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Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

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z

8.45 kN·m

15.6 kN·m

H

K

10.8 kN·m

Equivalent moments at the section that contains

points H and K.

y

x

• One moment component has a magnitude of (13 kN)(1.2 m) =

15.6 kN · m and acts about the +y axis.

• A second moment component has a magnitude of (13 kN)(0.65 m) =

8.45 kN · m and acts about the +z axis.

For the coordinate system used here, the longitudinal axis of the

pipe extends in the z direction; therefore, the moment component acting

about the z axis is recognized as a torque, while the components about the

x and y axes are simply bending moments.

Alternative method: The moments equivalent to the two loads at A can be

calculated systematically by the use of position and force vectors. The position

vector r from the section of interest to point A is r A = −0.65 m j + 1.2 m k.

The load at A can be expressed as the force vector F A = 13 kN i − 9 kN j.

The moment produced by F A can be determined from the cross product

M A = r A × F A :

i j k

MA = rA × FA

= 0 −0.65 1.2

13 −9 0

= 10.8 kN⋅ m i + 15.6 kN⋅ m j+ 8.45 kN⋅m

k

Section Properties

The outside diameter of the pipe is D = 200 mm, and the inside diameter is d = 176 mm.

The moment of inertia and the polar moment of inertia for the cross section are, respectively,

as follows:

π

I D d

64 [ ] π

= − = [(200 mm) − (176 mm) ] = 31,439,853 mm

64

4 4 4 4 4

π

J D d

32 [ ] π

= − = [(200 mm) − (176 mm) ] = 62,879,706 mm

32

4 4 4 4 4

Stresses at H

The equivalent forces and moments acting at the section of interest will be sequentially

evaluated to determine the type, magnitude, and direction of any stresses created at H.

Transverse shear stress is associated with the 13 kN shear force acting in the + x direction

at the section of interest. From Equation (9.10), the first moment of area at the centroid

for a hollow circular cross section is

y

1

Q D d

12 [ ] 1

= − = [( 200 mm) − ( 176 mm ) ] = 212,352 mm

12

3 3 3 3 3

The shear stress formula [Equation (9.2)] is used to calculate the shear

stress:

z

13 kN

H

3.659 MPa

K

x

3

VQ (13kN)(212,352 mm )(1, 000 N/kN)

τ = =

= 3.659 MPa

4

It (31, 439,853 mm )(200 mm − 176 mm)

y

Although shear stresses are associated with the 9 kN shear force that acts

in the −y direction, the shear stress at point H is zero.

648

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