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

a maximum intensity of (4.5 MPa − 1.5 MPa) = 3.0 MPa. The resultant
force acting on member (1) at section C is found from
321
RESuLTANT FORCES PROduCEd
by bENdINg STRESSES
2 2
1
2 2
Resultant FC
= (1.5 N/mm )(6,000 mm ) + (3.0 N/mm )(6,000 mm )
2
= 18,000 N = 18kN
The resultant forces caused by the bending stresses on member (1)
are shown in Figure 9.3c. Notice that the resultant forces are not
equal in magnitude. Why are these resultant forces unequal? The
resultant force on section C is larger than the resultant force on section
B because the internal bending moment M C is larger than M B .
The resultant forces F B and F C will be equal in magnitude only
when the internal bending moments are the same on sections B and C.
Is member (1) of beam segment BC in equilibrium? This portion
of the beam is not in equilibrium, because ∑F x ≠ 0. How much additional
force is required to satisfy equilibrium? An additional
force of 6 kN in the horizontal direction is required to satisfy equilibrium
for member (1). Where must this additional force be
located? All normal stresses acting on the two vertical faces (b–c and
e–f ) have been considered in the calculations of F B and F C . The bottom
horizontal face c–f is a free surface that has no stress acting on
it. Therefore, the additional 6 kN horizontal force required to satisfy
equilibrium must be located on horizontal surface b–e, as shown
in Figure 9.4. This surface is the interface between member (1) and
member (2). What is the term given to a force that acts on a surface
that is parallel to the line of action of the force? The 6 kN horizontal
force acting on surface b–e is termed a shear force. Notice that
this force acts in the same direction as the resultants of the bending
stress—that is, parallel to the x axis.
What lessons can be drawn from this simple investigation?
In those beam spans in which the internal bending moment is not
constant, the resultant forces acting on portions of the cross section
will be unequal in magnitude. Equilibrium of these portions can be
satisfied only by an additional shear force that is developed internally
in the beam.
In the sections that follow, we will discover that this additional
internal shear force required to satisfy equilibrium can be developed
in two ways. The internal shear force can be the resultant of shear
stresses developed in the beam, or it can be provided by individual
fasteners such as bolts, nails, or screws.
y
x
(2)
1.350 kN·m (1)
2.025 kN·m
B
C
(a) Internal bending moments
3.0 MPa (C)
y
x
1.0 MPa (T)
3.0 MPa (T)
y
x
F B = 12 kN
4.5 MPa (C)
a
d
(2)
1.5 MPa (T)
b
(1)
e
c
f
B
C 4.5 MPa (T)
(b) Bending stresses
b
(1)
e
c
f
F C = 18 kN
B
C
(c) Bending stress resultant forces
FIGURE 9.3 Moments, stresses, and forces acting on
beam segment BC.
y
x
F B = 12 kN
b
c
B
6 kN
(1)
e
f
C
FIGURE 9.4 Free-body diagram of member (1).
F C = 18 kN