Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Question
Chapter 3, Problem 3.3.6P
To determine
(a)
The design strength using load and resistance factor design (LRFD) method.
To determine
(b)
The allowable strength using allowable strength design (ASD) method.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Question 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 # 4
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 flat bar used as a tension member is connected to a gusset plate. Size of weld is 6 mm, with an ultimate tensile strength of 185 MPa. E 70 electrodes Thickness of plate = 6.25 mm. Use ASD.
Thickness of gusset plate = 9.5 mm.
Fy = 248 MPa
Fw = 0.60(485)
Fu = 400 MPa
Questions:
The connected parts are of A-36 steel and assumed that the tensile strength of the member is adequate:
a) Determine the strength of the connection per mm considering the design shear strength of the 6 mm fillet weld.
b) Determine the base metal shear strength per mm.
c) Determine the available strength of the connection.
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 DL
Chapter 3 Solutions
Steel Design (Activate Learning with these NEW titles from Engineering!)
Ch. 3 - Prob. 3.2.1PCh. 3 - Prob. 3.2.2PCh. 3 - Prob. 3.2.3PCh. 3 - Prob. 3.2.4PCh. 3 - Prob. 3.2.5PCh. 3 - Prob. 3.2.6PCh. 3 - Prob. 3.3.1PCh. 3 - Prob. 3.3.2PCh. 3 - Prob. 3.3.3PCh. 3 - Prob. 3.3.4P
Ch. 3 - Prob. 3.3.5PCh. 3 - Prob. 3.3.6PCh. 3 - Prob. 3.3.7PCh. 3 - Prob. 3.3.8PCh. 3 - Prob. 3.4.1PCh. 3 - Prob. 3.4.2PCh. 3 - Prob. 3.4.3PCh. 3 - Prob. 3.4.4PCh. 3 - Prob. 3.4.5PCh. 3 - Prob. 3.4.6PCh. 3 - Prob. 3.5.1PCh. 3 - Prob. 3.5.2PCh. 3 - Prob. 3.5.3PCh. 3 - Prob. 3.5.4PCh. 3 - Prob. 3.6.1PCh. 3 - Prob. 3.6.2PCh. 3 - Prob. 3.6.3PCh. 3 - Select an American Standard Channel shape for the...Ch. 3 - Prob. 3.6.5PCh. 3 - Use load and resistance factor design and select a...Ch. 3 - Select a threaded rod to resist a service dead...Ch. 3 - Prob. 3.7.2PCh. 3 - Prob. 3.7.3PCh. 3 - Prob. 3.7.4PCh. 3 - Prob. 3.7.5PCh. 3 - Prob. 3.7.6PCh. 3 - Prob. 3.8.1PCh. 3 - Prob. 3.8.2PCh. 3 - Prob. 3.8.3PCh. 3 - Prob. 3.8.4PCh. 3 - Prob. 3.8.5P
Knowledge Booster
Similar questions
- The double T-beam is fabricated by welding the three plates together as shown.  Part A Determine the shear stress in the weld necessary to support a shear force of V=78 kN�=78 kN.arrow_forwardCan u solve me step by step please . thank uarrow_forwardConsidering 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.arrow_forward
- Please answer all the questions with a complete solution. thank you.arrow_forwardPlease answer the problem attached image.(using Nscp 2015) thank youarrow_forwardTopic: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.arrow_forward
- Please do right answer and calculation.arrow_forwardSituation 5. The angular section shown below is welded to a 12 mm gusset plate. Both materials are A36 steel with Fy = 250 MPa. The allowable tensile stress is 0.6Fy. The weld is E80 Electrode and 12 mm thickness. INNOVATIONS Properties of L 150x90x12: y = 50 shear stress of weld = 0.3Fu A = 2750 Allowable REVIEW INNOVATIE a K ➜ www A. 234 KN B. 349 KN b 13. What is the value of P without exceeding the allowable tensile the angle? C. 382 kN p. 413 kN 14. Find required length of the weld based on shear? A. 280 mm C. 300 mm D. 380 mm B. 320 mm 15. Find the required value of a? A. 108 mm B. 97.9 mm D. 185 mm NEW INNOVATIONS REVIEW INNOVATIOf REVIEW NEW INNOVATIONSarrow_forward7. Use Allowable Strength Design – ASD 8. Use Load and Resistance Factor Design – LRFD Answer everything. Thank you!arrow_forward
- 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 100x100x10mm angle is to be welded to a gusset plate. The angle carries a load of 200 kN applied along its centroidal axis which is 28.7 mm above the short leg as shown in the figure. Use an 8 mm fillet weld with a minimum tensile strength Fu=483.33 MPa. Fv = 0.30 Fu Questions: a) Determine the length of a transverse fillet weld along the edge of the angle in order to avoid eccentricity of loading. b) Determine the length of side fillet weld required at the heel. c) Determine the length of side fillet weld required at the toe.arrow_forwardWT12 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.arrow_forwardFor the L8X6X3/4 angle shown below, determine the design tensile strength considering gross yielding and net fracture. The angle is A36. Welds are longitudinal and transverse.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Steel Design (Activate Learning with these NEW ti...Civil EngineeringISBN:9781337094740Author:Segui, William T.Publisher:Cengage Learning
Steel Design (Activate Learning with these NEW ti...
Civil Engineering
ISBN:9781337094740
Author:Segui, William T.
Publisher:Cengage Learning