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

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Shear Flow Formula

The shear flow formula rewritten as

y

5 in.

(typ)

500 lb

q

VQ

= (b)

I z

and the fastener force–spacing relationship

qs ≤

nV f f

(c)

500 lb

x

A B C D

5 in.

1,000 lb

7 ft 7 ft

500 lb

will be employed to determine the shear force V f

produced in the lag screws of the built-up beam.

Appropriate values for the terms appearing in

these equations will now be developed.

Beam internal shear force V: The shear-force

and bending-moment diagrams for the simply

supported beam are shown. The V diagram

reveals that the internal shear force has a constant

magnitude of V = 500 lb throughout the entire

span of the beam.

V

M

−500 lb

−3,500

lb.ft

First moment of area, Q: Q is calculated for the

portion of the cross section connected by the lag screw. Consequently, Q is calculated

for the flange board in this situation:

8 in.

y

Q = (8 in.)(2 in.)(2.5 in.) = 40 in. 3

Fastener spacing interval s: The lag screws are installed at 5 in. intervals along the

span; therefore, s = 5 in.

2.5 in.

z

2 in.

8 in.

Shear flow q: The shear flow that must be transmitted from the stem to the flange

through the fastener can be calculated from the shear flow formula:

2 in.

3

VQ (500 lb)(40 in. )

q = = = 68.8 lb/in.

(d)

4

I 290.667 in.

z

Notice that the result obtained in Equation (d) from the shear flow formula is identical to

the result obtained in Equation (a). While the shear flow formula provides a convenient

format for calculation purposes, the underlying flexural behavior that it addresses may not

be readily evident. The preceding investigation using an FBD of the beam at C may help

to enhance one’s understanding of this behavior.

Fastener shear force V f : The shear force that must be provided by the fastener can be

calculated from the fastener force–spacing relationship. The beam is fabricated with one

lag screw installed in each 5 in. interval; it follows that n f = 1, and we have

qs ≤

nV f f

qs (68.8 lb/in.)(5 in.)

∴ Vf

= = = 344 lb perfastener

Ans.

n 1fastener

f

351

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