If the service dead load of the simply supported beam including its weight is 1.5 k/ft, consider L (span of the beam) to be 24 feet. Find the maximum service live load that the shown beam can carry if fy= 60 ksi, fc'= 4 ksi, As= 4#8 and A's 2#6. Check if the section is acceptable by the ACI. Does the compression steel yield at the ultimate conditions? 21.5" O O O O 14" * I 2.5"
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- USE NSCP 2010 A simply supported beam is reinforced with 4 – 28 mm ø at the bottomand 2 – 28 mm ø at the top of the beam. Steel covering to centroid ofreinforcement is 70 mm at the top and bottom of the beam. The beamhas a total depth of 400 mm and a widthof 300 mm. fc’ = 30 MPa, fy = 415 MPa. Balanced steel ratio ρb = 0.031.Compute the ultimate moment capacity of the beam in kN-m. Usereduction factor of 0.90A beam has a width of 300 mm and an effective depth of 480 mm strength Pc= 34 MPa. Steel yield strength fy=415 MPa. The section is to be reinforced for tension only and the effective concrete cover is 70 mm. Deadload moment is 140 kN-m and the liveload moment is 180 kN-m. Use 2010 NSCP with t= 0.005. A. What is the governing steel ratio? B. What is the maximum number of bars needed for a 25mm diameter reinforcements? C. What is the minimum number of bars needed for a 25mm diameter reinforcements?STRUCTURAL STEEL DESIGN (SHOW FULL SOLUTION) COLUMNS A W 410 x 74 of A36 steel with Fy = 248 MPa has a length of 7.2m is used as a column with both ends fixed. 1. Compute the allowable compressive stresss. 2. Compute the allowable compressive load that the column could carry. 3. Compute the Euler’s critical buckling load.
- The compound bar carries axial loads as shown. Determine the maximum allowable value of P if the change in length of the entire bar is limited to 0.08 inch and the working stresses listed in the table are not to be exceeded. Steel 2 ft Bronze 4 ft Aluminum 3 ft 2P 3P 4P A (in.?) E (psi) a. (psi) Steel 0.75 30 x 10 20000 12 x 10 10 x 10 Bronze 1.00 18000 Aluminum 0.50 12000in 21- ő A steel beam is reinforced by three steel rods. The modulous of elasticity for concrete is 20Gpa and for steel is 200Gpa. Maximum allowable stress for concrete is 9.45Mpa and for steel is 140Mpa. Calculate the largest allowable positive WBT bending moment in the beam. иа mo maod motinu od o boblow need sved Ha bus 400mm aidT (alonnerlo ert to rigiow stongi)/4000E of Trup 200-0 signs or botuesom sWnworla es berloss M bris V zsra erli owob stiW maod sdi to mergeib insm brol bus borlism tuo sallimill nollor 10 nige over on The modulus of elasticity for concrete is 20Gpa and for the steel is 200 Gpa. Allowable stress for concrete is 9Mpa and for steel is 140Mpa. Calculate the maxium allowable bending momemt that can be applied per metal rod of the concrete slab. 150mm 200mm D=16mm 140 mm -22mm Diameter TOPABA t 50mm T 140 mm 140 mm 140 mmQ2: Verify that the steel section W12 x 106 can be used for the cantilever beam shown in the Figure (2). Check for flexural strength and deflection only. The beam has lateral support at the ends. The service uniform distributed load includes, wp = 6 k/ft dead load and w₁ = 3 k/ft live load. Use steel A992 and C₁ = 1. A -10 ft- B Fig. (2)
- Determine the safe load of the column section shown, if it has a yield strength of 25 MPa. E = 200000 MPa. Use NSCP Specifications. Fyz248 mpa Properties of Channel Section d = 305 mm t₂ = 7.2 mm A = 3929 mm² t₁ = 12.7 mm Ix=53.7 x 10mm¹ x = 117 mm Properties of W 460 x 74 A = 9450 mm² b = 190 mm ly= 1.61 x 10 mm x = 17.7 mm tw = 9.0 mm rx = 188 mm ry = 41.9 mm d = 457 mm tr = 14.5 mm Ix = 333 x 10 mm Iy = 16.6 x 10mm* 7.21 When the height of column is 6 m. When the height of column is 10 m. Assume K= 1.0 457 CIVIL ENGINEERING- STEEL DESIGNQuestion 8 1 p The rectangular doubly reinforcement stress concrete block with the arrangement of reinforcement of 2N28 bars on top and 3N28 bars on bottom. The modulus of elasticity are Ec=23,500MPa and Es = 200,000MPa. Dead load is 18KN/m and live load is 12KN/m. f'c = 25MPa 800 TE 730 N12 ligs (fsv.t-500MPa) Ast = 3N28 350 Span = 10m Figure 3 Calculate the concrete stresses in tension and compression of the reinforcement concrete beam shown in Figure 3. Compression concrete stress Choose ] Tension Concrete stress Choose]3. A rectangular beam has the following properties: Width, b = 400 mm Effective depth, d 620 mm Tension bars, 10 pcs 28-mm-diameter Compression bars, 3 pcs 25-mm-diameter d' = 70 mm fy = 415 MPa f'c = 22 MPa Determine the design strength of the beam and the safe service live load if the service dead load is 320 kN-m.
- QUESTION: A W10x112 column rest on a base plate and on a 400 mm x 400 mm concrete pedestal. The pedestal andbase plate are of the same dimensions. Yield strength of steel Fy = 248MPa, effective length of column Le=W/4 m. Concrete strength fc = W MPa. Use ASD. (Hint Le = kL) a.What is the allowable axial capacity of the column in kN?b.What is the allowable bearing capacity of the pedestal in kN?C.What is the minimum required thickness of the base plate rounded off to nearest safe 5 mm? Reference for datas W= 25 mm , 25 MPaW/2 = 12.5 mW/4 =6.25mX=475 KNY= 155 KN, 155 KPa, 175 mmZ=60×+Z=535mmSituation #2: A rectangular beam has the following properties: Width, b = 400 mm 415 MPa Effective depth, d = 620 mm 22 MPa Tension bars, 10 pcs 28-mm-diameter Compression bars, 3 pcs 25-mm-diamter d' = 70 mm fy= = fc' = Determine the design strength of the beam and the safe live load if the service dead load is 320 kN.m1. A steel column 10 m long is fabricated from a W and C section arranged as shown. Determine the safe compressive load. Fy = 248 MPa, E = 200 GPa. Use AISC/NSCP Specs. C 310 x 45 y1=x=17 y2 d1 10 m W 200 x 46 a) Both ends of column are fixed b) Both ends of column are hinged c) One end fixed, the other end hinged Use design values of k. C 310 x 45 A = 5690 mm? d = 305 mm bf = 80 mm tf = 12.7 mm tw = 13 mm Ix = 67.3x10° mm ly = 2.12x10° mm x = 17 mm W 200 x 46 A = 5860 mm? d = 203 mm bf = 203 mm tf = 11 mm tw = 7.2 mm Ix = 45.5x10° mm ly = 15.3x10° mm