Homework 10-2: Compute the net area of each of the given members. Use standard-size bolt holes for all problems. PL 4 X 10 3/4 in bolts (Typical) O PL 3/4X10 Bolts are =中
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- The bolted connection shown in Figures is connected with 3/4-in-diameter bolts in standard holes. The plate material is A36 steel. By using strength level design (LRFD), find the available tensile strength of the plates. u=0.8 (5)-¾" bolts 4" C12× 20.7 | 4½" ,3" , 4½" |Two plates each with thickness t = 10 mm are bolted together with 6 - 16 mm diameter bolts forming a lap connection. Plate (A36): Fy = 248 Mpa FU = 400 MPa bolts are A325 in standard holes with threads excluded from shear planes. Considering Tension Yielding, compute the the design strength for LFRD. Use NSCP 2015 Round your answer to 2 decimal places.3. Determine Rn based on bolt limit states. Slip critical connection is used for design and Grade 10.9 M22 Bolts with Class A surface. Check minimum edge distance and spacing with TSSDC 2016. All members are $355. There is no filler between plates and Du=1.0. Fy= 355 MPa; Fu=510 MPa; E=200000 MPa. All units are in mm. $R.=? 100 100 O 60 Geset Plate 75 Plate Gusset Plate $R.=? *20*201 Plate
- Question No.3: Determine the maximum tensile force, Pu, that can be applied to the connection shown in figure below based on the bolt strength; assume ASTM A36 steel for the following bolt types in standard holes: Check every limit state. a. 1-in.-diameter A325N, b. 7/8-in.-diameter A490X. 3" 3" 1½" /" plate ½P, P ½P. /" plateFind the available strength of the S-shape shown in the figure. The holes are for 3/4-inch- diameter bolts. Use A36 steel. Use LRFD Use ASD (1 (2) 3½" I S15 × 50 $23/4" CIVIL ENGINEERING- STEEL DESIGN 1½" 12" e b С d Answer VUZMGLThe butt connection shows 8-22 mm diameter bolts spaced as shown below. P- 50 100 50 50 100 50 16 mm +HHHH 40 80 40 12 mm Steel strength and stresses are: Yield strength, Fy = 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Based on the gross area of the plate. Based on the net area of the plate. Based on block shear strength. Bolt hole diameter = 25 mm Calculate the allowable tensile load, P, under the following conditions:
- The butt connection shows 8-22 mm diameter bolts spaced as shown below.Use NSCP2001 Steel strength and stresses are: Yield strength, Fy 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Bolt hole diameter = 25 mm Calculate the allowable tensile load, P, under the following conditions: 1Based on the gross area of the plate. 2Based on the net area of the plate. 3Based on block shear strength. 50 100 50 50 100 50 40 80 40 12 mm 16 mmFor the cross section shown, calculate the tensile capacity of the steel plate. Assume that the bolt hole is 1/16 inch larger than bolt diameter. Plate 1/2" x 11" .T 7/8" dia. bolt, typ. ASTM A325NQUESTION 5 3. check the capacity of the shear connection. A992: Fy=50ksi, Fu=65ksi A36: Fy=36ksi, Fu=58ksi effective bolt hole diameter = bolt hole + 1/16" beam properties: tw=0.295 in. Fexx=70ksi what is the bolts shear capacity in kips? (2 decimal places) 0.5"| 1.5" 1.5" 5/16 _ 4-3/40 A325 SSLT Bolts 5/16
- Determine the maximum factored LRFD tensile force capacity in tension only (no block shear). The angle is ASTM A36 steel. X = 0.7 Y = 3.8 Z=1/2 Round your answer to 2 decimal places. Your Answer: Incorrect The answer is 77.15 ± 1%. 1/2" X" bolts 1½" Gusset plate -L3 X 3 X Z Tu IA bolted connection shown is made up of 2L150 x 100 x 9.5 mm. The angle is made of A-36 steel and the bolts have diameter of 22 mm. Properties of 2L150 x 100 x 9.5 mm: A = 4660 mm2 t = 9.5 mm Fy = 248 MPa Fu = 400 MPa Properties of 20 mm A-325 bolts: Fu = 400 MPa 50 50 50 50 50 50,50 힝힝이어 Gusset Plate 9.5mm 100mm Gusset Plate 100mm 31.25 62.5 T 56.25 a. Nominal dimension of hole diameter, dn = 25 mm b. Determine the ultimate shear capacity (kN) of the connection. c. Determine the ultimate tearing capacity (kN) of the connection if the shear lag factor, U = 0.75.Find øR, considering bolt limit states only. (S275 Steel, fy = 2804 kgf/cm2, fu = 4385 kgf/cm2, Bearing type connection 8.8 M20 bolts, ti = 15 mm, t2 = 18 mm) PE >P 50 mm 80 mm 80 mm 50 mm
