Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 3, Problem 3.33PFS
To determine
The LRFD design strength and the ASD allowable strength of the given section,
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Determine the lengths of the side fillet welds required at the heel and toe of the angle shown.
Allowable shearing stress through the throat of each weld is 124 MPa.
L 150 x 150 x 13
L1
P1
108
T= 250 kN
42
13
150 mm
P2+
L2
gusset plate
150
A WT6x15 shape made from A36 steel will be welded along its flanges to a gusset plate.
Determine the LRFD available strength, φRn, for gross tension yield of the WT shape.
Determine the required weld length, l, on each side of the WT in order to exceed the LRFD available strength for gross tension yield of the WT shape. Assume a weld size of w=5/16 inch and E70XX electrodes.
Based on the weld length, compute the available LRFD strength, φRn, for tensile rupture of the WT shape.
The welded girder shown in the accompanying illustration has an external shear VD = 300 k and VL = 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.
Chapter 3 Solutions
Structural Steel Design (6th Edition)
Ch. 3 - Prob. 3.1PFSCh. 3 - Prob. 3.2PFSCh. 3 - Prob. 3.3PFSCh. 3 - Prob. 3.4PFSCh. 3 - Prob. 3.5PFSCh. 3 - Prob. 3.6PFSCh. 3 - Prob. 3.7PFSCh. 3 - Prob. 3.8PFSCh. 3 - Prob. 3.9PFSCh. 3 - Prob. 3.10PFS
Ch. 3 - Prob. 3.11PFSCh. 3 - Prob. 3.12PFSCh. 3 - Prob. 3.13PFSCh. 3 - Prob. 3.14PFSCh. 3 - Prob. 3.15PFSCh. 3 - Prob. 3.16PFSCh. 3 - Prob. 3.17PFSCh. 3 - Prob. 3.18PFSCh. 3 - Prob. 3.19PFSCh. 3 - Prob. 3.20PFSCh. 3 - Prob. 3.21PFSCh. 3 - Prob. 3.22PFSCh. 3 - Prob. 3.23PFSCh. 3 - Prob. 3.24PFSCh. 3 - Prob. 3.25PFSCh. 3 - Prob. 3.26PFSCh. 3 - Prob. 3.27PFSCh. 3 - Prob. 3.28PFSCh. 3 - Prob. 3.29PFSCh. 3 - Prob. 3.30PFSCh. 3 - Prob. 3.31PFSCh. 3 - Prob. 3.32PFSCh. 3 - Prob. 3.33PFSCh. 3 - Prob. 3.34PFSCh. 3 - Determine the LRFD design strength and the ASD...Ch. 3 - Prob. 3.36PFSCh. 3 - Prob. 3.37PFS
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- Determine the load-carrying capacity of the 2 PL 5/16 x 3 tension member shown based on the limit states of weld/base metal (tension member only) capacity. The weld is 3/16 inch E70 and the plate is grade A36. PL ½x 6 2 PL 16 x 3 3"> The capacity of the two plates (2 PL 5/16 x 3) tension member is closet to: DELLarrow_forward9) what is the weld capacity of the shear connection in kips? (2 decimal places) 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=70ksiarrow_forwardDetermine the LRFD design strength and the ASD allowable strength of sections given. Neglect block shear.arrow_forward
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- B1 USING NEW NSCP CODES, PLEASE ANSWERarrow_forwardProblem 3. An un-equal leg angle L8x4x2 made of ASTM A242 steel (Fy = 50 ksi, Fu = 70 ksi) is used as a tension member and connected to a gusset plate as shown in Figure 2 (a to d). ASABINCAT SALLALALALALALALALAL w₁ Weld * Gusset Rate 3 (c) Longitudinal and transverse welds O 18x4x2 Gusset Plate Steel Type: A242 18x4x2 bolts: 3/4 dia. Steel Type: A242 T (d) Bolted connection (d) Calculate factored fracture strength of the tension member in Figure 2(c). (e) Calculate factored fracture strength of the tension member in Figure 2(d). (f) What is the maximum permissible length for this tension member?arrow_forwardCompute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate using welds 12 inches long, as shown in the figure. The service live load is three times the dead load. Use A992 steel. USE LRFD ONLY, no block shear will occur. WT12 x 38 Longitudinal welds 11.2 in? y = 3.0 in. Given: Properties of WT12 × 38: Ag = Use A992 Steel: F, = 50 ksi Fu = 65 ksi bf = 8.99in. %3D %3D LL = 3 DL %3D %3D tw y = centroidal distance bf C. What is the Governing Ultimate Tensile Capacity based on Net Fracture Round your answer to 3 decimal places.arrow_forward
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