BILKENT UNIVERSITY
MECHANICAL ENGINEERING
ME443 - MACHINE ELEMENTS
WEEK 3
Design of Shafts
2024 Fall
Dr. Levent Subaşı
(Based on the lecture notes of Prof. Dr. Orhan Yıldırım)
UNCLASSIFIED
Let’s assume that loading at a section of a shaft are given as below
DESIGN OF SHAFTS
Introduction
A «shaft» is a rotating member, usually of circular cross section, used to transmit
power or motion. It carries elements such as gears, pulleys, flywheels, etc. Shafts
are typically subjected to a mean torsional stress (due to a constant torque) and
alternating bending stress (due to reversed bending load).
M
T
F
(2D element)
Design steps for a shaft: (assuming design objective is known)
∙ Material selection (generally steel alloys)
∙ Geometric layout
∙ Stress and strength (Static strength, Fatigue strength)
∙ Deflection and rigidity (Bending deflection, Torsional deflection, Shear deflection)
∙ Vibration due to natural frequency (Critical speed)
𝜎𝑥 =
𝑑
𝑐= ,
2
Only static & fatigue strength shall be considered in this course
𝜏𝑥𝑦 =
Shaft Design for Static Loads
Stresses are evaluated only for critical locations. Critical locations are usually:
• on the outer surface
• where bending moment is large
• where torque is present
• where stress concentrations exist
UNCLASSIFIED
𝑀𝑐 𝐹
+
𝐼
𝐴
𝑐=
𝜋𝑑 4
𝐼=
,
64
𝑇𝑐
𝐽
𝑑
,
2
I: Moment of inertia
𝜋𝑑 2
𝐴=
4
𝜎𝑥 =
32𝑀 4𝐹
+
𝜋𝑑 3 𝜋𝑑 2
J: Polar moment of inertia
𝜋𝑑 4
𝐽=
32
𝜏𝑥𝑦 =
16𝑇
𝜋𝑑 3
Use stress concentration factors and revise the formulas for critical locations by
multiplying M with Kt, F with Kt and T with Kts.
1/6
2/6
According to MSST:
Shaft Design for Fatigue Loads
𝜎𝑥 − 𝜎𝑦 2
+ 𝜏𝑥𝑦 2
2
Remember: 𝜏𝑚𝑎𝑥 =
Assuming a solid shaft with round cross section with F=0 (and expanding the
geometry terms c, I and J), the stresses due to bending and torsion can be
calculated as:
σy term drops
𝑆𝑦
16
=
2𝑛 𝜋𝑑 3
𝐾𝑡 𝑀 2 + 𝐾𝑡𝑠 𝑇 2
+ 𝐾𝑡𝑠 𝑇 2
1
3
Alternating
Mean
2
𝜎𝑥2 + 3𝜏𝑥𝑦
𝜎′ =
𝑑
𝐾𝑡 𝑀 + 𝐾𝑡 𝐹
8
2
UNCLASSIFIED
32
𝜎𝑎 ′ =
𝜋𝑑 3
𝐾𝑓 𝑀𝑎
𝜎𝑚 ′ =
2
2
𝜎𝑥𝑚
+ 3𝜏𝑥𝑦𝑚
𝜎𝑚 ′ =
32
𝜋𝑑 3
𝐾𝑓 𝑀𝑚
2
2
3
+ 𝐾𝑓𝑠 𝑇𝑎
4
2
+
2
3
𝐾𝑓𝑠 𝑇𝑚
4
Solving for the critical diameter:
3
+ 𝐾𝑡𝑠 𝑇 2
4
32𝑛 1
𝑑=
𝜋 𝑆𝑒
In case F=0, critical diameter can be solved in closed form:
32𝑛
𝜋𝑆𝑦
𝜎𝑎 ′ =
2
2
𝜎𝑥𝑎
+ 3𝜏𝑥𝑦𝑎
𝜎𝑎 𝜎𝑚 1
+
=
𝑆𝑒 𝑆𝑦 𝑛
𝑆𝑦
32
=
𝑛
𝜋𝑑 3
𝑑=
16𝑇𝑚
𝜋𝑑 3
Using Soderberg Criteria:
All y and z terms drop
Compare with strength:
𝑆𝑦
𝑛=
𝜎′
𝜏𝑥𝑦𝑚 = 𝐾𝑓𝑠
𝜎𝑥𝑚 = 𝐾𝑓
Writing the Von Mises stress:
Similarly, according to DET:
Remember:
32𝑀𝑚
𝜋𝑑 3
Mean
16𝑇𝑎
𝜋𝑑 3
2
𝑑
𝐾𝑡 𝑀 + 𝐾𝑡 𝐹
8
In case F=0, critical diameter can be solved in closed form:
32𝑛
𝑑=
𝜋𝑆𝑦
𝜏𝑥𝑦𝑎 = 𝐾𝑓𝑠
𝜎𝑥𝑎 = 𝐾𝑓
Compare with strength (using a safety factor):
(𝑆𝑦 /2)
𝑛=
𝜏𝑚𝑎𝑥
32𝑀𝑎
𝜋𝑑 3
Alternating
𝐾𝑡 𝑀 2 +
3
𝐾 𝑇 2
4 𝑡𝑠
𝐾𝑓 𝑀𝑎
2
2
3
1
+ 𝐾𝑓𝑠 𝑇𝑎 +
4
𝑆𝑦
𝐾𝑓 𝑀𝑚
2
2
3
+ 𝐾𝑓𝑠 𝑇𝑚
4
1
3
1
3
3/6
4/6
Average shear stress:
Using Goodman Criteria:
𝜎𝑎 𝜎𝑚 1
+
=
𝑆𝑒 𝑆𝑢𝑡 𝑛
𝑇
𝜏𝑎𝑣 = 𝑅
𝑏𝐿
Solving for the critical diameter:
𝑑=
32𝑛 1
𝜋 𝑆𝑒
𝐾𝑓 𝑀𝑎
2
+
3
𝐾 𝑇
4 𝑓𝑠 𝑎
2
+
1
𝑆𝑢𝑡
𝐾𝑓 𝑀𝑚
2
+
3
𝐾 𝑇
4 𝑓𝑠 𝑚
2
1
3
𝜏𝑎𝑣 ≤ 𝜏𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒,𝑘𝑒𝑦 =
Shaft Component – Key
𝑆𝑠𝑦,𝑘𝑒𝑦
𝑛
Ssy: Shear yield strength
Contact stress:
Used for securing the rotating elements and to transmit torque
shaft
Shear area
𝑇
𝜎= 𝑅
ℎ
𝐿
2
Contact area
𝜎 ≤ 𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒,𝑘𝑒𝑦 =
hub
𝑆𝑦,𝑘𝑒𝑦
𝑛
Failure of keys is by either direct shear or bearing stress (contact stress)
𝜎 ≤ 𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒,𝑠ℎ𝑎𝑓𝑡 =
𝜎 ≤ 𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒,ℎ𝑢𝑏 =
Sy: Yield strength
𝑆𝑦,𝑠ℎ𝑎𝑓𝑡
𝑛
𝑆𝑦,ℎ𝑢𝑏
𝑛
Check for the weakest one!
UNCLASSIFIED
5/6
6/6