Please help with what's in red. 3- Shaft ABC is composed of two segments: segment AB, whichis hollow-core; and segment BC, which is solid. Each end isattached to circular plate, which is fixed to the surroundings viafive 8 mm diameter bolts. If the permissible shear stress on thebolts is 150 MPa, what is the maximum allowable value of torqueT applied at point B?End viewholes for 8mm boltsIs this wherethe It is comingfrom?T10%FBDTAKEIs thiswhereneget the & from?¹8+)=9274 mmtJABJBC = (10²)=15708 mm4фавт фесhollow-core0.5mTLGJPerspective View»TET=0:-TA+T+Tc=0TAB 500ФАВ=G9274$BC = TBC 1000G15708Pleaseadd stepsas to howthe finalansweris foundIs I'm assuming this is from the 20mm?-How do wefor one bott failing in shear P=50.27 (150)=7540N get thisTend=5 Pr = 5(7540) 20= 150,800 N·mm (= 150.8 KN-m)solidIm20mmT=TACTA=0.4585T✔✓ ControlsTc=-0.5415Tを=T-T-What happenedTA 500 + (TA-T) 1000 to the B & BC ?=O9274157⁰8150.80.4585 04585 328.9 KN-mm

Answered: Image /qna-images/answer/29ec9751-3c31-401c-a33e-557e710c5267.jpg

Answered: 3- Shaft ABC is composed of two…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Please help with what's in red.

3- Shaft ABC is composed of two segments: segment AB, which
is hollow-core; and segment BC, which is solid. Each end is
attached to circular plate, which is fixed to the surroundings via
five 8 mm diameter bolts. If the permissible shear stress on the
bolts is 150 MPa, what is the maximum allowable value of torque
T applied at point B?
End view
holes for 8mm bolts
Is this where
the It is coming
from?
T
10%
FBD
TAKE
Is this
where
neget the & from?
¹8+)=9274 mmt
JAB
JBC = (10²)=15708 mm4
фавт фес
hollow-core
0.5m
TL
GJ
Perspective View
»T
ET=0:-TA+T+Tc=0
TAB 500
ФАВ=
G9274
$BC = TBC 1000
G15708
Please
add steps
as to how
the final
answer
is found
Is I'm assuming this is from the 20mm?
-How do we
for one bott failing in shear P=50.27 (150)=7540N get this
Tend=5 Pr = 5(7540) 20= 150,800 N·mm (= 150.8 KN-m)
solid
Im
20mm
T=TA
C
TA=0.4585T✔✓ Controls
Tc=-0.5415T
を=T-T
-What happened
TA 500 + (TA-T) 1000 to the B & BC ?
=O
9274
157⁰8
150.8
0.4585 04585 328.9 KN-mm

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Im still confused since we didn't end up with the same final values as the photo. Its showing TA = 0.5415 but in the photo TA is set to 0.4585. Please help me out to understand

фавт Фес=0
TA 500+ (TA-T) 1000
9274 15708
TA = =
0.4-58sT
Tc=-0.5415T
T=TA
O
✓ controls
150.8
0.4585 04585=3289 KN-mm
=

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