A WT8×50 (A992) is attached to a 12x3/4 inch plate through its flange with 7/8 inch Ø bolts at a spacing of 3 inches and placed in 2 rows as shown. Determine the shear lag factor (consider Case 2 and Case 7) and effective net area of the WT. Considering yielding and net section rupture, determine the design strength by LRFD and the allowable strength by ASD. -WT8×50

A WT8×50 (A992) is attached to a 12x3/4 inch plate through its flange with 7/8 inch Ø bolts at a spacing of 3 inches and placed in 2 rows as shown. Determine the shear lag factor (consider Case 2 and Case 7) and effective net area of the WT. Considering yielding and net section rupture, determine the design strength by LRFD and the allowable strength by ASD. -WT8×50

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter5: Beams

Section: Chapter Questions

Problem 5.12.2P

See similar textbooks

Related questions

Q: Compute its design strength based on rupture in kiloNewtons..* 412 485 453 389 438 This is a…

A: Given, Fu = 400 MPa Fy = 248 MPa D = 16 mm Dh = D + 2

Q: 60 b P P 150 100 100 100

Q: For the tension member shown is made of 12 mm x 240 mm plate attached to gusset plate, Fy = 248 MPa,…

A: Tensile strength is defined as the “resistance to lengthwise stress, measured by the greatest load…

Q: 4. An UPE 300 is to be used as a tension member and it is bolted to a 10mm thick gusset plate with…

A: Tension member UPE 300 bolted connection. Fy = 275 N/mm2 Fu = 430 N/mm2 Ag= 5660 mm2 tw= 9.5 mm

Q: Problem 2. Using 22 mm diameter bolt, compute: a. t for Palone b. t for moment alone c. Critical…

A: Tension member connection and splice: It subjects the bolts to powers that will in general shear the…

Q: A 3/4 in (thickness) X 5 in (height) steel plate is connected to another one through two bolts of…

Q: The tension member is fastened togother using two bolts, one on each side of the member as shown.…

Q: Determine the allowable tensile force P which can be appllied to the plates shown. The AISC…

A: Diameter of bolt, Thickness of plate, Width of plate, To find: Allowable tensile force.

Q: he bolt is made of a material havin llowable shearing stress of 55.44 M

A: Given:- Allowable tensile stress=140.44MPa Allowable shearing stress=55.44MPa Axial load=17kN…

Q: A PL40 mm X 250 mm (smaller member) is connected to a gusset plate (bigger member) as shown. The…

Q: 1. A tension member is made of 90 mm thick steel plates 225 mm wide by a lap joint made with three…

A: Given data, Thickness of steel plate (t) = 90mm Width of the plate (B) = 225 mm 3 rows of bolts…

Q: A Channel Cóx13 is used as a tension member. The holes are for 5/8 inch in diameter of bolts. Using…

Q: For the tension member shown, determine the total service load (DL + LL) if LL = 4DL. Use A36 steel…

Q: 2. For the figure shown, calculate the net area of the 10 x 200 mm plate. The plate is connected at…

A: We are given chain pattern of boltingNet area for chain pattern of bolting is given by An = Ag -…

Q: The angle bracket shown, 75 mm x 75 mm x 15 mm thick, is 200 mm long. It is connected to the 16 mm…

Q: Six bolts are used in the connection between the axial member and the support. The ultimate shear…

A: Allowable stress in bolt is given by ; σ = Ultimate shear strength Factor of safety σ = 3453 = 115…

Q: The bolted connection shown must support an applied load of P = 5300 kN. The average shear stress in…

A: The values and diagram given in the question are as follows: Applied load(P)=5300 kN or 5300*103 N…

Q: (a)If T = 200KN, compute the shearing stress on the bolts. (b)If the allowable shearing stress on…

Q: A W14X120 is used as a tension member in atruss. The flanges of the member are connected to a gusset…

Q: design tensile strength

A: Load and resistance factor design (LRFD): - It is one of the design methods for designing any type…

Q: The tension member shown in Figure 3.4-2 is a PL 5/8 x 10, and the steel is A36. The bolts are…

A: Member Steel Bolts are diameter.

