7laha solutions

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Solutions for 7laha problems

Let us remember semester problems and know their solutions.

1- Equilibrium problem (source similar to prob3-41 statics book)

•       A continuous cable of total length 4m is wrapped around small pulleys at A,B,C and D if each block has mass =15.6Kg .you should know that the springs are un stretched when d=2m and K for the springs =500 N/M.

•      At different masses and vertical spring .Can the cable take this form?

•     Consider α=60deg,θ=30deg ,a and b are known and the weight at pulley D is lighter than the weight at pulley B.

•     If your answer is yes find the mass at pulley D in terms of ∆1 and∆2.

•     If your answer is No say why?

•      THE Answer is No, the cable cannot take this form

•      When we put different weights at the pulleys the spring should change its position because the tension always equal.

  

Truss problem (6-52/53 statics book)

a-Determine the force in members KJ,NJ,ND and CD

•         Hint : use sections aa and bb

b-Determine the force in members JI and DE of the k truss.

•         Final answer for part b :Fji=10.67KN   FDE=10.67KN

•         Hint : use sections aa and bb

 

Stress and strain problem (similar to problem 2-57 mechanics of material)

•      A brass link (W=100mm, αb=20.9*(10^-6)/C , Eb=105 G pa )and a steel rod its center line at the  center of the link as shown (Es=200 G pa , αs=11.7*(10^-6)/C ).Intial temp =20 C , final temp=45  C.

•      Determine the final length of steel rod.

Solutions:

Twist and shear problem

Cylindrical Composite shaft (steel and brass) 

 Steel core (G=77Gpa) 40mm diameter and 5mm brass (39Gpa)

Jacket the shaft consist from two parts

Connected by bolts as shown

The right part (L =1.2m) is fitted with spring (K=133.3N.m/Rad) and the left part (L=.8 m) is fitted with fixed thin steel angle (.127*.127*.0095 m) length =.3 m

If a worker made a force=70N by a tool (L=.2m) on the left sec of the square box of bolts, the bolts permit a2 deg rotation of one flange with respect to the other. Note that the relative rotation angle does not depend on if the ends of parts are fixed or not.

Determine

1-    The max shear stress on the steel core and the brass jacket in the right part.

The corresponding angle of twist in the steel angle

SOLUTION:

T=F*L=70*.2=14N.m

J=π(c^4)/2

Js=π(.02^4)/2=2.5*(10^-7) m^4

Jb=π((.025^4)-(.02^4))/2 =3.6*(10^-7)m^4

Φr=.03Rad

Φ=T*L/GJ

Φr= -((10^-2)*14*.8/(2.5*77)+(3.6*39))+

((10^-2)*T*1.2/(2.5*77)+(3.6*39))

(a)

Tright=(.03+.33)*333*100/1.2=10N.m

Tspring=133.3*.03=4N.m

NOTE THAT WE PUT TORSION SPRING WITH THIS CONSTANT TO PREVENT THE MOTION after .03 rad.

So Torsion on the right part can be got from

14-100*.03=10N.M

T on the right part will divided on the steel core and the brass

Tsteel/Tbrass=Jbrass/Jsteel=1.4

TSteel+Tbrass=10N.m

Tsteel=5.83N.m  T Brass=4.1N.m

Shear =T*C/J

Max shear on steel core =5.83*.02/2.5*(10^-7)=0.46Mpa

Max shear on brass jacket=4.1*.025/3.6*(10^-7)=0.28Mpa

(b)

On The lift part T=14N.m

J=b*(t^3)/3

=2*.127*(.0095^3)/3 =7.2*10^-8 m^4

Φ=T*L/G*J =14*.3/7.2*770  =7.5*10^-4 Rad