A short RCC circular column having gross diameter of 450 mm is provided with 6 nos. 20 mm diameter, Fe500 steel bars with effective cover of 50 mm. M25 concrete is used and column is provided with spiral ties of 6 mm diameter. The minimum volume of spiral reinforcement per unit length of the column is_ cm³ (round off to two decimal places).
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- Question Eight tensile bars with a diameter of 16mm, equally divided into Fwo layers, are developed, C=50mm Spacing between Stirrups = 150 mm center to Center and spacing be tween longitudinal bars is 80 mm if Atr=230 mm? [leastof Cs& cb], Fy = Fyt = 350 M Pa. use Fé = 25,35 and 45 MPa and assume Atr is Satis fied with ACI & BS, Find length of development vequired. > Subject: Advanced Concvete Design, %3DA 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0. Determine the design tensile strength of the section based on yielding of the gross area. Answer: |0 O O О оAnswer 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 in
- Question j=169" R=2.1" ;f 4- 2.81 d $ 100 08 96 07 h £=0.5" * # >P d=7.3" Material is ASTM A36. Plate is fixed 511 by 4 screws of {"$. using the information provided, calculate the ultimate stress (factored by LRFD) for shear fracture and tensile rupture of block pattern passing through points the shear 3-10-9-4..SITUATION II: Calculate the permissible tensile load P 350 under the following conditions: Fy-248 MPa; Fu=600 MPa. The diameter of the bolts is 16 mm. 716 100 50 50 100 100 50 50 Allowable stresses: Tensile stresses on gross area: 0.6Fy Shear stress of the bolts: Fv=130MPA Bearing stress of the bolts: Fp=1.2Fu 7. Based on shear capacity of bolts. a. 156.83 kN C. 475.21 kN b. 1105.92 kN d. 450.78 k 8. Based on the bearing capacity of bolts. a. 156.83 kN C. 475.21 kN b. 1105.92 kN d. 450.78 kN 9. Based on the yielding of plates. a. 156.83 kN C. 475.21 kN b. 1105.92 kN d. 450.78 kNNeed complete and elaborate solution A.Using ASD, check the adequacy of the base plate (PL400x350x19 mm) centered on a W12X152 column as shown. It is resting on a concrete pedestal that is 450 mm x 400 mm. The loads are PDL = 600kN and PLL = 320 kN. Use A-36 steel and concrete fc’=20 MPa. B. Using LRFD, PL400x350 mm base plate, W12x120 column, 500x500 mm pedestal, PDL = 1200 kN and PLL = 800 kN. Use A-50 steel and concrete fc’ = 30 MPa. Determine the required base plate thickness.
- Determine the capacity of the details shown below.A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksio) what is the block shear rupture (U-shaped) capacity of flange plate of the flange connection in kips? (2 decimal places)Q2. The joint connection shown in Figure Q2 is subjected to a medium term tensile load under service class C24. The joint comprises of six 3mm diameter x 150mm long wire nails. If the characteristic density of the timber is 340kg/m³, determine the load carrying capacity P1 and P2 of the connection and slip per nail P2 P 45 mm thick Figure Q2 75 mm thick P1 P1 P 100 mm thickA W1422 acts compositely with a 4-inch-thick floor slab whose effective width b is 90 inches. The beams are spaced at 7 feet 6 inches, and the span length is 30 feet. The superimposed loads are as follows: construction load = 20 psf, partition load = 10 psf, weight of ceiling and light fixtures = 5 psf, and live load = 60 psf, A992 steel is used, and fc=4 ksi. Determine whether the flexural strength is adequate. a. Use LRFD. b. Use ASD.
- SITUATION 14: A PL 300 x 20 mm is to be connected to two plates of the same material with half the thickness by 25 mm Ø rivets as shown in the figure. The rivet holes have a diameter 2 mm larger than the rivet diameter. The plate is A36 steel with Fy= 250 MPa, allowable tensile stress of 0.60Fy and allowable bearing stress of 1.35Fy. The rivets are A502, Grade 2, hotdriven rivets with allowable shear stress of 150 MPa. 25 mm Ø rivets + + PL 300x20 P/2 P P/2 PL 300x10 43. Which of the following most nearly gives the max. load in kN that can be applied to the connection without exceeding the allowable shear stress in the rivets? a. 675 KN b. 490 KN c. 598 KN d. 790.5 kN 44. Which of the following most nearly gives the max. load in kN that can be applied to the connection without exceeding the allowable bearing stress between the plates and the rivets? a. 490 KN b. 675 KN c. 598 KN d. 790.5 kN 45. Which of the following most nearly gives the max. load in kN that can be applied to the…cd efgh ij O klmn u op V 0 0 0 t Р What is the line for which the section will rupture in tension? All holes are the same diameter. m n j h dW 10x45 is connected to two plates with two lines of ¾” bolts as shown below. Determine the tensile strength of the system if Fy = 50 ksi and Fu = 65 ksi.