Question 172L5x3x7mm tension momber is connocted to a gussot plate by 16-mm-diameter bolts as shown in the figure, Given that e= 30 mm, s= 50 mm, Fy =345 MPa and Fu=402 MPa,caicukato the following2 L5 x3x % LLBD(two angles, long legsback-to-back)A242 stoel60 mme (am)(mn)e (mm)55 Not area in shoar for a single angle under block shear lailue, Arivmm?56. Not area in tension for a single anglo under.block shoar failuto Arit =57. Norninal block shear strongth in kN of the double angle Use Ups 10, PnINPUT ALL ANSWERS AS WHOLE NUMBER

Solution: Given: Section: 2 L5x3x1/4 LLB d = 5 in = 127 mm bw = 3 in = 76.2 t = 1/4 in = 6.35 mm…

Question 17 2L5x3x7mm tension momber is connocted to a gussot plate by 16-mm-diameter bolts as shown in the figure, Given that e= 30 mim, s= 50 mmI, Fy =345 MPa and Fu=402 MPa, caicukato ihe tollowing 2 L5 x3x % LLBD (two angles, long legs back-to-back) A242 steel 60 mm e (am) (mn) e (mm) 55 Not area in shoar for a single angle under block shear failue, Ariv 56. Not area in tension for a single anglo under. block shoar failuto Arit = 57. Norninal block shear strongth in kN of the double angle Use Ups "10, Pn INPUT ALL ANSWERS AS WHOLE NUMBER 11

Question 17 2L5x3x7mm tension momber is connocted to a gussot plate by 16-mm-diameter bolts as shown in the figure, Given that e= 30 mim, s= 50 mmI, Fy =345 MPa and Fu=402 MPa, caicukato ihe tollowing 2 L5 x3x % LLBD (two angles, long legs back-to-back) A242 steel 60 mm e (am) (mn) e (mm) 55 Not area in shoar for a single angle under block shear failue, Ariv 56. Not area in tension for a single anglo under. block shoar failuto Arit = 57. Norninal block shear strongth in kN of the double angle Use Ups "10, Pn INPUT ALL ANSWERS AS WHOLE NUMBER 11

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Please provide solution for this question with diagram
Give me right solution according to the question. Note : Don't copy from other websites. I have other website solution. A 620×230 UB 125 tension member of S355 steel is connected through both flanges by 30 mm bolts (in 25 mm diameter bolt holes) in four lines, two in each flange. Check the member for a design tension force of Nt,Ed = 3000 kN.
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.
Find øR, considering bolt limit states only. (S275 Steel, fy = 2804 kgf/cm2, fu = 4385 kgf/cm2, Bearing type connection 8.8 M20 bolts, ti = 15 mm, t2 = 18 mm) PE >P 50 mm 80 mm 80 mm 50 mm
4. The tension member shown below is a C 310 x 30.8 of A572M gr 345 steel materials. Will it safely support a service dead load of 265 kN and a service live load of 555 kN. L 50 102.5 102.5 50 65 65 65 O O 22 veny diameter bolts C310 x 30.8 Iso A36M stool matorial and select a double anglo tonsion member to rosist
5) Calculate P the allowable tensile load for the butt joint shown. Each A325-N high-strength bolt has a diameter of 0.9 inches. The plates are made of ASTM A36 steel. If the applied load is 200 kips, is this connection adequate? If not, what needs to be changed so that this connection can handle an applied load of 200 kips? P- ФФ P ! 12"x " P. P.
Please answer Letter d , e and f only
An A36 tension member shown in the figure is connected with three 19 mm diamter bolts. Diameter of hole = 22mm. Properties of 150 mm x 100 mm x 12.5 mm angle A = 3065 mm2 y = 50.55 mm Fy = 248 MPa Fu = 400 MPa Determine the design strength due to yielding in the gross section. (kN)
Two plates each with thickness t = 10 mm are bolted together with 6 - 16 mm diameter bolts forming a lap connection. Plate (A36): Fy = 248 Mpa FU = 400 MPa bolts are A325 in standard holes with threads excluded from shear planes. Considering Tension Yielding, compute the the design strength for LFRD. Use NSCP 2015 Round your answer to 2 decimal places.
The plate splice shown in the figure below has 9 – 3/4 inch diameter A325-N type bolts (Group A, N type) on each side. Based on the shear strength of bolts only, the maximum Pu that can be applied to the splice is equal to: Group of answer choices 202 kips 322 kips 161 kips 405 kips
Subject: Steel design- Bolted Steel Connection *Use NSCP 2015 formula/guide to solve this problem *Use Handwritten
Use ASD to determine the width of the 5/8 inch thick A36 plate subjected to tension as shown below. The dead load effect is PD=85 kip while the live load effect is PL=80 kip. Two rows of 5/8 inch standard bolts will be used for the connection and the plate width can be manufactured in 1 inch increments. Assume that the larger connected plate does not fail. Consider the limit states of gross tensile yield and tensile rupture. Find the required tension strength, Pa, for the controlling ASD load combination. Determine the width necessary for the plate to resist gross tensile yield. Determine the width necessary for the plate to resist tensile rupture. Report the required plate width.