3.6-5 Use load and resistance factor design and select an American Standard Channel shape to resist a factored tensile load of 180 kips. The length is 15 ft, and there will be two lines of s-in. diameter bolts in the web, as shown in Figure P3.6-5. Estimate the shear lag factor U to be 0.85. (In a practical design, once the member and bolt layout are selected, the value of U could be computed and the member design could be revised if neces- sary.) Use A36 steel. Bolt lines FIGURE P3.6-5

3.6-5 Use load and resistance factor design and select an American Standard Channel shape to resist a factored tensile load of 180 kips. The length is 15 ft, and there will be two lines of s-in. diameter bolts in the web, as shown in Figure P3.6-5. Estimate the shear lag factor U to be 0.85. (In a practical design, once the member and bolt layout are selected, the value of U could be computed and the member design could be revised if neces- sary.) Use A36 steel. Bolt lines FIGURE P3.6-5

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 5.6.3P

See similar textbooks

Similar questions

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.
A double-angle shape is shown in the figure. The steel is A36, and the holes are for 2-inch-diameter bolts. Assume that A₂ = 0.75An. Ae 1 2 Determine the design tensile strength for LRFD. Determine the allowable strength for ASD. CIVIL ENGINEERING - STEEL DESIGN AL Section 2L5 x 3 x 516 LLBB
A double-angle shape is shown in the figure. The steel is A36, and the holes are for 1/2-inch-diameter bolts. Assume the Ae = 0.75An. Section 215 x 3 x %6 LLBB Determine the design tensile strength for LRFD. Select one: а. 66 b. 156 c. 78 d. 132
The 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:
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
4. An UPE 300 is to be used as a tension member and it is bolted to a 10mm thick gusset plate with bearing type 8-M20 8.8 bolts as shown in the figure below. a. Check all spacing and edge distance requirements. b. Compute the allowable tensile strength of the channel UPE 300. Don't consider shear strength of the bolts and bearing strength of the bolt holes. [all dimensions are in mm] Material both for gusset plate and UPE 300: S275 F-275N/mm² Fu=430N/mm² UPE 300 Ag = 5660 mm² tw 9.5 mm x = 28.9 mm 70 160 70 + 10mm thick gusset plate +50+ 80 -80-80-50 o +50+80 -80 80-50 UPE 300
The given plate below with width of 200 mm andthickness of 16 mm is to be connected to two plates ofthe same width with half the thickness by 20 mmdiameter rivets as shown. The rivet hole is 2 mm greaterthan the rivet diameter. Allowable tensile stress on netarea is 0.6Fy. Allowable bearing stress is 1.35Fy. Use a501 plate and a502 gr2 rivet a. Determine maximum load P without exceeding allowable tensile stress on plate b. Determine maximum load P without exceeding allowable shear stress on rivets c. Determine maximum load P without exceeding allowable bearing stress between plates and rivets
A Channel C6x13 is used as a tension member. The holes are for 5/8 inch in diameter of bolts. Using A36 steel with Fy=36 ksi and Fu=58 ksi. The gross area of C6x13 is 3.81 in? and Assume U =0.85 Using LRFD what is the design strength based on Tensile Rupture/Fracture 4 @ 2" a 1 1/2 Ob 3" Oc 1 1/2 " I d e C6 x 13 Round your answer to 3 decimal places.
Determine the design tensile strength of plate (200x8 mm) connected to 10-mm thick gusset using 20 mm bolts as shown in the figure, if the yield and the ultimate stress of the steel used are 250 MPa and 410 MPa, respectively. Add 1mm around the bolt for the hole. Use LRFD method. Plate 8-mm thick 2 3 40+ 30 301 T 200 mm Gusset 10-mm thick 3af 30 2_3 *40 40+ 50,54 +40
Problem # 6.0 For the tension member shown, compute the tensile design strength. Weld -6" PL 4X6 A36 steel
Compute the design strength (øPn) of the double angle tension member shown below for the following limit states: • tensile yielding in the gross section • tensile rupture in the net section • block shear rupture Identify the controlling limit state. Note: • The angles conform to ASTM A36 • LLBB = long legs back-to-back PL I" 11 (6) 7/B DIA. BOLTS z" 3" 2" -2L6x4x%2 LL BB
A double-angle shape is shown in the figure. The steel is A36, and the holes are for 1/2-inch-diameter bolts. Assume the Ae = 0.75An.a) Determine the allowable tensile strength for ASD.b) Determine the design tensile strength for LRFD.