For the R function shown below(Attachment):
1 answer:
Answer:
The given function will return an integer in term of j on each iteration of while loop. However, it is noted that this program has some syntax error. The complete code executed in an online R compiler with explanation is given below
Explanation:
def<-function(j){
k=0;
i=1;
while (i<j**3) #j raise to power 3
{
k<-k+1;#k=k+1
i <- i*2;#i=i*2
print(k);#
}
}
def(6)# if you pass the value 6 to function def(j)
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Answer:
I have solved the problem below. I hope it will let you clear the concept.
For any inquiries ask me in the comments.
Explanation:
Answer:
a) Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
b) principal stress :
б1 = ( бx + бy ) / 2 - ( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
Explanation:
Combined normal stress and shear stress sketches attached below
The terms in the sketch are :
бx = tensile stress in x direction
бy = tensile stress in y direction
Txy = y component of shear stress acting on the perpendicular plane to x axis
бn = Normal stress acting on the inclined plane EF
Tn = shear stress acting on the inclined plane EF
A) Normal and shear stresses on inclined plane
Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
B) principal and maximum shear stresses
principal stress :
б1 = ( бx + бy ) / 2 - ( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
