Mohr’s circle

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Stresses in thin circular cylinder and spherical shell

Stresses in thin circular cylinder and spherical shell

Thin Wall Pressure Vessels

Thin wall pressure vessels are in fairly common use. We would like to consider two specific types. Cylindrical pressure vessels, and spherical pressure vessels. By thin wall pressure vessel we will mean a container whose wall thickness is less than 1/10 of the radius of the container. Under this condition, the stress in the wall may be considered uniform.

We first look at a cylindrical pressure vessel shown in Diagram 1, where we have cut a cross section of the vessel, and have shown the forces due to the internal pressure, and the balancing forces due to the longitudinal stress which develops in the vessel walls. (There is also a transverse or circumferential stress which develops, and which we will consider next.)

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The longitudinal stress may be found by equating the force due to internal gas/fluid pressure with the force due to the longitudinal stress as follows:
P(A) = clip_image002(A’); or P(3.1416 * R2) = clip_image002[1](2 * 3.1416 * R * t), then canceling terms and solving for the longitudinal stress, we have:
clip_image002[2]= P R / 2 t ; where
P = internal pressure in cylinder; R = radius of cylinder, t = wall thickness

To determine the relationship for the transverse stress, often called the hoop stress, we use the same approach, but first cut the cylinder lengthwise as shown in Diagram 2.

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We once again equate the force on the cylinder section wall due to the internal pressure with the resistive force which develops in walls and may be expressed in terms of the hoop stress, clip_image004. The effective area the internal pressure acts on may be consider to be the flat cross section given by (2*R*L). So we may write:
P(A) = clip_image004[1](A’); or P(2*R*L) = clip_image004[2](2*t*L), then canceling terms and solving for the hoop stress, we have:
clip_image004[3]= P R / t ; where
P = internal pressure in cylinder; R = radius of cylinder, t = wall thickness
We note that the hoop stress is twice the value of the longitudinal stress, and is normally the limiting factor. The vessel does not have to be a perfect cylinder. In any thin wall pressure vessel in which the pressure is uniform and which has a cylindrical section, the stress in the cylindrical section is given by the relationships above.

Example A: A thin wall pressure vessel is shown in Diagram 3. It’s cylindrical section has a radius of 2 feet, and a wall thickness of the 1″. The internal pressure is 500 lb/in2. Determine the longitudinal and hoop stresses in the cylindrical region.

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Solution:We apply the relationships developed for stress in cylindrical vessels.
clip_image002[3]= P R / 2 t = 500 lb/in2 * 24″/2 * 1″ = 6000 lb/in2.
clip_image004[4]= P R / t = 500 lb/in2 * 24″/ 1″ = 12, 000 lb/in2.

Next we consider the stress in thin wall spherical pressure vessels. Using the approach as in cylindrical vessels, in Diagram 4 we have shown a half section of a spherical vessel. If we once again equal the force due to the internal pressure with the resistive force expressed in term of the stress, we have:

clip_image006

P(A) = clip_image007(A’); or P(3.1416 * R2) = clip_image007[1](2 * 3.1416 * R * t), then canceling terms and solving for the stress, we have:
clip_image007[2]= P R / 2 t ; where
P = internal pressure in sphere; R = radius of sphere, t = wall thickness
Note that we have not called this a longitudinal or hoop stress. We do not do so since the symmetry of the sphere means that the stress in equal in what we could consider a longitudinal and/or transverse direction.

Example B. In Example A, above, if the radius of the spherical section of the container is also 2 feet, determine the stress in the spherical region.

Solution: We apply our spherical relationship:
clip_image007[3]= P R / 2 t = 500 lb/in2 * 24″/2 * 1″ = 6000 lb/in2.
Thus in the container in Example A, the limiting (maximum) stress occurs in the in the cylindrical region and has a value of 12,000 lb/in2.

Pressure Vessels – Problem Assignment 1

1. A welded water pipe has a diameter of 8 ft. and a wall of steel plate 3/4 in. thick.  After fabrication, this pipe was tested under an internal  pressure of 230 psi.  Calculate the circumferential stress developed in the walls of the pipe. (14,700 psi)

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2. Calculate the tensile stresses developed (circumferential and longitudinal) in the walls of a cylindrical boiler 5 ft. in diameter with a wall thickness of 1/2 in.   The boiler is subjected to an internal gage pressure of 155 psi. (9300 psi, 4650 psi)

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3. Calculate the internal water pressure that will burst a 15 in. diameter cast-iron water pipe if the wall thickness is 1/2 in.  Use an ultimate tensile strength of 62,500 psi. for the pipe. (8333 psi)

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4. Calculate the wall thickness required for a 5 ft. diameter cylindrical steel tank containing gas at an internal gage pressure of 600 psi.  The allowable tensile stress for the steel is 17,500 psi. (1.03″)

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5. A spherical gas container, 50 ft. in diameter, is to hold gas at a pressure of 40 psi.  Calculate the thickness of the steel wall required.  The allowable tensile stress is 20,000 psi. (.3″)

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6. Calculate the tensile stresses  (circumferential and longitudinal) developed in the walls of a cylindrical pressure vessel.  The inside diameter is 15 in.   The wall thickness is 1/4 in.  The vessel is subjected to an internal gage pressure of 450 psi. and a simultaneous external axial tensile load of 45,000 lb. (13,500 psi, 10,570 psi)

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7. A thin wall pressure vessel is composed of two spherical regions and a cylindrical region is shown in Diagram 5.

