A bolted connection shown is made up of 2L150x 100 x 9.5 mm. The angle is made of A-36 steeland the bolts have diameter of 22 mm.Properties of 2L150 x 100 x 9.5 mm:A = 4660 mm2t = 9.5 mmFy= 248 MPaFu = 400 MPaProperties of 20 mm A-325 bolts: Fu = 400 MPa50 50 50 50 50 50,50힝힝이어Gusset Plate9.5mm100mmGusset Plate100mm31.2562.5T56.25a. Nominal dimension of hole diameter, dn = 25mmb. Determine the ultimate shear capacity (kN) ofthe connection.c. Determine the ultimate tearing capacity (kN)of the connection if the shear lag factor, U =0.75.

Draw the sectional diagram: Hole dia=25 mm

A bolted connection shown is made up of 2L150 x 100 x 9.5 mm. The angle is made of A-36 steel and the bolts have diameter of 22 mm. Properties of 2L150 x 100 x 9.5 mm: A = 4660 mm2 t = 9.5 mm Fy = 248 MPa Fu = 400 MPa Properties of 20 mm A-325 bolts: Fu = 400 MPa 50.50 힝힝어요 Gusset Plate 9.5mm 100mm Gusset Plate 100mm 31.25 62.5 56.25 a. Nominal dimension of hole diameter, dn = 25 mm b. Determine the ultimate shear capacity (kN) of the connection. c. Determine the ultimate tearing capacity (KN) of the connection if the shear lag factor, U = 0.75.

A bolted connection shown is made up of 2L150 x 100 x 9.5 mm. The angle is made of A-36 steel and the bolts have diameter of 22 mm. Properties of 2L150 x 100 x 9.5 mm: A = 4660 mm2 t = 9.5 mm Fy = 248 MPa Fu = 400 MPa Properties of 20 mm A-325 bolts: Fu = 400 MPa 50.50 힝힝어요 Gusset Plate 9.5mm 100mm Gusset Plate 100mm 31.25 62.5 56.25 a. Nominal dimension of hole diameter, dn = 25 mm b. Determine the ultimate shear capacity (kN) of the connection. c. Determine the ultimate tearing capacity (KN) of the connection if the shear lag factor, U = 0.75.

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

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter3: Tension Members

Section: Chapter Questions

Problem 3.3.7P

See similar textbooks

Similar questions

Bolted and Riveted Connections From the bolted connection shown, the diameter of bolts is 18mm0 with the hole diameter equal to 3mm bigger than the bolt, the angular section is 100 x 75 x 8 mm, with an area of 1340 mm2. Thickness of the gusset plate is 9mm. The gusset plate and angle arc A36 steel with Fy = 250 MPa and Fu = 400 MPa. Determine the tensile capacity of the connection based on gross area. Determine the tensile capacity of the connection based on effective net area using a reduction factor of 0.85. Determine the tensile capacity of the connection based on gross area.
A bolted connection shown consists of an angular section L 150mmx100mm x8mm. Bolt diameter = 18mm. Determine the effective net area of the section. (Ae). Properties of an L150x100x8: Area = 1,894 mm2 Fy = 248 MPa Fu = 400 MPa x = 23 mm y = 48 mm 50 50 50 50 50 50 150 40 www.AwNNO 60 100 50 P
The diagonal at the left to the connection is a double angle 90 mm x 90 mm x 8 mm, with area of 2700 mm?, bolted to the 8 mm thick gusset plate. Bolt diameter = 16 mm Bolt hole diameter = 18 mm Bolt bearing capacity, Fp = 480 MPa Bolt shear strength, Fv = 68 MPa Steel plate strength and stresses are as follows: Yield strength, Fy 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 0.60 Fy Allowable tensile stress on the net area = %3D 0.50 Fu Allowable shear stress on the net area = 0.30 Fu Bolt bearing capacity, Fp = 1.2 Fy Calculate the allowable tensile load, P(kN) under the following conditions: %3D
The connection shown is subjected to a tensile force of P = 100 kN. The allowable shear stress for the bolts is 100 MPa. Assume each bolt supports an equal portion of the load. Determine the required diameter of the bolts. a. 20.6 mm b. 25.2 mm c.35.7 mm d.17.8 mm choose letter of correct answer
4. Determine the design strength of the connection shown in the Figure below. Thebolts are 25 mm diameter A490 bolts with the threads not in the plane of shear.A36 steel is used (Fy = 250 MPa, Fu = 400 MPa).a. Compute the shear strength for all bolts.b. Compute the bearing strength for the tension member on all bolts.c. Compute the bearing strength for the gusset plate on all bolts.d. Compute the tensile strength of the tension member.e. Compute the design strength of the connection.
Problem. A bolted connection shown consists of two plates 300 mm x 12 mm connected by 4 - 22 mm diameter bolts. 12 mm bolts T12 mm te12 mm P 300 75 75 75 Edge distances = 75 mm %3D dnole for tensile and rupture = dp +3 mm %3D dnole for bearing strength for Lec = d + 1.5 mm Fy = 248 MPa Fu = 400 MPa %3D En 330 MPa %3D Use LRFD design method. Determine the design strength due to tensile rupture strength of plates. (kN)
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.
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.
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:
A76 x 76 x6 mm angular section shown is welded to an 8 mm thick gusset plate. The length L1 is 65 mm, L2 is 125 mm, and the cross sectional area of the angle is 929 sq.mm. Fy-248MPA and Fu -400MPA. Gusset Plate Alowable Sresse Alowable tensle stress igross ara)0.00 Alowable tereile stres net area)0SOF. Allowable shear stress (net ares)0.30F. Determine the value of P based on net area using a strength reduction coefficient of 85% O 138.24 KN O 140.14 kN O 157.93 kN O 304.00 N Next
Agui - 4550/2 =2276 two plates each with thickness t = 16 mm are bolted together with 6-22 mndia. bolts forming alap Joint - Bolt spacing are follows St=40mm S2 - 80mm, 53= 100mm effective bolt hole dia is 25mm A36 is used. Compute the allowable tensile strength t = 16 mm ob 27 40 60 80 40 +=16mm S1-40 S2-100 52-40 P 7 @ 4umm
sted ingid 250MM B k steel bar bolt AUR rigid bar 12 mm thick 8 mm Brass 350mm Jointd 16 mene thick 350mm JointB 250 mm Xg = 20x10% Eg = 90 G Pa -6 A, = 12x10/0 Es = 200 G Pa thickness> 16 mm 8 mm bolt %3D 12 mmi thick shear strength f bolf bearing streng th of holt= 100 la 50 MPa %3D Assume no failure will take place in steel or brass. Temperature chauge brass steel on Temperature change on Determine thai can be applied to system - the max