STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS

1. BIBIN CHIDAMBARANATHAN
THERMAL STRESSES IN
BARS
OF CIRCULAR
TAPPERING SECTION
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

2. THERMAL STRESSES IN BARS OF CIRCULAR TAPPERING SECTION
❖ Consider a circular bar of uniformly tapering
section fixed at its ends A and B and subjected to
an increase of temperature.
Let
❖ 𝐿 = length of the uniformly tapering circular rod
❖ 𝐷1= larger diameter of the rod
❖ 𝐷2 = diameter at the smaller end of the rod
❖ T= change in temperature
❖ 𝛼 =coefficient of thermal expansion.
A
𝐷1 𝐷2
𝐷𝑥
𝑥
L
B
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

3. ❖ As a result of increase in temperature, the bar AB will tend to expand. But since it is
fixed at both ends, therefore it will cause some compressive stress.
𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒅𝒖𝒆 𝒕𝒐 𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝜹𝒍 = 𝜶 𝑻 𝑳 → 𝒆𝒒𝒏 (𝟏)
A
𝐷1 𝐷2
𝐷𝑥
𝑥
L
B
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

4. Let
❖ P= load or force required to bring the deformed bar to the original length
❖ Decrease in length of circular bar due to load P
Decrease in length of circular bar due to load 𝜹𝒍 =
𝟒𝑷𝑳
𝝅 𝑬 𝑫𝟏 𝑫𝟐
→ 𝒆𝒒𝒏 (𝟐)
A
𝐷1 𝐷2
𝐷𝑥
𝑥
L
B
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

5. ❖ Equating equations (1) and (2)
𝛼 𝑇 𝐿 =
4𝑃𝐿
π E 𝐷1 𝐷2
𝑃 =
π E 𝐷1 𝐷2 𝛼 𝑇
4
𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑑𝑢𝑒 𝑡𝑜 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 = Decrease in length of circular bar due to load
𝑳𝒐𝒂𝒅 (𝑷) =
𝛑 𝐄 𝑫𝟏 𝑫𝟐 𝜶 𝑻
𝟒
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

6. 𝑃 =
π E 𝐷1 𝐷2 𝛼 𝑇
4
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑟𝑒𝑠𝑠(𝜎𝑚𝑎𝑥) =
𝑃
𝐴2
𝜎𝑚𝑎𝑥 =
π E 𝐷1 𝐷2 𝛼 𝑇
4 ×
𝜋
4
× 𝐷2
2 𝐴2 =
𝜋
4
× 𝐷2
2
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒔𝒕𝒓𝒆𝒔𝒔(𝝈𝒎𝒂𝒙) =
𝜶 𝑻 𝑬 𝑫𝟏
𝑫𝟐
A
𝐷1 𝐷2
𝑥
L
B
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

7. Problem 03
A circular bar rigidly fixed at both ends is 1 m long and tapers uniformly from 20 cm
diameter at one end to 10 cm diameter at the other. Find the maximum stress in the bar, if
its temperature is raised through 50°C. 𝐸 = 2 × 105 Τ
𝑁 𝑚 𝑚2 and 𝛼 = 12 × Τ
10−6 ° 𝐶
𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂:
𝑻𝒐 𝒇𝒊𝒏𝒅:
𝐿 = 1 𝑚 = 1000 𝑚𝑚
𝐸 = 2 × 105 Τ
𝑁 𝑚 𝑚2
maximum stress in the bar 𝝈𝒎𝒂𝒙 =?
𝛼 = 12 × Τ
10−6 ° 𝐶
𝐷1 = 20 c𝑚 = 200 𝑚𝑚
A
𝐷1 𝐷2
𝐷𝑥
𝑥
L
B
𝐷2 = 10 c𝑚 = 100 𝑚𝑚 𝑇 = 50°𝐶
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

8. 𝑃 =
π E 𝐷1 𝐷2 𝛼 𝑇
4
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑟𝑒𝑠𝑠(𝜎𝑚𝑎𝑥) =
𝑃
𝐴2
𝜎𝑚𝑎𝑥 =
π E 𝐷1 𝐷2 𝛼 𝑇
4 ×
𝜋
4
× 𝐷2
2
𝐴2 =
𝜋
4
× 𝐷2
2
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒔𝒕𝒓𝒆𝒔𝒔(𝝈𝒎𝒂𝒙) =
𝜶 𝑻 𝑬 𝑫𝟏
𝑫𝟐
A
𝐷1 𝐷2
𝑥
L
B
𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

9. Maximum stress (𝜎𝑚𝑎𝑥) =
𝛼 𝑇 𝐸 𝐷1
𝐷2
𝐸 = 2 × 105 Τ
𝑁 𝑚 𝑚2
𝛼 = 12 × Τ
10−6 ° 𝐶
𝐷1 = 20 c𝑚 = 200 𝑚𝑚
𝑇 = 50°𝐶
Maximum stress (𝝈𝒎𝒂𝒙) = 𝟐𝟒𝟎 𝑵/𝒎𝒎𝟐
𝜎𝑚𝑎𝑥 =
12 × 10−6 × 50 × 2 × 105 × 200
100
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
𝐷2 = 10 c𝑚 = 100 𝑚𝑚
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

10. Thank You
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY