Problem from Mech Design 1 class 1 a Sketch the stress eleme

Problem from Mech Design 1 class.

1.

a) Sketch the stress elements at pts E and F as designated in class and determine the magnitudes

and directions for all stresses acting on them.

b) Determine the principal stresses and maximum shear stress at E and F. State whether the max

shear stress isin plane or out of plane.

c) Which point will fail first?

Din in diameter S 15in

Solution

solution:

1)here force F is acting at end of that rod and it is given by

F=Fxi+Fyj+Fzk=75i-200j+100k

2)where from centre of rod at plane of E and F we have position vector upto force as

R=6i+0j-5k

3)so moment about origin is given by

M=F*R

on taking cross product we get moment as

M=1000i+975j+1200k

4)here bending stress is given by

S=M*y/I

where

Iyy=Izz=(pi/64)*d^4=.04908 in4

Ixx=.09817 in4

5)so for point E bending stress is given by

Sbxe=Mx*x/Ixx=0 as x=0

Sbye=My*y/Iyy=975*.5/.04908=9931.26 psi

Sbze=Mz*z/Izz=0 as Z=0

6)bending stresses for point FR is given by

Sbxf=Mx*x/Ixx=0

Sbyf=My*y/Iyy=0

Sbzf=Mz*z/Izz=12223.09 psi

7)where direct tensile and shear stresses is given by

Stx=Fx/Ax=(Fx*4/pi*d^2)=95.49 psi

Sty=Stz=0

shear stress is given by

Tx=0

Ty=Fy/Ay=(Fy*4/pi*d62)=-254.64 psi

Tz=Fz/Az=(Fz*4/pi*d^2)=127.32 psi

8)so resultant normal and shear stress at point E is given by

Se=95.49i+9931.26j+0k

te=0i-254.64j+127.32k

for point F

Sf=95.49i+0j+12223.09k

tf=0i-254.64j+127.32k

9)here principle normal and shear stress for point E is given by

characteristic equation is

S^3-AS^2+BS-C=0

A=Sx+Sy+Sz

B=Sx*Sy+Sy*Sz+Sx+Sz-(txy)^2-(tyz)^2-(txz)^2

C=Sx*Sy*Sz+2Txy*Tyz*Txz-Sx(Tyz)^2-Sy(Txz)^2-Sz(Txy)^2

on solving this for point E and F we get principle stresses as root of equation

S1=Smax=9937.8 psi

S2=(Sx+Sy+Sz)-Smax-Smin=181.6 psi

S3=Smin=-92.6 psi

T1=S2-S3/2=137.1 psi

T2=S1-S3/2=5015.2 psi

T3=S1-S2/2=4878.1 psi

so we get maximum shear stress is in plane

for point F

S1=Smax=12229.7 psi

S2=(Sx+Sy+Sz)-Smax-Smin=94.2 psi

S3=Smin=-5.4 psi

T1=S2-S3/2=49.8 psi

T2=S1-S3/2=6117.6 psi

T3=S1-S2/2=6067.8 psi

here maximu shear stress is in plane

10) as maximum stresses are at point F than point E hence point F will fail first

so we get maximum shear stress is in plane