4050doubleangle3550508 mmgusset plateO Based on bolt capacity in shear.O Based on bolt bearing capacity.® Based on block shear strength of thedouble angle. The diagonal at the left to the connection is adouble angle 90 mm x 90 mm x 8 mm, witharea of 2700 mm?, bolted to the 8 mm thickgusset plate.Bolt diameter = 16 mmBolt hole diameter = 18 mmBolt bearing capacity, Fp = 480 MPaBolt shear strength, Fv = 68 MPaSteel plate strength and stresses are asfollows:Yield strength, Fy 248 MPaUltimate strength, Fu = 400 MPaAllowable tensile stress on the grossarea = 0.60 FyAllowable tensile stress on the net area =%3D0.50 FuAllowable shear stress on the net area =0.30 FuBolt bearing capacity, Fp = 1.2 FyCalculate the allowable tensile load, P(kN)under the following conditions:%3D

Answered: Image /qna-images/answer/2ffefe28-9c00-41df-ac20-29a2b938f19e.jpg

Answered: O Based on bolt capacity in shear. O…

O Based on bolt capacity in shear. O Based on bolt bearing capacity. ® Based on block shear strength of the double angle.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Two steel plate tension members have been connected using 0.72” diameter bolts arranged in an equally-spaced four by four square formation.Total plate self-weight is specified as 912 pounds.Design of the elements adhered to the set of values that are twice as much as the minimum requirements. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 54ksi and the min required edge distance as 0.125ft. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load(excluding the self-weight)and live load assuming live load is half as much as dead load including the self-weight of the plates.
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
an angle bar 100x100x11 mm tension member is connected with 20-mm-dia bolts as shown in the fig. both legs of the angles are connected. use fy=248 MPa and Fu=400 MPa. 1. determine the nominal strength for tensile yielding in the gross section. a. 515.59 kN b. 429.67 kN c. 545.60 kN d. 390.10 kN 2. determine the effective net area of member a. 1573 mm^2 b. 1632 mm^2 c. 1605 mm^2 d. 1815 mm^2 3. determine the nominal strength for tensile rupture in the net area. a. 642 kN b. 730 kN c. 554 kN d. 650 kN
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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. plate: fy = 241 mpa fu = 4141 mpa rivet: fv = 152 mpa 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
Two steel plate tension members have been connected using 0.72 inch diameter bolts arranged in an equally-spaced 9 by 3 rectangular formation. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 51.2 ksi, clear distance for each edge bolt as 0.12 in, and the clear distance for the other bolts as 1.072 in. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load and live load assuming live load is half as much as dead load. Consider ASD.
STAGGERED CONNECTIONS: A PLATE WITH WIDTH OF 400 mm AND THICKNESS OF 12 mm IS TO BE CONNECTED TO A PLATE OF THE SAME WIDTH AND THICKNESS BY 34 mm DIAMETER BOLTS, AS SHOWN IN THE FIGURE. THE HOLES ARE 2 mm LARGER THAN THE BOLT DIAMETER. THE PLATE IS A36 STEEL WITH YIELD STRENGTH Fy = 248 MPa. ASSUME ALLOWABLE TENSILE STRESS ON NET AREA IS 0.60Fy. IT IS REQUIRED TO DETERMINE THE VALUE OF b SUCH THAT THE NET WIDTH ALONG BOLTS 1-2-3-4 IS EQUAL TO THE NET WIDTH ALONG BOLTS 1-2-4. a. CALCULATE THE VALUE OF b IN MILLIMETERS. b. CALCULATE THE VALUE OF THE NET AREA FOR TENSION IN PLATES IN SQUARE MILLIMETERS. c. CALCULATE THE VALUE OF P SO THAT THE ALLOWABLE TENSILE STRESS ON NET AREA WILL NOT BE EXCEEDED.
The five-bolt connection shown must support an applied load of P = 330 kN. If the average shear stress in the bolts must be limited to 255 MPa, determine the minimum bolt diameter that may be used in the connection.
For the clevis connection shown, the shear stress in the 0.287-in.-diameter bolt must be limited to 29 ksi. Determine the maximum load P (in kips rounded to the nearest hundredths) that may be applied to the connection.
Pls answer thank you! A flange bolt coupling has 10 steel coupling 1 inch diameter bolts evenly tighten around a 16.34 in. bolt circle. Determine the torque capacity of the connection if the allowable shearing stress in the bolt is 7.3 ksi?
9) 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=70ksi