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EPITACIO ALCANO III K. BSME -4 ME 416 PLATES

STRESSES AND STRAINS

1. The maximum torque that can be applied to a hollow circular steel shaft of 120 mm outside diameter
and 80mm inside diameter without exceeding a shearing stress of 80 MPa.

Given: Required: T=?

D= 120 mm

d= 80 mm

Ss= 80 MPa

Solution:
16𝑇𝐷 16𝑇(120 𝑚𝑚)
Ss = 𝜋(𝐷4 −𝑑4 ) = 80= 𝜋(1204 −804 )

T= 21781709.06 N-mm= 2,1781.71 N-m

2. A steel shaft is subjected to a constant torque of 2,260 N-m. The ultimate strength and yield strength
of the shafting material are 668 MPa and 400 MPa, respectively. Assume a factor of safety of 2 based on
the yield point and endurance strength in shear, determine the diameter of the shaft in stress.

Given: Required: D=?

T = 2,260N-m

St= 668 MPa

Sy= 400 Mpa

Fs= 2

Solution:
𝑆𝑦 16𝑇
= 𝜋𝑑3
𝐹𝑠

400 16(2260 𝑥 1000)

2
= 𝜋𝐷 3

D= 1.52 in

3. If the ultimate shear stress of the steel plate is 35,000 psi. What force is necessary to punch a 1.5 in
diameter hole in 1/8 thick plate?

Given: Required: F=?

Su = 35,000 psi

D= 1.5 in

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t= 1/8 in

Solution:
𝐴 𝜋𝑑𝑡
Su = 𝐹 = 𝐹

F = Su 𝜋𝑑𝑡

F= (35,000)(1.5)(1/8)

F = 20,616.70 lbs

4. If the ultimate shear stress of the steel plate is 35,000 psi , what force is necessary to punch a 1.9 in
diameter hole in a 1/8 in. thick plate?

Given: Required: F=?

Su = 35,000 psi

D= 1.9 in

t= 1/8 in

Solution:
𝐴 𝜋𝑑𝑡
Su = 𝐹 = 𝐹

F = Su 𝜋𝑑𝑡

F= (35,000)(1.9)(1/8)

F = 26,114.49 lbs

5. A copper column of annular cross-section has an outer diameter of 15ft. and is subjected to a force of
45, 000 lbs. The allowable compressive stress is 300 lb/ft2. What would be the wall thickness?

Given: Required: t=?

Sc = 300 lb/ft2

F = 45, 000 lbs

D = 15 ft

Solution:
4𝐹 4(45,000)
Sc = 𝜋(𝐷2 −𝑑2 ) = 300 = 𝜋(152 −𝑑2 )

d= 5.83 ft.

when d =5.83 ft.

15−5.83
t= 2
= 4.58 ft.

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6. What factor of safety is needed for a 2.5 in diameter shaft with an ultimate strength of 65,000 psi
transmit 45,000 lb-ft torque?

Given: Required: Fs=?

D= 2.5 in

Su = 65,000 psi

T = 45,000 psi

Solution:
𝑆𝑢
Sd= 𝐹𝑠
16𝑇
Ss = 𝜋(𝐷3 )

𝑆𝑦 16𝑇
=
𝐹𝑠 𝜋𝑑 3

𝜋𝑑 3 𝑆𝑢 𝜋2.53 (65,000)
Fs= 16𝑇
= 16(45,000)
= 0.3639

7. If the ultimate shear strength of a steel is 42,000 psi, what force is needed to punch a 0.75 in diameter
hole in a 0.625 in thick plate?

Given: Required: F=?

Su = 42,000 psi

D= 0.75 in

t= 0.625 in

Solution:
𝐴 𝜋𝑑𝑡
Su = =
𝐹 𝐹

F = Su 𝜋𝑑𝑡

F= (42,000)(0.75)(0.625)

F = 61,818.75 lbs

8. What force is required to punch a 30 mm diameter hole in a plate that is 35 mm thick? The shear
strength is 400 MPa.

Given: Required:

Ss =400 MPa F=?

D = 30 mm

T = 35 mm

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Solution:
𝐹 𝐹
Ss = 𝐴 = 𝜋𝑑𝑡

F = π(400)(30)(35)

F= 1,319.47 kN

9. Assume that a 400 mm diameter rivet joins the plate that are each 100 mm wide. The allowable stress
are 130 MPa for bearing in the plate material and 70 MPa for shear of rivet. Determine the minimum
thickness of each plate.

Given:

D = 40 mm

W = 100 mm

Sa = 130 MPa

Ss= 70 MPa

Solution:
𝜋
F= SSA = SS 4 𝑑2
𝜋
F = 70( 302 )
4

F = 1570π

F = SaAb

1570π = 130(30t)

t = 12.69 mm

10. A 400 mm-diameter pulley is prevented from rotating relative to 70-mm diameter shaft by a 80-mm
long key. If a torque t= 3.3Kn-m is applied to the shaft, determine the width if the allowable shearing
stress in the key is 60MPa.