Q: repare an excel file and to find a) area of the bolt, b) Average Shear Stress Bolt c) Bearing Area…

A: Hi! Thank you for the question. We are not allowed to share excel or other files but we can help you…

Q: 25 mm. 8000N 4000N 100 mm D A 4000N

Q: 4.002 Three bolts are used in the connection shown in the figure. The thickness of plate (1) is 15…

A: Given : thickness of plate 1 = 15 mm the ultimate shear strength of bolt 350 MPa The ultimate…

Question

3. A WT8X50 (A992) is attached to a 12x3/4 inch plate through its flange with 7/8 inch ø bolts at a
spacing of 3 inches and placed in 2 rows as shown. Determine the shear lag factor (consider
Case 2 and Case 7) and effective net area of the WT. Considering yielding and net section
rupture, determine the design strength by LRFD and the allowable strength by ASD.
-WT8×50
PL 12x/4'

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1 Introduction

VIEW

Step 2 Solution

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

Asap pls
Plate A, PL 16mm x 375mm, is connected to another plate B with 10- 22 mm o bolts as shown. The steel is A36 steel with Fy= 250 MPa and F,= 400 MPa. Assume that plate B has adequate strength. Use ASD. Plate B 375a 75 m 2 75mm Plate A 75 mm 75mm 150 50150 Compute the nominal strength based on yielding in KN.O Determine the allowable tensile strength of Plate A in KN.O Compute the allowable tensile strength at section a-1-2-2-4-c in KN.0
A 10mm x 150 mm plate of A36 steel (Fy = 248 MPa & Fu = 400 MPa) is used as a tension member. It is connected using 20 mm Ø bolts. U=1.0 a. Determine the tensile capacity based on yielding (ASD & LRFD) b. Determine the tensile capacity based on rupture (ASD & LRFD) c. Determine the tensile capacity based on block shear strength (ASD & LRFD) 75mm 37.5 75 75 75 87.5
WideFlange Section: W 16x67
3. A plate with width of 300mm and thickness of 20mm is to be connected to two plates of the same width with half the thickness by 24mm diameter bolts, as shown. The rivet holes have a diameter of 2mm larger than the rivet diameter. The plate is A36 steel with yield strength F,-248MPa and ultimate strength F,-400MPa. a. Determine the design strength of the section. b. Determine the allowable strength of the section 24mm 30mm
A PL40 mm X 250 mm (smaller member) is connected to a gusset plate (bigger member) as shown. The diameter of the holes are 25 mm. The pitch and gage of the holes are 50 mm and 75 mm, respectively. The yield strength of the steel is 260 MPa while the ultimate tensile strength of the steel is 400 MPa. Determine the design (LRFD) tensile strength of the tension member in kN. Neglect block shear. H FO E. • D A В
A PL40 mm X 250 mm (smaller member) is connected to a gusset plate (bigger member) as shown. The diameter of the holes are 25 mm. The pitch and gage of the holes are 50 mm and 75 mm, respectively. The yield strength of the steel is 260 MPa while the ultimate tensile strength of the steel is 400 MPa. Determine the design (LRFD) tensile strength of the tension member in kN. Neglect block shear. H G P P D B F C A E
A 16 mm thick tension member is connected by two 6.25 mm spliced plates as shown. The tension member carries a service loads of dead load of 110 kN and a live load of 100 kN. 40, 80 , 40 , 40, 80 40 1625 mm 16 mm 1625 mm Fy 248 MPa Fu = 400 MPa Diam. of bolts = 16 mm Fnv = 300 MPa O Determine the nominal strength for one bolt due to shear. O Determine the nominal strength for one bolt due to bearing strength of the connection. ® Determine the number of bolts required for the connection.
Material Strengths: Fy = 248 MPa, F = 400 MPa For the two lines of bolt holes, the pitch is the value that will give a net area DEFG equal to the one along ABC. The diameter of the holes is 22 mm. The thickness of the plate is 12 mm. Determine the LRFD design tensile strength based on yielding on gross area. D A to 50 1 ÓB 50 50 ic O 410 kN O 464 kN O 446 kN O 401 KN F IG Pitch Pitch KISS|Y
please solve 1.2
please solve 1.4
The A36 tension member is connected with 6- 16 mm diameter bolt. The member is made of Plate PL 4"x8" or PL 12mmx200mm. Net reduction factor = 0.85, Fy 248MPA, Fu = 400 MPa. Calculate the following using ASD and LRFD method. 1. Tensile yield strength of the section. 2. Minimum tensile strength PL ½x 8 O o o