Determine the LRFD design strength and the ASD allowable strength sections given. Neglect block shear.
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Determine the LRFD design strength and the ASD allowable strength sections given. Neglect block shear.
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- WT12 x 38 Longitudinal welds Given: Properties of WT12 x 38: A, = 11.2 in? y = 3.0 in. b = 8.99in. Use A992 Steel: F, = 50 ksi F = 65 ksi LL = 3 DL %3D %3D tw = centroidal distance bf A. Governing Ultimate Tensile Capacity based on Yielding of gross section Round your answer to 0 decimal places.Design the size and length of Fillet weld for the lap joint shown below, Use SMAW E70XX process, plates are A-36 steel? 90k LL 40k DL R-X7 5" Gusset P 8 90k LL 40k DLAn ISA 90 x 60 × 8 mm of steel of grade Fe410 is connected to a 12 mm gusset plate and is used as a tension member. Which of the following options are correct for length of welds 4 and l, to be provided on upper and lower edges of connected leg only as shown in figure? Assume size of weld as 6 mm. = 14.8 mm 60 mm For ISA 90 x 60 x 8 mm Ag = 1137 mm2, Czz = 29.6 mm, Cyy = 14.8 mm (Assume shop weld) for force P = 173.42 kN, length 1, is provided as for force P2 = 84.99 kN, length 12 is provided as -uw 06 - C. = 29.6 mm - wW 06 -
- Topic:Welded Connection - Civil Engineering -Steel Design *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem A tension member consists of a double angle section with long legs back to back. The angles are attached to a 9.5 mm thick gusset plate. Fu = 400 MPa Fy = 248 MPa for angular section. Fw = 480 MPa for 8 mm fillet weld. Reduction factor U = 0.80 Prop. of One Angle L 125m x 75m x 12.7 m A= 2419 mm2 y=44.45 mm Questions: a) Compute the design strength capacity of one angle. b) Compute the base metal shear strength (gusset plate) per unit length. c) Compute the length L1 and L2.Problem # 6.0 For the tension member shown, compute the tensile design strength. Weld -6" PL 4X6 A36 steel20. 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=70ksit) what is the weld capacity perpedicular to the beam of the flange connection in kips? (2 decimal places)
- Considering the following steel connection. The plates in Pink are 9mm steel plates. The middle plate (Yellow) is 18mm thick. The width of the plate is 100mm. The maximum allowable tension stresses on any of the plates is 100Mpa in Gross Area Yielding and 150 Mpa for Net Area or Tension Rupture. The bolts used are 8mm in diameter, the holes are 10mm in diameter, no need to add 1.6mm. The bolts allow a maximum of 280 Mpa of shear. Determine the maximum allowable "P" of the connection in kN.The connection shows a PL10X200 loaded in tension and welded to a gusset plate 10mm thick. Calculate the following, assuming A992 steel is used. Use U =0.85. The weld thickness is ómm and the electrode is E70 (485MPA). PL 10 X 200 350mm What is the allowable weld connection strength (kN)?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=70ksi i) what is the weld capacity of the shear connection in kips? (2 decimal places)
- The welded girder shown in the accompanying illustration has an external shear V₁ = 300 k and V₁ = 350 kat a particular - section. Determine the fillet weld size required to fasten the plates to the web if the SMAW process is used. E70. Use LRFD and ASD. PL1 x 16 I Web x 48 PL1 x 16Question 6 Two plates of Steel A72 Gr 50 are connected with each as shown in Figure 5. E70XX Electrodes were used and 7/16 in fillet weld were made by SMAW process. What is the design strength of the weld by LRFD and ASD? E10 in - 10in P, or P. - PLx 10 Figure # 4Problem 2: What is the U value for these cases? 1. Plate connected using bolts (fasteners) only. 2. Plate using transverse welds only 3. 3/4 in thick plate using longitudinal welds only where the length of the weld (L=10 in) and L = 1.25 W where W is the width of the plate. 4. Single angle where both legs are bolted to the support. 5. Single angle L8x8x1 with longitudinal welds (L = 9 in) on one leg 6. Channel C10x20 connected using one bolt only in the web 7. Single angle connected on one leg using transverse and longitudinal welds.