to the roof surface, minimum roof live load is 0.6 kPa. Consider the roofing to provide full lateral support of the purlins. Fy = 345 mPa. Use the NSCP 2015 specifications. 2. 5m 2.5m 2. 5m roofing TRUSS TOP CHORD Wuy, Truss Spacing = 6m Wux 1 Purlins C150x12.5 Purlin Load Wt kN/m mm^2 mm C150x12.5 0.12 11400 12.7 152 5.1 d tw by SECTION Ix Sx rx ly sy Zy ry Zx mm mm mm^4 mm^3 mm mm^4 mm^3 mm mm^3 mm^3 mm mm 48 8.7 143 1100 112 48.4 378 65.2 1230 574 O 4.35 kN/m 5.56 kN/m O 25.02 kN.m

Prepare the Bar bending schedule of a simply supported R.C.C. Lintels from the following specification: Size of lintel 300mm widex 200mm depth. Main bars in tension zone of Fe 250(grade I) 3 bars of 16mm dia., one bar is cranked through 45 degree at 170 mm from each end 2 No. anchor bars at top 8mm dia. Two legged stirrups@150mm c/c of 6mm dia. through out. Clear span of the lintel is 1150mm. Bearing on either side is 150mm

Which of the following best gives the purlin load wuy? See loaded purlin. The figure shown has trusses 6m apart with midpoint sag rods and Purlins C150x12.5 spaced 2.5m on center. The roof truss is inclined 1 vertical to 2 horizontal. The weight of roofing materials is 0.9 kPa (includes self wt.), wind load is 0.8 kPa perpendicular to the roof surface, minimum roof live load is 0.6 kPa. Consider the roofing to provide full lateral support of the purlins. F, = 345 mPa. Use the NSCP 2015 specifications. 2. 5m 2.5m 2. 5m roofing Wuy TRUSS TOP CHORD Truss Spacing = 6m Wux Purlins C150x12.5 Purlin Load Wt A kN/m mm^2 mm C150x12.5 0.12 11400 12.7 152 5.1 SECTION dtw by Ix Sx rx ly mm mm^4 mm^3 mm mm^4 mm^3 mm mm^3 mm^3 48 8.7 143 1100 112 48.4 378 65.2 1230 574 Sy ry Zx| Zy mm mm mm O 4.35 kN/m 5.56 kN/m O 25.02 kN.m

A plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.

If the beam in Problem 5.5-9 i5 braced at A, B, and C, compute for the unbr Cb aced length AC (same as Cb for unbraced length CB). Do not include the beam weight in the loading. a. Use the unfactored service loads. b. Use factored loads.

A beam must be designed to the following specifications: Span length = 35 ft Beam spacing = 10 ft 2-in. deck with 3 in. of lightweight concrete fill (wc=115 pcf) for a total depth of t=5 in. Total weight of deck and slab = 51 psf Construction load = 20 psf Partition load = 20 psf Miscellaneous dead load = 10 psf Live load = 80 psf Fy=50 ksi, fc=4 ksi Assume continuous lateral support and use LRFD. a. Design a noncomposite beam. Compute the total deflection (there is no limit to be checked). b. Design a composite beam and specify the size and number of stud anchors required. Assume one stud at each beam location. Compute the maximum total deflection as follows: 1. Use the transformed section. 2. Use the lower-bound moment of inertia.

A column section b- 400 mem, h - 600 mm is reinforced with longitudinal bars Ast - 12-020 mm equally distributed along all faces. Lateral ties are composed of 12 me rectangular hoops with clear cover equal to 40 mm and 12 mm cross ties in both directions. Concrete yield strength f'c - 27 MPa and steel yield strength, fy - 415 MPa. Calculate the nominal axial strength of the column in kN. O 6,985 O 5,588 O 5,938 O 5,239