00:01
This question we have to find the resultant shear force developed in the vertical segment ab and we also have to plot the intensity of the shear stress across the cross -sectional area.
00:13
So from the cross -sectional area we can find the moment of inner shear which is 0 .182 times 10 to the power of minus 3 meter to a power of 4.
00:26
And we also can find q at point c, which is y, bar dad, ad, for point c, and that is 0 .1, time 0 .05, time 0 .15, that is 0 .75 times 10 to minus 3 cubic meter.
00:47
And qd, which is sigma y bar, a dad.
00:57
Time 0 .05, 0 .15 plus 0 .0 .125 time 0 .35 times 0 .025.
01:10
That is 0 .86, 10, minus 3, meter to the power of 3.
01:19
So, from tau equal vq over it, have that tau.
01:27
At the thickness .05 meter is v is v max that is 130 to the power of 3 and q is .75 divided by i, time t and that is 10 .7 mpa and that is 10 .7 mpa and equals 0 .35 meter and that will be 1 .53 megabascar.
02:15
We have that tau -d is v.
02:22
Xx2d right by i.
02:28
Time t and that is 1 .76 megapascar.
02:42
So we have that a dash is 0 .05175 for section ad, a .b, sorry.
02:59
And we have that...