Given: Required:

D = 300 mm b=?

T= 3.3 kn

Solution:

T = 0.003F

F = 1100 kn

A = 80b

T= 60 MPa

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Thus,

F(1,000) = 60(80b)

B = 229.17mm

11. A 3 in. diameter east iron rod is subjected to an axial compressive load of 12,000 in-lb. Determine
the maximum normal stress.

Sy= 0
−(12,000)(8)
Sx= = -7,000 psi
𝜋(3)2

(2,500)(1.5)(32)
Txy= = 122 psi
𝜋(3)4

12. If the force of the steel is 60,717.70 lbs. What is the ultimate shear stress is necessary to punch a 1.5
in. diameter hole in a 1/8 thick plate.

Given: solution:
𝐹 𝐹 (30,717.70)
Su = 𝐴 = 𝜋𝑑𝑡 1 Su= 1
𝜋(1.5)( )
8

Required: Su = 39,000 psi

Su=?

13. If the ultimate shear strength of a steel is 30,000 psi, what force is necessary to punch a 2 in.
diameter hole in a 1/8 in. thick plate?

Given:

Su= 35,000 psi

D = 1/9 in

t = 1/8 in

Solution:
𝐹 𝐹
Su= =
𝐴 𝜋𝑑𝑡

F = πdtSu

F= π(2)(1/8)(30,000)

F= 23,561.94 lbs.

14. If the force of the steel is 60,717.70 lbs. What is the ultimate shear stress is necessary to punch a 1.5
in. diameter hole in a 1/8 thick plate.

Given: solution:

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𝐹 𝐹 (30,717.70)
Su = 𝐴 = 𝜋𝑑𝑡 1 Su= 1
𝜋(1.5)( )
8

Required: Su = 39,000 psi

Su=?

15. If the ultimate shear stress of the steel plate is 35,000 psi. What force is necessary to punch a 1.5 in.
diameter hole in a 1/8 thick plate?

Given: Solution:
𝐹 𝐹
Su= 35,000 psi Su = =
𝐴 𝜋𝑑𝑡 1

D = 1.5 in. F = πdt Su

t= 1/8 in. F= π(1.5in)(1/8in)(35,000psi)

Required: F= ? F= 20,616.70 lbs.

16. What factor of safety is needed for a 2.5 in. diameter shaft with an ultimate strength of 65,000 psi
transmits a 45,000 lb-ft torque?

Given: Solution:
𝑆𝑢
D= 2.5 in. Sd= , Sd = Ss
𝐹𝑠

16𝑡 𝜋𝑑 3 𝑆𝑢
T= 65,000 psi Ss= 𝜋𝑑3 , Fs= 16𝑡
1
𝜋(2.5)3 (65,000)( )
12
Required: Fs= = 0.37
16(45,000)

Fs=?

17. If the ultimate shear strength if a steel is 43,000 psi, what force is needed to punch a 0.85 in.
diameter hole in a 0.625 in. thick plate?

Given: Required:

Su = 43,000 psi F=?

D = 0.85 in

t = 0.625 in

Solution:
𝑆𝑢
Sd= 𝐹𝑠 , Sd = Ss

F = πdtSu

F= π(0.85)(.625)(43,000)

F= 71,765.76 lbs

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18. If the ultimate shear stress of the steel plate is 35,000 psi. What force is necessary to punch a 1.5 in.
diameter hole in a 1/8 thick plate?

Given: Solution:
𝐹 𝐹
Su= 35,000 psi Su = 𝐴 = 𝜋𝑑𝑡 1

D = 1.5 in. F = πdt Su

t= 1/8 in. F= π(1.5in)(1/8in)(35,000psi)

Required: F= ? F= 20,616.70 lbs.

19. If the ultimate shear stress of the steel plate is 35,000 psi. What force is necessary to punch a 1.5 in.
diameter hole in a 1/8 thick plate?

Given: Solution:
𝐹 𝐹
Su= 35,000 psi Su = 𝐴 = 𝜋𝑑𝑡 1

D = 1.5 in. F = πdt Su

t= 1/8 in. F= π(1.5in)(1/8in)(35,000psi)

Required: F= ? F= 20,616.70 lbs

20. The maximum torque that can be applied to a hollow circular steel shaft of 120 mm outside
diameter and 80mm inside diameter without exceeding a shearing stress of 80 MPa.

Given: Required: T=?

D= 120 mm

d= 80 mm

Ss= 80 MPa

Solution:
16𝑇𝐷 16𝑇(120 𝑚𝑚)
Ss = 𝜋(𝐷4 −𝑑4 ) = 80= 𝜋(1204 −804 )

T= 21781709.06 N-mm= 2,1781.71 N-m

SHAFTINGS

1. If the ultimate shear stress of the steel plate is 35,000 psi. What force is necessary to punch a 1.5 in.
diameter hole in a 1/8 thick plate?

Given: Solution:
𝐹 𝐹
Su= 35,000 psi Su = 𝐴 = 𝜋𝑑𝑡 1

D = 1.5 in. F = πdt Su

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