   # Two sections of steel drill pipe, joined by bolted flange plates at B, arc subjected to a concentrated torque 4000 kip-in. at x = 3 ft, and a uniformly distributed torque t 0 = 50 kip-ft/ft is applied on pipe BC . Let G = 11,800 ksi and assume that pipes AB and BC have the same inner diameter, d = 12 in. Pipe AB has a thickness t A B = 3/4 in., and pipe BC has a thickness t B C = 5/8 in. Find the reactive torques at A and C and the maximum shear stresses in each segment. ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347 ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 3, Problem 3.8.9P
Textbook Problem
413 views

## Two sections of steel drill pipe, joined by bolted flange plates at B, arc subjected to a concentrated torque 4000 kip-in. at x = 3 ft, and a uniformly distributed torque t0= 50 kip-ft/ft is applied on pipe BC. Let G = 11,800 ksi and assume that pipes AB and BC have the same inner diameter, d = 12 in. Pipe AB has a thickness tAB= 3/4 in., and pipe BC has a thickness tBC= 5/8 in. Find the reactive torques at A and C and the maximum shear stresses in each segment. To determine

The reaction torque at A and the reaction torque at B .

The twist angle at the center of the bar.

### Explanation of Solution

Given information:

The concentrated torque is 400kipft/ft , diameter of shaft is 12in , diameter of the right shaft is 25mm , length is 3m and the distributed torque is 150kipft/ft , thickness of left pipe is 34in and the thickness of right pipe is 58in .

Write the expression for polar moment of inertia of left bar.

IPAB=π32[(d+2tAB)4d4]........ (I)

Here, polar moment of inertia of the left bar is IPAB and the thickness of the left bar is tAB .

Write the expression for polar moment of inertia of right bar.

IPBC=π32[(d+2tBC)4d4] ........ (II)

Here, polar moment of inertia of the right bar is IPBC and the thickness of the right bar is tBC .

Write the equilibrium Equation for torques.

TATC50LBC+4000=0 ........ (III)

Here, torque at A is TA , torque at C is TC and the length between B and C is LBC .

Write equilibrium equation for AD .

Here, torque at D is TD .

Write equilibrium equation for DB .

TA+4000+TDB=0TDB=TA4000........ (V)

Write the equilibrium Equation for BC .

TA+400050(Lx)+TBC=0........ (VI)

Here, torque in segment BC is TBC , length is L and the distance from left end is x .

Write the expression for angle of twist for AD .

Here, angle of twist is ϕD , modulus of rigidity is G and length of section AD is LAD .

Write the expression for angle of twist for B .

ϕB=TDBLDBGIPAB ....... (VIII)

Here, angle of twist is ϕB , modulus of rigidity is G and length of section DB is LDB .

Substitute TA4000 for TDB in Equation (VIII).

ϕB=(TA4000)LDBGIPAB ....... (IX)

Write the expression for angle of twist for BC .

ϕBC(x)=LABxTA7600+50ςGI PBCdς=1GIPBC[TA7600ς+25ς2]LABx=1GIPBC((TAx7600x+25x2)(TALAB7600LAB+25LAB2)) ....... (X)

Write the compatibility equation for angle of twist.

ϕD+ϕB+ϕC=0…... (XI)

Here, angle of twist for C is ϕC .

Write the expression for maximum shear stress in AB .

τmax,AB=Tmax,ABIPAB×d+2tAB2........ (XII)

Here, maximum shear stress is τmax,AB .

Write the expression for maximum shear stress in BC .

τmax,BC=Tmax,BCIPBC×d+2tBC2 ........ (XIII)

Here, maximum shear stress is τmax,BC.

Calculation:

Substitute 12in for d and 34in for tAB in Equation (I).

IPAB=π32[(12in+2× 3 4in)4(12in)4]=π32(12479.0625in4)=1225.13in4

Substitute 12in for d and 58in for tBC in Equation (II).

IPBC=π32[(12in+2× 5 8in)4(12in)4]=π32(10086.19141in4)=990.21in4

Substitute 4ft for LBC in Equation (III).

TATC50×4ft+4000=0TA+TC=1600........ (XIV)

Substitute 6ft for L in Equation (VI).

TA+400050((6ft)( 12in 1ft)x)+TBC=0TBC=TA7600+50x......... (XV)

Substitute 3ft for LAD , 11800ksi for G and 1225.13in4 for IPAB in Equation (VII).

ϕD=TA×3ft11800ksi×1225.13in4=TA×(3ft)( 12in 1ft)11800ksi×1225

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