00:01
Hello students, in first part to find the tensile stress at extreme fiber at mid section is determined by the expression so the bending moment mid section this will be given by 4p.
00:25
Now substituting the value so we get 4 into 16 that is equal to 64 in terms of it so 64 multiplied by 12 so we get 768 in inch.
00:43
Now the bending stress is given by the expression that is sigma is equal to m upon i into y m is the bending moment at the section i is the moment of inertia of section about neutral axis and y is the distance.
01:07
Now substituting the values so i in terms of standards so this will be 310 and y is equal to 11 .94 we have to half this so this will be equal to 5 .97 inch.
01:26
Now substituting the values here so we get 768 divided by 310 into 5 .97 after solving this we get the bending stress that is 14 .79.
01:42
Now for second part in second part for a simply supported beam with the two point loops equally spaced so the deflection at center so this will be written with the expression that is p upon 24 e i 3l square minus 4a square.
02:11
Now here p is the point loop l is the total span of beam and a is the spacing between loops.
02:18
Now substituting the values before let's write what we have given so p will be equal to 16 and a is given as 4 that is an inch 48 inch and l is given as 12 feet that is 144 inch and l is given as 12 feet that is equal to 144 inch.
02:46
We have already written l value now i is given as 310 and a will be equal to 10 ,000.
02:59
Now substituting the values in the expression so we can write the deflection at the center will be equal to 16 into 48 divided by 24 into 10 ,000 into 310 with 3 into 144 square minus 4 into 48 square.
03:31
Now further solving this we get the deflection value that is 0 .47 inch.
03:40
Now for third part in third part we find the minimum shear stress in the web at the support so the shear stress at the point in the section that is given the expression so this will be equal to this expression...