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A part of a bracket in the seat assembly of a bus is shown in Figure P5-8 The load vanes from 1450 to 140 N as passengers enter and exit the bus The bracket is made from AISI 1020 hot-rolled steel. Determine the resulting design factor

Question-Answer › Category: Machine Design › A part of a bracket in the seat assembly of a bus is shown in Figure P5-8 The load vanes from 1450 to 140 N as passengers enter and exit the bus The bracket is made from AISI 1020 hot-rolled steel. Determine the resulting design factor
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GME asked 10 months ago

problem 5.19 part of a bracket in the seat assembly of a bus is shown in Figure P5-8. The load varies from 1450 to 140 N as passengers enter and exit the bus. The bracket is made from AISI 1020 hot-rolled steel. Determine the resulting design factor. Load on bracket 175 mm 275 mm Bracket Simple supports 450 mm 12 mm 60 mm Section A

A part of a bracket in the seat assembly of a bus is shown in Figure P5-8 The load vanes from 1450 to 140 N as passengers enter and exit the bus The bracket is made from AISI 1020 hot-rolled steel. Determine the resulting design factor

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Mazurek Gravity answered 10 months ago

 

for AIST 1020 hat rolled steel Stress factor Reliability factor cm - maternal St= 1 CR= 0.81 factor o l . to first find To fi
CS=0.9913 Now wsing Table of carbon of alloy steels propertion - Sy= 207 Mpq & s4= 379 MPg. Now wing the tentle strength gro
Now wing Goodmans critena Cat. 6g) + 6m - sel sy N . - Noro, to find om & 6o case as below. we will consider too Case During
A pornt not A any vertical force is present just left to - MAZO MB = RA CAB) = 88 6x 1754103 MB = ls Nm Me=o Case 1 Pusing ex
Similar 2103) MB = RA (AR)= 85. (x ( M - SNM Mead from case ① & 2 Mmon - IN Mmax nm section modulus sa BH3 = 60X193 144o mm²
6min Amin (1440X10-9) = 10.42X 106 pq onio - 10.42 MP Mean stress om 6max t omin. 2 107.64 + 10.42 2 6m = 59.03 MP9 Alternati
Assuming lt=1 (18 48.61) + 101.0785 59.03 379 20.4809 + 0.157= 0.63665 - 1 N N- 0-63165 FN= 1.5707
 

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Sazid Ahmad answered 7 months ago

 
 

Find the safety factor using Good mans relation as, κ.σ. σ ur find the endurance limit usin g, surface factor is, D, 0.808Vh
Applying the equations of equilibrium, for the max imum load of 1450 N -R, (0.45)+1450 (0.175) 0 R. -563.889 N R +R1450 R. =
Cra, + σ min-107.7 + 10.4 - 59.IMPa σα-G. 7,-107.7-59.1-48.6 MPa find the safety factor as, 1 59.1 (1)48.6 N 379 101 N = 1 .5

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