Strength of Materials I EGCE201 กำลังวัสดุ 1

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Transcript Strength of Materials I EGCE201 กำลังวัสดุ 1

Strength of Materials I
EGCE201 กำลังวัสดุ 1
Instructor:
ดร.วรรณสิ ริ พันธ์อุไร (อ.ปู)
ห้องทำงำน: 6391 ภำควิชำวิศวกรรมโยธำ
E-mail: [email protected]
โทรศัพท์: 66(0) 2889-2138 ต่อ 6391
www.egmu.net/~civil/wonsiri
Shearing Stress in Multiple Shafts
Static equilibrium must be satisfied.
The vector of all applied torques = 0
TA+TB+TC+TD=0
The shearing stress at any point in the shaft is a function of the
Internal torque on a plane containing the point. The maximum
Stress is computed from
Internal torque
Radius of the shaft
Tc
 max 
J
The polar moment of inertia
The shearing stress at a point on shaft
through which section 1 passes can only
be defined once the internal torque is
determined.
From the free-body diagram, the internal torque
at the section
The shearing stress is given by
T1c1

J1
The shearing stress at a point on shaft
through which section 2 passes can only
be defined once the internal torque is
determined.
From the free-body diagram, the internal torque at the section
The shearing stress is given by
T2 c2

J2
Angle of Twist in Multiple Shafts
For this shaft sections AB, BC,
and CE will each have
different
- Internal torque (FBD)
- Polar moment of inertia
- Shear Modulus
- Length
The angle of twist of the end of a shaft consisting of
N sections is expressed as
n
Ti Li
 
i 1 Gi J i
Relative Rotation
fixed
In some situations both ends of
a shaft rotate. The angle of twist
is the angle through which one end
rotates w.r.t the other.
A/ B
Both rotates
Shaft AB
TL
  A  B 
JG
Shearing Stress on inclined planes
Shearing Stress on inclined planes
Statically Indeterminate Shafts
• Both ends of the shaft are built in, leading
to two reaction torques but one has only
one moment equilibrium equation.
• The compatibility equation is the relative
rotation!
• See example
Assuming a counterclockwise
torque is positive, summing
moments about the axis of the
shaft results in
TA  TC  T
The total angle of twist of the
shaft must be 0 since both ends
are fixed.
  AB  BC  0
โจทย์ตวั อย่ำงทบทวนควำมเค้นเฉื อนภำยใต้ภำระบิด
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