MODULE TAXATION 2

Room 206 JPD Building 1955 CM Recto Avenue, Manila Telepone !um"e#$ %02& 516 '559 ()Mail$ mega#evie*+200 mega#evie* +200 -.aoo/com

Review  MD<(

2  TR(!T TR(!T4 4  MAT MAT(RA (RA

TATCA3 TA TCA3 !D(RM!AT( M(M B(R PROBLEM 1  A horizontal rigid bar of negligible mass, 2 m long hinged at A, is supported by steel able B! at the enter and steel able "E at the right end# $he ables are of e%ual length & '#(m and e%ual ross setional A & 1)'mm2# A load P& )'*+ is applied '#m from the right end# Assume that the yielding stress of eah able is 2-'MPa and modulus of elastiity E & 2'' .Pa# a# "alul "alulat ate e the the stress stress of able able B!# B!# b# "alul "alulat ate e the the defor deformat mation ion of abl able e "E# "E# # "alul "alulat ate e the fator fator of safe safety ty in eah eah able# able# T4(RMA D(RMAT!

δ t 

=

α L∆T 

T 7 hange in temp#

7 thermal oeffiient

Po*e# T#an;mitted P 7 T ω  7 T 2  πf   πf  0here$ 0here$ T 7 To#?ue@

ω

7 angula# velocit. @> 7 >#e?uenc.

PROBLEM  A tor%ue of magnitude $ & ) *+=m is applied at end A of the omposite shaft sho0n# :f . &  .Pa for steel and 2 .Pa for the aluminum a# !etermine the ma;imum ma;imum shearing stress in the the steel ore b# !eterm !etermine ine the the ma;imu ma;imum m and minimu minimum m sheari shearing ng stress stress in the aluminum >a7et "# "alulate the angle angle of t0ist at A

PROBLEM 2  A steel steel rod / m long is seured bet0een bet0een t0o 0alls# "ompute "ompute the stress stress in $he rod 0hen the temperature rises to -' O"# $he ross setional setional area of  the rod is 1''' mm 2, 8 & 11# um  m o"# a# Assuming the 0alls are rigid and b# Assuming Assuming yields yields a distane distane of '#''mm '#''mm 0hen the temperature temperature rises# T4!  :A(D C3!D(R Tangential t#e;;

σ T 

=

P4(RCA 4(

ongitudinal t#e;;

 ρ D 2t 

σ L =

 ρ D

σ L

)t 

PROBLEM  $he shaft sho0n in the figure is made of bronze and steel and is fi;ed at both ends# :t is sub>eted to a onentrated tor%ue at the >untion# $he steel segment is 1#- meter long#

t#e;;

 ρ D )t 

=

PROBLEM / 3MA4 2'11 "E BOAR! E5AM6  A 0ater tan7 /m in diameter and  m high i s made from a steel ha8ing a thi7ness of 12mm# a# 9hen the the tan7 tan7 is filled filled 0ith 0ith 0ater 0ater,, determ determine ine the the irumfer irumferentia entiall stress# b# !ete !eterm rmin ine e the the long longit itudi udina nall stre stress ss at the the bott bottom om of the the tan7 tan7 0hen it is filled 0ith 0ater# # :f the ir irum umfe fere rent ntia iall stre stress ss is limit limited ed to - MPa, MPa, 0hat 0hat is the ma;imum height of 0ater to 0hih the tan7 may be filled# PROBLEM ) $he 0all thi7ness of a ) ft diameter spherial tan7 is -1 in# a# "alulate "alulate the allo0a allo0able ble internal internal pressure pressure if the stress stress is limited limited to (''' psi#

b#

"alulate the bursting fore in the shell#

TR!  is produed produed 0hen there there is twisting in the member member due to tor%ues in opposite rotation applied at the ends#

a#

!etermine !etermine the angle of t0ist t0ist at the interfae interfae of the the t0o t0o materials, in degrees# b# !etermine !etermine the the ratio ratio of the length length of steel steel to to the length length of bronze bronze suh that the allo0able allo0able stress in eah material material is reahed reahed simultaneously# # !eterm !etermine ine the reat reation ion at the the left left end if the the ma;i ma;imum mum tor%ue tor%ue is applied to the shaft, in *ilo+e0tons 4(AR : τ

T =

TORSIONAL / SHEARING STRESS

τ

2tA m

=

q  t 

? 7 ;ea# >lo*

τ

= ;t#e;;

PROBLEM  !etermine the shear flo0 of the gi8en thin 0alled tubes aused by the gi8en tor%ues#

D CRC<AR 4AT

 7

ma=

16T Tr  7 0here< # 0here< # 7 #adiu; o> o> te ;olid ;a>t 3  J πd πR

J7

4

4

7

2

πD

4(CA PR!

32

4: CRC<AR 4AT

σ= ma=

7

TR min

J

0here< R 7 out;ide #adiu; 7 D 7 out;ide diamete# J7

(

4

4

π R -r 2

Angle o> T*i;t

Tr  J

7

D 

# 7 in;ide #adiu; 7

2

d 7 in;ide diamete#

)

(

4

4

π D -d 7

32

)

 4m − 1  4m − 4 + 

16 PR

πd 

3

0 615  .

m

 

σ=

16 PR

πd 

3

 1 + 

  4 R  d 

3

d 2

δ =

64 PR n

Gd 

4

PROBLEM (  A -#2- m long rigid bar of negligible mass is supported by t0o idential springs at its third points B and " and at hinge at the left end A# $he properties of the springs are< n& 2' turns, d & 1- mm, ! & 1-' mm# :f a mass of -' 7ilograms is applied at the free end !< a# !etermine the ratio of the fore fore e;erted e;erted by spring " to spring spring B# b# Assuming light spring, alulate the ma;imum shearing stress in the   spring # "alulate the minimum shearing stress in the spring #