For the system shown in Figure (1), the bolt in C is...

For the system shown in Figure (1), the bolt in C is subjected to a single shear and all other bolts are double shear. The diameter of bolt at C is 15 mm, diameter of bolt at D is 10mm and diameter of bolt A is 20mm. The allowable shear stress of bolt C is 200 MPa, allowable shear stress of bolt D is 150 MPa and allowable shear stress of bolt A is 100 .MPa If the diameter of bars AB, BD and BC is 25 mm and allowable normal stress is 125 MPa determine the safe value of the load (w) that can be applied based on the allowable shear stress of bolts at A,D, C and allowable .normal stress of bars AB, BD, BC 6800 N Figure (1) 1 m w N/m 1.25 m 2 m

Transcript

00:01 Today we will discuss about this question.

00:03 In this question we are given that the diameter of bolt at c that is equals to 15 mm.

00:13 The diameter of bolt at d that is equals to 10 mm.

00:20 The diameter of bolt at a that is equals to 20 m.

00:26 Now allowable shear stress, to bolt c show here we can write allowable the shear stress on c that is equal to 200 megapascal now allowable shear stress on d that is equals to 150 mp .a and the stress on a that is equals to 100 mpm and here the diameter of bags that is equals to ab that is equals to bd that is equals to bc that is equals to 25 mm and here allowable normal stress that is equals to 125 megapaacal so here first of all in this question we need to find weight of or the load of load that can be applied based on the allowable shear stress bolts at a d c that is equal to question mark so here first of all from the figure here we can write we can write c of c all that is equals to 200 newton per mm square c of d that is equals to 150 newton per mm square that means a shear dress and c of a that is equals to 100 newton per mm square and sigma all that is equals to 125 newton per mhm square.

02:43 So here diameter of a that will be denoted as we can denoted it as da that is equals to 20 mm d that is equals to 10 mm now dc that is equals to 15 mm and capital a a that is equals to 100 pi m m square here it is a area so capital ad that is equal to 25 pi m m m square now, ac, that is equals to 56 .25mm.

03:20 And here we can write 10 theta that is equal to 1 divided by 1 .25.

03:26 So that implies that theta that is equal to 38 .66 degree.

03:31 Now, we will apply summation of m .e.

03:35 That is equals to 0.

03:36 So therefore, 1 divide by 2, multiplied by w, multiplied by 2, multiplied by 2, multiplied by 2 divided by 2, multiplied by 2 plus 680 .00 multiplied by 2 .25.

03:48 So that is equal to hc multiplied by 1.

03:51 So that implies that 4 w divided by 3 plus 153 -0 that is equals to hc.

04:00 So therefore hc that is equals to 1 -53 -0 plus 4 divided by 3w.

04:06 That will be the hc.

04:09 Now we know that summation of ff, f, f, that is equals to 0 so that implies that he minus thc that is equal to 0 that implies that he that is equals to hc...

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