Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 1, Problem 1.8.13P

A steel riser pipe hangs from a drill rig located offshore in deep water (see figure). Separate segments are joined using bolted flange plages (see figure part b and photo). Assume that there are six bolts at each pipe segment connection. Assume that the total length of the riser pipe is L = 5000 ft: outer and inner diameters are d2= l6in.and d1= 15 in.; flange plate thickness t1= 1.75 in.; and bolt and washer diameters are db= 1.125 in..and dW. = 1.875 in., respectively.

(a) If the entire length of the riser pipe is suspended in air. find the average normal stress a in each bolt, the average bearing stress abbeneath each washer, and the average shear stress t through the flange plate at each bolt location for the topmost bolted connection.

(b) If the same riser pipe hangs from a drill rig at sea. what are the normal, bearing, and shear stresses in the connection? Obtain the weight densities of steel and sea water from Table I-1. Appendix I. Neglect the effect of buoyant foam casings on the riser pipeChapter 1, Problem 1.8.13P, A steel riser pipe hangs from a drill rig located offshore in deep water (see figure). Separate

(a)

Expert Solution
Check Mark
To determine

The stresses on a steel riser pipe suspended in air.

Answer to Problem 1.8.13P

The average normal stress in each bolt is 69500psi_ .

The bearing stress for each washer is 39100psi_ .

The average shear stress at each bolt location in flange is 21000psi_ .

Explanation of Solution

Given information:

Steel riser pipe length is 5000ft , inside pipe diameter is 15in , outside pipe diameter is 16in , flange thickness is 1.75in diameter of the bolt is 1.125in, and diameter of washer is 1.875in .

Write the expression for average normal stress in each bolt for steel riser pipe suspended in air.

  σb=Ap×LP×ρPAb×n  .........(I)

Here, the area of the riser pipe is Ap , the length of pipe is LP , weight density of pipe is ρP , the area of a bolt is Ab, and number of bolts are n .

Write the expression for the area of pipe.

  Ap=π4(d22d12)  .........(II)

Here, the area of pipe is Ap , the inner diameter of pipe is d1 and the outer diameter of pipe is d2 .

Write the expression for the area of a bolt.

  Ab=π4(db2)   .........(III)

Here, the diameter of bolt is db .

Write the expression for bearing stress for each washer.

  σw=Ap×LP×ρPAw×nw  .........(IV)

Here, the area of a washer is Aw and number of washers are nw .

Write the expression for the washer area.

  Aw=π4(db2dw2)  .........(V)

Here, the diameter of bolt is db and the diameter of washer is dw .

Write the expression for average shear stress at each bolt location in flange.

  τ=Ap×LP×ρPtf×dw×nw   .........(VI)

Here, the thickness of flange is tf .

Calculation:

Substitute 16in for d2 and 15in for d1 in Equation (II).

  Ap=π4( ( 16in )2 ( 15in )2)=π4(256 in2225 in2)=24.347in2

Substitute 1.125in for db in Equation (III).

  Ab=π4(1.125in)2=(0.7857×1.265625)in2=0.994in2

Substitute 5000ft for Lp , 0.2836lb/ft3 for ρP , 0.994in2 for Ab , 24.347in2 for Ap and 6 for n in Equation (I).

  σb=24.347 in2×5000ft×0.2836 lb/in30.994 in2×6=24.347 in2×5000ft( 12in 1ft )×0.2836 lb/in30.994 in2×6=414288.552lb0.994 in2×6=69500lb/in2( 1psi 1 lb/in 2 )

  =69500psi

Substitute 1.875in for db and 1.125in for dw in Equation (V).

  Aw=π4( ( 1.875in )2 ( 1.125in )2)Aw=0.785(3.515625 in21.265625 in2)Aw=1.7671in2

Substitute 5000ft for Lp , 0.2836lb/in3 for ρP , 1.7671in2 for Aw , 24.347in2 for Ap and 6 for nw in Equation (IV).

  σw=24.347 in2×5000ft×0.2836 lb/in31.7671 in2×6=24.347 in2×5000ft( 12in 1ft )×0.2836 lb/in31.7671 in2×6=39100( lb/in2× 1psi 1 lb/in 2 )=39100psi

Substitute 5000ft for Lp , 0.2836lb/in3 for ρP , 1.875in for dw , 24.347in2 for Ap and 6 for nw and 1.75in for tw in Equation (IV).

  τ=24.347 in2×5000ft×0.2836 lb/in31.75in×1.875in×6=24.347 in2×5000ft( 12in 1ft )×0.2836 lb/in31.75in×1.875in×6=414288.552lb1.75in×1.875in×6=21000lb/in2( 1psi 1 lb/in 2 )

  =21000psi

Conclusion:

The average normal stress in each bolt is 69500psi_ .

The bearing stress for each washer is 39100psi_ .

The average shear stress at each bolt location in flange is 21000psi_ .

(b)

Expert Solution
Check Mark
To determine

The stresses on a steel riser pipe hanging from a drill rig at sea water.

Answer to Problem 1.8.13P

The average normal stress in each bolt is 21000psi_ .

The average bearing stress beneath each washer is 34000psi_ .

The average shear stress at each bolt location in flange is 21000psi_ .

Explanation of Solution

Write the expression for average normal stress in each bolt for a steel riser pipe hanging from a drill rig in sea water.

  σb=Ap×LP×ρNPAb×n  .........(VII)

Here, the net weight density is ρNP .

Write the expression for average bearing stress beneath each washer.

  σw=Ap×LP×ρNPAw×nw  .........(VIII)

Here, the area of a washer is Aw and number of washers are nw .

Write the expression for average shear stress at each bolt location in flange.

  τ=Ap×LP×ρNPtf×dw×nw   .........(IX)Calculation:

Substitute 5000ft for Lp , 0.28lb/in3 for ρP , 0.994in2 for Ab , 24.347in2 for Ap and 6 for n in Equation (VII).

  σb=24.347 in2×5000ft×0.28 lb/in30.994 in2×6=24.347 in2×5000ft( 12in 1ft )×0.28 lb/in30.994 in2×6=60400lb/in2( 1psi 1 lb/in 2 )=60400psi

Substitute 5000ft for Lp , 0.28lb/in3 for ρNP , 1.7671in2 for Aw , 24.347in2 for Ap and 6 for nw in Equation (VIII).

  σb=24.347 in2×5000ft×0.28 lb/in30.994 in2×6=24.347 in2×5000( 12in 1ft )×0.28 lb/in30.994 in2×6=34000lb/in2( 1psi 1 lb/in 2 )=34000psi

Substitute 5000ft for Lp , 0.28lb/in3 for ρNP , 1.875in for dw , 24.347in2 for Ap and 6 for nw and 1.75in for tw in Equation (IX).

  τ=24.347 in2×5000ft×0.28 lb/in31.75in×1.875in×6=24.347 in2×5000ft( 12in 1ft )×0.28 lb/in31.75in×1.875in×6=18300( lb/in2× 1psi 1 lb/in 2 )=21000psi

Conclusion:

The average normal stress in each bolt 21000psi_ .

The average bearing stress beneath each washer 34000psi_

The average shear stress at each bolt location in flange 21000psi_ .

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Chapter 1 Solutions

Mechanics of Materials (MindTap Course List)

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