Question

Determine the largest axial load P which may be safely supported by a flat steel bar consisting of two portions, both 10 mm thick and respectively 40 and 60 mm wide, connected by fillets of radius r = 10 mm.

Assume the material is brittle(%EL < 5%), has an ultimate strength of 246 MPa, and a factor of safety of 1.1 is required.

Answer 1

55.908 KN

Step-by-step explanation:

The ratio D/d=60/40=1.5

The ratio r/d=10/40=0.25

From the curve attached as missing part of the question and using the above two ratios we get concentration factor, k=1.6

Therefore,
`\sigma_(ave)=\frac {\sigma_(max)}{k}=\frac {\sigma_(max)}{1.6}`

Since factor of safety, FS is given by

`FS=\frac {Ultimate strength}{allowable strength}`

Allowable strength=
`\frac {Ultimate}{FS}=\frac {246}{1.1}=223.63 Mpa`

Substituting 223.63 Mpa for
`\sigma_(max)` then for d=40 the stress is given by
`\frac {223.63}{1.6}=139.77`

Also,
`stress=\frac {P}{A}` hence

`P=stress* A=40*10*139.77= 55908 N`

P=55.908 KN