The larger spherical region has a radius of 3 ft and a wall thickness of 1.25″. The smaller spherical region has a radius of 2.5 ft and a wall thickness of 3/4″. The cylindrical region has a radius of 2 ft and a wall thickness of 1/2″.

Determine the axial and hoop stress in the cylindrical region, and the wall stresses in the two spherical regions. Which stress is the largest? (1: 5,760 psi, 2: 19,200 psi, 9600 psi, 3: 8,000 psi)

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8. A thin wall pressure vessel is composed of two spherical regions and is shown in Diagram 4. The larger spherical region has a radius of 3 ft , and the smaller spherical region has a radius of 2.5 ft. The vessel is made of steel with an allowable tensile stress of 24,000 lb/in2

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If we wish the maximum stress in both spherical sections to at the allowable stress, Determine wall thickness for each spherical region.(.45″, .375″)

solved examples_shaft

Torsion: Deformation – Angle of Twist

Another effect of applying an external torque to a shaft is a resulting deformation or twist as the material is stressed. The resulting shaft deformation is expressed as an Angle of Twist of one end of the shaft with respect to the other.

In Diagram 1 we have shown a section of a solid shaft. An external torque T is applied to the left end of the shaft, and an equal internal torque T develops inside the shaft. Additionally there is a corresponding deformation (angle of twist) which results from the applied torque and the resisting internal torque causing the shaft to twist through an angle, phi, shown in Diagram 1.

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The angle of twist may be calculated from:
clip_image002= T L / J G ; where
T = the internal torque in the shaft
L = the length of shaft being “twisted”
J = the polar moment of inertia of the shaft
G = the Modulus of Rigidity (Shear Modulus) for the material, for example for steel and brass we have, G steel = 12 x 106 lb/in2, G brass = 6 x 106 lb/in2.

We will now look at an example of determining the angle of twist in a shaft. In Diagram 2a we have shown a solid steel circular shaft with an external torque of 1000 ft-lb. being applied at each end of the shaft, in opposite directions.

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The shaft has a diameter of 1.5 inch. We would like to determine the angle of twist of end B with respect to end A.

To find the angle of twist we first determine the internal torque in the shaft. We cut the shaft a distance x feet from the left end, and make a free body diagram of the left section of the shaft – shown in Diagram 2b. From the free body diagram, we see that the internal torque must be 1000 ft-lb. to satisfy rotational equilibrium.

We next apply the Angle of Twist formula: clip_image002[1]= T L / J G ; where
T = 1000 ft-lb. = 12,000 in-lb.
L = 2 ft. = 24 inches
J = polar moment of inertia = (clip_image004/32) d4 for a solid shaft = (3.1416/32) (1.54in4) = .5 in4.
G steel = 12 x 106 lb/in2
Then,
clip_image002[2]= T L / J G = (12,000 in-lb.* 24 in) / (.5 in4 * 12 x 106 lb/in2) = .048 radians = 2.75o.
The angle of twist will have units of radians, and in this problem is clockwise with respect to end A as shown in Diagram 3.

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We also take a moment to calculate the maximum shear stress in the shaft, just out of interest in it’s value.
clip_image006= T r / J = 12,000 in-lb. * .75 in./.5 in4. = 18,000 lb/in2.

Shear Stress  – Example 1

Part I. In Diagram 1a we have shown a solid shaft with what we will call a driving external torque of 1000 ft-lb. at end A, and a load torque of 1000 ft-lb. at end B. The shaft is in equilibrium. We would like to determine the maximum transverse shear stress in the shaft due to the applied torque.

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To solve, we first need to determine the internal torque in the shaft. We cut the shaft a distance x from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 1b. Where we cut the shaft there is an internal torque, which in this case must be equal and opposite to the torque at end A for equilibrium. So for this shaft the value of the internal torque is equal to the value of the externally applied torque.

We next simply apply the torsion formula for the shear stress: clip_image006[1]= T r / J; where:
T is the internal torque in that section of the shaft = 1000 ft-lb = 12,000 in-lb.
r = the radial distance from the center of the shaft to the point where we wish to find the shear stress. In this problem the outer edge of the shaft since that is where the transverse shear stress is a maximum; r = 1 in.
J = polar moment of inertia = (clip_image004[1]/32) d4 for a solid shaft = (3.1416/32) (24in4) = 1.57 in4.
So, clip_image006[2]= T r / J = 12,000 in-lb. * 1 in./1.57 in4. = 7,640 lb/in2.
This is the Maximum Transverse (and longitudinal) Shear Stress in the shaft.

