Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 280 MPa, and a factor of safety of 4.3 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 475 kN. Answer: dmin = mm

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 280 MPa, and a factor of safety of 4.3 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 475 kN. Answer: dmin = mm

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

Similar questions

Determine the design tensile strength of the 12 in. x 1/2 in. steel plate shown in the figure. The bolts are 3/4 in. diameter. The steel is A572 Gr. 50. Check yielding and fracture. Check Block Shear. T 3in. 73im 13in 1 3in t
A steel plate is to be attached to a support with three bolts. The cross-sectional area of the plate is 800 mm² and the yield strength of the steel is 260 MPa. The ultimate shear strength of the bolts is 570 MPa. A factor of safety of 1.67 with respect to yield is required for the plate. A factor of safety of 4.0 with respect to the ultimate shear strength is required for the bolts. Determine the minimum bolt diameter required to develop the full strength of the plate. Note: consider only the gross cross-sectional area of the plate-not the net area. Support Plate ↓ P
1. A tension member is made of 90 mm thick steel plates 225 mm wide by a lap joint made with three (3) rows of 20mm. Use LFRD specifications ASTM A36 (Fy=138 MPa, Fu=220 MPa, compute the following: a. Capacity of the section based on the yielding of the gross area Critical effective net area b. c. Capacity of the section based on the tensile fracture of the critical net area. Dia 20mm DIAMETER OF BOLTS P
CS Scanned with CamScanner 1. A tension member is made of 90 mm thick steel plates 225 mm wide by a lap joint made with three (3) rows of 20mm. Use LFRD specifications ASTM A36 (Fy=138 MPa, Fu-220 MPa, compute the following: a. Capacity of the section based on the yielding of the gross area Critical effective net area b. c. Capacity of the section based on the tensile fracture of the critical net area. } 1 1 1 F 1 P
A plate with width of 420 mm and thickness of 13 mm is to be connected to a plate of the same width and thickness by 30 mm diameter bolts, as shown in the Figure 1. The holes are 3mm larger than the bolt diameter. The plate is A36 steel with yield strength Fy = 248MPa. 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 equl to the net width along bolts 1-2-4. W = 420 mm t = 13 mm a = 64 mm c = 104 mm d = 195 mm Bolt Diameter = 30 mm Holes Diameter = 30 mm + 3 = 33 mm ?? = 248 MPa Allowable Tensile Stress on ?? = 0.60Fy a. Calculate the vaue 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.
Subject: Steel design- Bolted Steel Connection *Use NSCP 2015 formula/guide to solve this problem *Use Handwritten A plate having a width of 400 and a thickness of 20mm is to be connected to two plates with thickness of 12mm by four rivets. The allowable shear stress for the rivets is 150 MPa and for bearing is 1.35 Fy. If the yield strength and the ultimate strength is 248 MPa and 400 MPa respectively. Determine the Following: a.) Required diameter of rivets w/o exceeding the aloowable stress in the shear. b.) Required diameter of rivets w/o exceeding the allowable stress in bearing. c.) Required diameter of rivets w/o exceeding the allowable stress in tension.
The bolt is made of a material having an allowable tensile stress of 140.44 MPa and an allowable shearing stress of 55.44 MPa. The bolt will be tightened so that the axial load will be 17 KN. Find the required thickness of the bolt head if the thread is 0.25 mm deep.
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
H.W (2): A 37.5 mm round bar has been machined from AISI 1050 cold-drawn round bar Su =689 MPa. This part is to withstand a fluctuating tensile load varying from 0 to 72.5 kN. Because of the design of the ends and the fillet radius, a fatigue stress-concentration factor of 1.85 exists. Find the factor of safety F.S.
A single unequal angle 100 x 75 x 6 is connected to a 10 mm thick gusset plate at the ends with six 16 mm diameter bolts to the 100 mm of leg to transfer tension as shown in figure. What will be the block shear strength (in kN) of the angle section assuming that the yield and the ultimate stress of steel used are 250 MPa and 410 MPa respectively? ISA (100 × 75 x 6) -T 16 mm o bolt 10 mm 40 mm 40 mm| 40 mm 40 mm 40 mm 40 mm 10 mm
Six bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 345 MPa, and a factor of safety of 3.0 is required with respect to fracture. Determine the minimum allowable bolt diameter D required to support an applied load of P = 315 kN. 14.01 mm 17.05 mm 17.97 mm 12.35 mm 10.87 mm
Compute the nominal shear strength of an M107.5 of A572 Grad 65 steel.