Concrete Reinforcement Width of section Reinforcement strengths. fcu:= 30-MPa fy:= 450-MPa b:= 280 mm 2 d:= 510-mm A, 2410-mm´
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- 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' = 25MPa 800 IDE 730 N12 ligs (fsv.f-500MPa) Ast = 3N28 350 Span = 10m Figure 3 Calculate the combination load for the reinforcement concrete beam shown in Figure 3. Unit: KN/m with two decimal. 1.2G+1.5Q=wAnswer the following for the section at Point D Only Calculate the distributed load "w" that: Will cause the section crack Will cause the reinforcement to yield. Material Properties: F'c = 5000 psi Fy = 60000 psi Es = 29000000 psi Ln = 27 ft L wl₂² 16 wl,² 14 CD L wl,2 vl₁² 10 11 win² 16 h: 28 in A=4 in² b=14 in n d: 25 inFind the ultimate moment of resistance for the rectangular section reinforced as shown below. material strengths: Concrete Reinforcement Width of section Reinforcement fcu= 30-MPa fy:= 450-MPa b:= 280 mm d:= 510 mm d':= 50-mm 2 A, 2410-mm A's:= 628-mm 2 b A', Hi As
- The details of the cantilever beam are shown below. Use f’c = 28 MPa, fy = 415 MPa, and fyt = 275 MPa. Notes:• All dimensions are in millimeters.• Column reinforcement not shown for clarity• Column and beam are neither exposed to weather nor contact with the ground.• Analyze as singly reinforce beam only.• Neglect the weight of the beam.• Concrete is normal weight and reinforcements are galvanized. What is the flexural capacity of the beam at face of support in kNm? What is the shear capacity of the beam at face of support in kN? What is the maximum Pu (kN) that can be applied at the free end of the beam considering flexure and shear? What is the required minimum cut length (mm) of each stirrup? Round your answer to nearest safe25 mmProblem 1. The composite beam shown below carries a cantilevered load of 10 kN. The beam consists of one 30 x 124 mm plate and four 12 x 50 mm plates. They are pinned together at 120 mm intervals with round pins. The pin material has a shear strength of 159 MPa. Compute the minimum acceptable diameter for the pins. O O O O O O O -0 O 0- O 1000 mm Do O -120 mm (typ) O O O P = 10 KN 30 x 124 mm 12 x 50 mm (typ)Question 2A composite bar is made up of a steel strip 2 mm thick and 12 mm wide, sandwichedbetween two aluminium strips each 3 mm thick and 12 mm wide. The three strips areheld together by two rivets of 5 mm diameter each.2.1 Make a neat detailed sketch of the assembly.2.2 If the three strips are fastened together at 15 0C, determine the shear stress set up inthe rivets when the temperature is raised to 21 0C.Note: The resistance of a rivet in double shear may be taken as 1.75 times the resistancein single shear.(For steel: E = 200 GPa and α = 11×10 -6/ 0C)(For aluminium: E = 70 GPa and α = 22×10-6/0C)
- The Composite section &hown, is subjectuol to a sagging bevn diny mom ent of Loo KNom, Determine the maximum tensite oind Compressive Dlress in each material. Es / Et = 20 Tio 220 ns: (Ieg.)tim = i, 665.2 xlo mm 1. >Hm S.SIMPA, (6c)Hirm. = 4.4M Pn, (ot)st.=134.1Mpa,ldc)st,=16596 M pa. 6kel I20 %3D %3D rio| 130A simply supported beam is reinforced with 4-ø28 mm at the bottom and 2-ø20 mm at the top of the beam, Concrete covering to centroid of reinforcement is 70 mm at the top and 64 mm at the bottom of the beam. The beam has a gross depth of 450 mm and gross width of 300 mm. fc'=28 MPa, fy=415 MPa. Assume bars laid out in single layer. Calculate the following if the limitin tensile steel strains is 0.004 for a ductile failure: Depth of the neutral axis from the extreme concrete compression fiber to the nearest whole number = _____________mm Design strength of the beam section to the nearest whole number =____________ kN-m Maximum service uniform live load over the entire span in addition to a DL = 20 kN/m (including the weight of the beam) if it has a span of 6 m = _____________ kN/m (to the nearest whole number)06. The composite bar, firmly attached to unyielding supports, is initially stress-free. What maximum axial load P can be applied if the allowable stresses are 10 ksi for aluminum and 18 ksi for steel? Steel Aluminum A = 1.25 in.? E = 10 x 106 psi A = 2.0 in.2 E = 29 x 106 psi SUMMARY OF ANSWERS P kips -15 in: -12 in.- CS Scanned with CamScanner
- c) Find the stresses in the concrete and the reinforcement for the following applied moments: (i) – M1= 35kN.m and (ii) – 120kN.m Cross-section: width height Modulus of elasticity: concrete reinforcement Reinforcement: depth area b:= 300-mm h := 520-mm Ec:= 30-GPa Es:= 200-GPa d:= 460-mm As 1470-mm 2 h d bFill in the blanks: Determine the required tension steel area of the T beam with given properties below. Width of flange bf= 500 mm Width of web bw = 400mm Thickness of flange tf 140 mm Effective depth d = 380 mm Effective concrete covering d' = 75 mm Compressive strength of concrete fc' = 30 MPa Yield stress of steel bar fy = 300 MPa Mu = 448 kN-m As = mm2 CS Scanned with CamScanner2. A column HP 14 x 102 of A572 Gr. 55 steel has a length of 15 ft is fixed at both ends. Compute the design compressive strength for LRFD and the allowable compressive strength for ASD. (Steel section properties are provided in the next page) ASTM Designation A572 Gr. 42 Gr. 50 Gr. 55 Gr. 60⁰ Gr. 65⁰ Yield Stress (ksi) 42 50 55 60 65 Fu Tensile Stressa (ksi) 60 65 70 75 80