Part II
We now would like to consider the case where the shaft is not solid, but a hollow shaft with an outer diameter of 2″ and an inner diameter of 1″, as shown in Diagram 2a. We still apply the same driving and load torque, and still have the same value of the internal torque, as is shown in Diagram 2b.

clip_image008

We next apply the torsion formula for the shear stress for the hollow shaft:
clip_image006[3]= T r / J; where we observe that all the values are the same as in part one, except for the value of J, the polar moment of inertia.
T is the internal torque in that section of the shaft = 1000 ft-lb = 12,000 in-lb.
r
= the radial distance from the center of the shaft to the point where we wish to find the shear stress. In this problem the outer edge of the shaft since that is where the transverse shear stress is a maximum; r = 1 in.
J
= polar moment of inertia = (3.1416/32) [do4 – di4] for a hollow shaft = (3.1416/32) [(24in4) – (1.04in4)] = 1.47 in4.
So, clip_image006[4]= T r / J = 12,000 in-lb. * 1 in./1.47 in4. = 8,150 lb/in2.

This then is the Maximum Transverse (and longitudinal) Shear Stress in the hollow shaft.

Shear Stress  – Example 2

In Diagram 1 we have shown a solid compound shaft with what we will call the driving external torque of 1600 ft-lb. acting at point B, and load torque of 400 ft-lb. at end A, 900 ft-lb. at point C, and 300 ft-lb. at end D. Notice that the shaft is in rotational equilibrium. We would like to determine the maximum transverse shear stress in each section of the shaft due to the applied torque.

clip_image009

To solve, we first need to determine the internal torque in each section of the shaft. We cut the shaft a distance 0′ < x < 1′ from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 2. Where we cut the shaft there is an internal torque, which in this case must be equal and opposite to the torque at end A for equilibrium. So for this shaft the value of the internal torque is equal to the value of the externally applied torque.

clip_image010

We next apply the torsion formula for the shear stress: clip_image006[5]= T r / J; where:
T is the internal torque in that section of the shaft = 400 ft-lb. = 4,800 in-lb.
r = the radial distance from the center of the shaft to the point where we wish to find the shear stress. In this problem r is to the outer edge of the shaft since that is where the transverse shear stress is a maximum; r = .5 in.
J = polar moment of inertia = (clip_image004[2]/32) d4 for a solid shaft = (3.1416/32) (14in4) = .098 in4.
So, clip_image006[6]= T r / J = 4,800 in-lb. * .5 in./.098 in4. = 24,500 lb./in2.

This value is the Maximum Transverse Shear Stress in the shaft section AB, and it falls in a reasonable range for allowable shear stresses for metals.

We now determine the internal torque in the next section of the shaft. We cut the shaft a distance 1′ < x < 3′ from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 3. Where we cut the shaft there is an internal torque, and by mentally summing torque, we see that in order to have rotational equilibrium we must have an internal torque in section BC of 1200 ft-lb. acting in the direction shown.

clip_image011

We apply the torsion formula for the shear stress once again: clip_image006[7]= T r / J; where:
T is the internal torque in that section of the shaft = 1,200 ft-lb. = 14,400 in-lb.
r = the radial distance to the outer edge of the shaft since that is where the transverse shear stress is a maximum; r = 1 in.
J = polar moment of inertia = (clip_image004[3]/32) d4 for a solid shaft = (3.1416/32) (24in4) = 1.57 in4.
So, clip_image006[8]= T r / J = 14,400 in-lb. * 1 in./ 1.57 in4. = 9,170 lb./in2.

This then is the Maximum Shear Stress in shaft section BC. We note that even though the internal torque is much larger in section BC as compared to section AB, because of the size of the shaft in section BC, the shear stress is much lower in BC.

We now determine the internal torque in the next section of the shaft. We cut the shaft a distance 3′ < x < 4′ from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 4. Where we cut the shaft there is an internal torque. From the Free Body Diagram we see that in order to have rotational equilibrium we must have an internal torque in section CD of 300 ft-lb. acting in the direction shown.

clip_image012

We apply the torsion formula for the shear stress: clip_image006[9]= T r / J; where:
T is the internal torque in that section of the shaft = 300 ft-lb. = 3,600 in-lb.
r = the radial distance from the center of the shaft to the outer edge of the shaft since that is where the transverse shear stress is a maximum; r = .25 in.
J = polar moment of inertia = (clip_image004[4]/32) d4 for a solid shaft = (3.1416/32) (.54in4) = .0061 in4.
So, clip_image006[10]= T r / J = 3,600 in-lb. * .25 in./ .0061 in4. = 147,500 lb./in2.

This is the Maximum Shear Stress in shaft section CD. We note that this is much larger than the ultimate shear stress most metals, thus this section of the shaft would fail – a larger diameter is needed to carry the torque.