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

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The directions and intensities of the shear stress in the inverted-tee shape are shown

in the accompanying sketch. Note that the shear stress in the tee stem is distributed parabolically

while the shear stress in the flange is distributed linearly. At the junction of the

stem and the flange, the shear stress intensity is cut in half as shear flows outward in two

opposing directions.

19.46 MPa

30.67 MPa

33.72 MPa

7.80 MPa 15.60 MPa

31.20 MPa

15.60 MPa 7.80 MPa

ExAmpLE 9.8

A beam with the cross section shown is subjected to a vertical

shear force V = 65 kN. The location of the centroid appears on the

sketch, and the moment of inertia of the cross section about the z

centroidal axis is I = 2,532,789 mm 4 . Determine the shear stress in

the cross section at point a.

Plan the Solution

To calculate the shear stress at a, we must cut an FBD through the

cross section so that the shear stress at a is exposed. We will consider

three different FBDs that could be used to determine the

quantity Q necessary to calculate the shear stress from the formula

τ = VQ/It.

30.5 mm

74.5 mm

b

c

a

d

6 mm

50 mm

z

50 mm

180 mm

y

V

y

6 mm

6 mm

99 mm

105 mm

SolutioN

Approach 1

The first FBD that we will consider is obtained by making two

vertical cuts through the cross section. One vertical cut will be

made at point a, and a second vertical cut will be made on the y

centroidal axis, which is an axis of symmetry for the cross section

considered here.

The FBD exposed by these two cuts is shown in the accompanying

figure. The pieces of the cross section have been separated

in the figure to help illustrate the shear stresses on the interior

surfaces. The directions of the shear flow and the shear stresses

on the various surfaces are also indicated. Notice that shear stress

on the vertical plane of symmetry is zero.

The shear stress at point a will be calculated from the formula

τ = VQ/It. the shear force V and the moment of inertia I

to be used in the formula are the same for any FBD that we

might consider.

27.5 mm b a

c

d

Vertical plane of symmetry

b

c

d

τ = 0

a

z

V

τ = 0

z

6 mm

y

V

x

365

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