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

Plan the Solution
Whenever a cross section includes discrete fasteners (such as nails, screws, or bolts), the
shear flow formula [Equation (9.12)] and the related fastener force–spacing relationship
[Equation (9.13)] will be helpful in assessing the suitability of the fasteners for the
intended purpose. To determine the shear force acting in the fasteners, we must first identify
those portions of the cross section which are held in place by the fasteners. For the
basic tee-shaped cross section considered here, it is evident that the top flange board is
secured to the stem board by the lag screws. If the entire cross section is to be in equilibrium,
the resultant force acting in the horizontal direction on the flange board must be
transmitted to the stem board by shear forces in the fasteners. In the analysis that follows,
a short length of the beam equal to the spacing interval of the lag screws will be considered
to determine the shear force that must be supplied by each fastener to satisfy equilibrium.
V=
500 lb
M=
2,500 lb.in.
12.90 psi
C
55.91 psi
Compressive C
bending stress
344 lb
350
Tensile
bending stress
30.10 psi
y
y
Tensile
bending stress
30.10 psi
12.90 psi
C
y
5 in.
5 in.
500 lb
D
500 lb
V f
Shear force
from
fastener
5 in.
D
500 lb
D
x
x
x
SolutioN
Free-Body Diagram at C
To better understand the function of the fasteners, consider a free-body diagram
(FBD) cut at section C, 5 in. from the end of the overhang. This FBD
includes one lag screw fastener. The external 500 lb concentrated load creates
an internal shear force V = 500 lb and an internal bending moment
M = 2,500 lb · in. acting at C in the directions shown.
The internal bending moment M = 2,500 lb · in. creates tensile bending
stresses above the neutral axis (i.e., the z centroidal axis) and compressive
bending stresses below the neutral axis. These key normal stresses
acting on the flange and the stem can be calculated from the flexure formula
and are labeled in the figure.
The approach outlined in Section 9.2 can be used to compute the
resultant horizontal force created by the tensile bending stresses acting on
the flange. The resultant force has a magnitude of 344 lb, and it pulls the
flange in the −x direction. If the flange is to be in equilibrium, additional
force acting in the +x direction must be present. This added force is
provided by the shear resistance of the lag screw. With that force denoted
as V f , equilibrium in the horizontal direction dictates that V f = 344 lb.
In other words, equilibrium of the flange can be satisfied only if 344 lb
of resistance from the stem flows through the lag screw into the flange.
The magnitude of V f determined here is applicable only to a 5 in. long
segment of the beam. If a segment longer than 5 in. were considered, the
internal bending moment M would be larger, which in turn would create
larger bending stresses and a larger resultant force. Consequently, it is
convenient to express the amount of force that must flow to the connected
portion in terms of the horizontal resistance required per unit of beam
span. The shear flow in this instance is
344 lb
q = = 68.8 lb/in.
(a)
5in.
The preceding discussion is intended to illuminate the behavior of a builtup
beam. A basic understanding of the forces and stresses involved in this
type of flexural member facilitates the proper use of the shear flow formula
[Equation (9.12)] and the fastener force–spacing relationship [Equation
(9.13)] to analyze and design fasteners in built-up flexural members.