A steel plate 375mm x 16mm thick is bolted by four bolts placed symmetrically to form a single lap joint. The bolts are spaced horizontally at a distance of 140mm and vertically at a distance of 150mm. If the diameter of bolt is 20 mm, calculate the design tensile strength in KN based on net section fracture considering a straight vertical path of failure Use Fy=248MPa and Fu=400MPa a) 1339.2 b) 1569.6 c)2092.8 d) 1488

A steel plate 375mm x 16mm thick is bolted by four bolts placed symmetrically to form a single lap joint. The bolts are spaced horizontally at a distance of 140mm and vertically at a distance of 150mm. If the diameter of bolt is 20 mm, calculate the design tensile strength in KN based on net section fracture considering a straight vertical path of failure Use Fy=248MPa and Fu=400MPa a) 1339.2 b) 1569.6 c)2092.8 d) 1488

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

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter1: Introduction

Section: Chapter Questions

Problem 1.5.3P: A tensile test was performed on a metal specimen having a circular cross section with a diameter 0....

See similar textbooks

Similar questions

The data in Table 1.5.3 were obtained from a tensile test of a metal specimen with a rectangular cross section of 0.2011in.2 in area and a gage length (the length over which the elongation is measured) of 2.000 inches. The specimen was not loaded to failure. a. Generate a table of stress and strain values. b. Plot these values and draw a best-fit line to obtain a stress-strain curve. c. Determine the modulus of elasticity from the slope of the linear portion of the curve. d. Estimate the value of the proportional limit. e. Use the 0.2 offset method to determine the yield stress.
The double - riveted joint connects two plates which are each 3.0 mm thick, and the rivets have a diameter of 4 mm., if the failure strength of the rivets in shear is 370 N/mm2 and the ultimate tensile strength of the plate is 465 N/mm?, determine the necessary rivet pitch b (mm) if the joint is to be designed so that failure due to shear in the rivets and failure due to tension in the plate occur simultaneously. Calculate also the joint efficiency n (%). 3 mm 4 mm diameter
STRENGTH OF MATERIAL(SIMPLE STRESS) PLEASE INCLUDE FBD 100% GUARANTEE UPVOTE
PROBLEM 1) An aluminum bar carries the axial loads at the positions shown. If E=70GPA, compute the total deformation of the bar. Assume that the bar is suitably braced to prevent buckling. 0.4m D 10KN 0.8m 0.4m B 5KN 0.6m AL 20KN What is the deformation &pE in mm? A=800 mm² A=1,200 mm²
One bolt 34 in-16 UNF × 2 12 in. SAE grade 5 is subjected to a P load of 6 kip at atension joint. The initial stress is Fi = 25 kip. The stiffness of the bolt and the joint are kb = 6.50and km = 13.8 Mlbf / in, respectively.a) Determine the preload and service load stresses on the bolt. Compare them withthe minimum SAE test strength of the bolt.
A 20-mm-thick plate is bolted to the flange of a T-beam to form a 200 mm 20 mm built-up beam. If the beam is subjected to a vertical shear force of V = 200 kN, determine (a) the maximum shear stress in the beam, 10 mm 300 mm (b) the required spacing s of the bolts. The diameter of each bolt is 19 mm, and the allowable average shear stress in the bolts is 140 MPa. 10 mm
2) A solid rod with the given dimension shown in Fig. Axial loads are applied at the positions indicated. Find the f a) The max Compressive stress in the system P-60 kN Aluminum A=400 mm b) The max Tensile stress in the system P-60 kN Steel A = 500 mm² Bronze A-200 mm² 4P 2.5 m 1.5 m 2.0 m 2P
The light rigid bar ABCD shown Is pinned at C and connected to two vertical rods. The bar was initially horizontal, and the rods were stress-free before the load P = 20 KN is applied. E200 CPa E1m 2.0 m 0.6m 1.5 m VP-20 KN Aluminum E-70 GPa A=900mm2 L-1.5 m What is the axial stress at steel rod in MPa?
A hole is to be punched through a steel plate of 8 mm thickness. What is the least diameter of hole which can be punched, if the steel punch can be worked to a compressive stress of 800 N/mm² and the ultimate shear strength is 300 N/mm²? A 1.2 mm B 12 mm C 21 mm D 2.1 mm
Determine the design tensile strength of the plate (200 × 100) mm connected to a 12 mm thick gussets, using 20 mm bolts shown below, if the yield and ultimate stress of the steel used are 250 MPa and 420 MPa respectively, fy= 250 MPa fu= 450 MPa
2. A steel plate is 360 mm wide and 20 mm thick with four bolts hole into the place as shown in the figure. Compute the following: a. Critical net area required by the NSCP specs. b. Max. critical net area required by the NSP specs. Note that, Max. net area is 85% of the critical net area Capacity of the joint if the allowable tensile stress is 0.75Fy. Use A36 steel Fy=248 MPa 25 mm= DIAMETER OF BOLTS 45 45mm fmmy C. 90mm 90mm 90mm Comm P
L1 = 2.02 m %3D L2 = 1.21 m H= 3.03 m Width of plate hanger = 40.04 mm Thickness of plate hanger = 10.01 mm %3D Bolt diameter = 20.02 mm %3D Allow shear stress of the bolt 68.08 MPa %3D Allow bearing stress of the bolt = 240.04 MPa LI L2 W 1. What is allowable stress in the hanger based on the bolt capacity in double shear at D