A steel cantilever beam, as shown in Fig. 6.21, is subjected...

A steel cantilever beam, as shown in Fig. 6.21, is subjected to a transverse load at its end that varies from 45 N up to 135 N down as the axial load varies from 110 N (compression) to 450 N (tension). Determine the required diameter at the change of section for infinite life using a factor of safety of 2. The strength properties are as follows: Ultimate strength = 550 MPa Yield strength = 470 MPa Endurance limit = 275 MPa The stress concentration factors for bending and axial loads are 1.44 and 1.63 respectively, at the change of cross-section. Take size factor = 0.85 and surface finish factor = 0.9.

Transcript

00:01 Hello students, to determine the diameter at the change of section of infinite life using a factor of safety of 2 we are given that sigma u which is the ultimate strength is 550 mpa and then yield strength sigma y is given as 470 mpa.

00:25 Endurance limit sigma e is given as 275 mpa.

00:31 Stress concentration factor kb is given as 1 .44.

00:37 Stress concentration over axial loads is given as 1 .63.

00:42 Size factor ks is given as 0 .85 and the surface finish factor is given as 0 .9 and factor of safety fos is given as 2.

00:55 So to design the design equations we will use are first the bending stress for bending stress sigma b is equal to m times k times ks divided by zd cube and for axial stress sigma a will be equal to f times k times ks times divided by ad square.

01:26 So here m is the maximum bending moment, f is the axial load, k is the stress concentration factor, z is the section modulus, a is the cross sectional area and d is the diameter.

01:38 We will calculate the required diameter using both equations and select the larger value to ensure both bending and axial stresses meets the safety requirements.

01:47 So let's first calculate the bending moment using the given loads.

01:51 So m is equal to 135 n times 0 .2 d plus 450 n times d must be equal to 27 d plus 450 d.

02:08 So that is equal to 477 d.

02:11 Next we will calculate the section modulus z and cross sectional area.

02:18 So z is equal to pi times d cube by 32 and a will be equal to pi times d square by 4.

02:30 This is d square by 4.

02:33 Now let's calculate the bending stress sigma b.

02:35 Sigma b will be equal to 477 d times 1 .44 times 0 .85 divided by pi d cube by 32 times d cube.

02:51 So that will give you 8265 .44 by d raised to 4.

03:02 Now we can calculate the axial stress sigma a...

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