= 3. A circular shaft is subjected to pure torsion as shown. The torque is created by applying load P to a cable as shown. Notice that the shaft is hollow over much of its length. a) What is the maximum allowable P given that Tau 200MPa? Assume normal stresses are not critical here. b) What is the displacement of the loading point when P is the maximum value? Assume the cable does not change length. your he Fig.3 Вост Locm →→→→→→→ SUPPORT 110cm RADIUS 10cm

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Chapter2: Loads On Structures
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Please explain the questions marked in red! Having trouble understanding where the values are from &/or how they were converted. Thanks in advance 

3. A circular shaft is subjected to pure torsion as shown. The torque is created by applying load P to a cable as shown. Notice that
the shaft is hollow over much of its length. a) What is the maximum allowable P given that Tau = 200MPa? Assume normal stresses
are not critical here. b) What is the displacement & of the loading point when P is the maximum value? Assume the cable does not
change length.
Voem
Fig.3
Вост
Locm
where
do
SUPPORT
what did
we do to get
2010 MPa?"
JOCM
TJ
a) 7 → T = Pr = II } ₂₁
10cm RADIUS
What is
G?
10cm
where
do we
get 1.0
✓
we get Tuollow = √((4_b²)=√(5904) mm ²
T/2
Lwhat value is b4?
Hollow section is critical
Paul
Isolid=C=10"mm" why to the 4th power?
• How do we get 5904?
ris radius of wheel
b) 8=rd=100mm 8
0= Hollow + Osolid
200 MPa I (5904)mm ²4
(10cm) (10mm)
concert to ma
TL Hollow Tholid
THollow G Isolid G {
+
1855N
?T= Pr = 1.855×10³ N·mm
1.855×105N.mm
800mm
200mm
+
1.0 x 110³ N/mm² (17 (15904) mm" I x 10"mm"
8=100mm (0.184)= 18.4 mm
-what does
this do
0.184 radians
min
Transcribed Image Text:3. A circular shaft is subjected to pure torsion as shown. The torque is created by applying load P to a cable as shown. Notice that the shaft is hollow over much of its length. a) What is the maximum allowable P given that Tau = 200MPa? Assume normal stresses are not critical here. b) What is the displacement & of the loading point when P is the maximum value? Assume the cable does not change length. Voem Fig.3 Вост Locm where do SUPPORT what did we do to get 2010 MPa?" JOCM TJ a) 7 → T = Pr = II } ₂₁ 10cm RADIUS What is G? 10cm where do we get 1.0 ✓ we get Tuollow = √((4_b²)=√(5904) mm ² T/2 Lwhat value is b4? Hollow section is critical Paul Isolid=C=10"mm" why to the 4th power? • How do we get 5904? ris radius of wheel b) 8=rd=100mm 8 0= Hollow + Osolid 200 MPa I (5904)mm ²4 (10cm) (10mm) concert to ma TL Hollow Tholid THollow G Isolid G { + 1855N ?T= Pr = 1.855×10³ N·mm 1.855×105N.mm 800mm 200mm + 1.0 x 110³ N/mm² (17 (15904) mm" I x 10"mm" 8=100mm (0.184)= 18.4 mm -what does this do 0.184 radians min
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