08_shear_built

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404 Chapter 7 transverse shear
7
Fig. 7–13
dx
dx
M � dM
M
(a)
dx
dF
F � dF
(b)
t
F
A¿
*The use of the word “flow” in this terminology will become meaningful as it pertains 
to the discussion in Sec. 7.4.
7.3 Shear Flow in built-up 
MeMberS
Occasionally in engineering practice, members are “built up” from several 
composite parts in order to achieve a greater resistance to loads. An 
example is shown in Fig. 7–13. If the loads cause the members to bend, 
fasteners such as nails, bolts, welding material, or glue will be needed to 
keep the component parts from sliding relative to one another, Fig. 7–2. 
In order to design these fasteners or determine their spacing, it is necessary 
to know the shear force that they must resist. This loading, when measured 
as a force per unit length of beam, is referred to as shear flow, q.*
The magnitude of the shear flow is obtained using a procedure similar 
to that for finding the shear stress in a beam. To illustrate, consider 
finding the shear flow along the juncture where the segment in Fig. 7–14a
is connected to the flange of the beam. Three horizontal forces must act 
on this segment, Fig. 7–14b. Two of these forces, F and F + dF, are the 
result of the normal stresses caused by the moments M and M + dM,
respectively. The third force, which for equilibrium equals dF, acts at the 
juncture. Realizing that dF is the result of dM, then, like Eq. 7–1, we have
dF =
dM
I LA′y dA′
The integral represents Q, that is, the moment of the segment’s area A′
about the neutral axis. Since the segment has a length dx, the shear flow, 
or force per unit length along the beam, is q = dF>dx. Hence dividing 
both sides by dx and noting that V = dM>dx, Eq. 6–2, we have
q =
VQ
I
 (7–4)
Here
q = the shear flow, measured as a force per unit length along the beam
V = the shear force, determined from the method of sections and the 
equations of equilibrium
I = the moment of inertia of the entire cross-sectional area calculated 
about the neutral axis
Q = y′A′, where A′ is the cross-sectional area of the segment that is connected
to the beam at the juncture where the shear flow is calculated, and y′ is 
the distance from the neutral axis to the centroid of A′
 
Fig. 7–14
7.3 shear Flow in Built-up memBers 405
7
Fastener Spacing. When segments of a beam are connected by 
fasteners, such as nails or bolts, their spacing s along the beam can be 
determined. For example, let’s say that a fastener, such as a nail, can 
support a maximum shear force of F (N) before it fails, Fig. 7–15a. If 
these nails are used to construct the beam made from two boards, as 
shown in Fig. 7–15b, then the nails must resist the shear flow q (N>m) 
between the boards. In other words, the nails are used to “hold” the top 
board to the bottom board so that no slipping occurs during bending. 
(See Fig. 7–2a.) As shown in Fig. 7–15c, the nail spacing is therefore 
determined from
F (N) = q (N>m) s (m)
The examples that follow illustrate application of this equation.
Other examples of shaded segments connected to built-up beams by 
fasteners are shown in Fig. 7–16. The shear flow here must be found at 
the thick black line, and is determined by using a value of Q calculated 
from A ′ and y′ indicated in each figure. This value of q will be resisted by 
a single fastener in Fig. 7–16a, by two fasteners in Fig. 7–16b, and by three 
fasteners in Fig. 7–16c. In other words, the fastener in Fig. 7–16a supports 
the calculated value of q, and in Figs. 7–16b and 7–16c each fastener 
supports q>2 and q>3, respectively.
_
y ¿
A¿
AN
(a) 
N A
(b)
A¿
A¿
AN
(c)
_
y ¿
_
y ¿
Fig. 7–16
Important poInt
 • Shear flow is a measure of the force per unit length along the axis 
of a beam. This value is found from the shear formula and is used 
to determine the shear force developed in fasteners and glue 
that holds the various segments of a composite beam together.
(a)
F
F
A
N
V
�
(b)
�
�
�
�
(c)
F
�
Fig. 7–15
406 Chapter 7 transverse shear
7
The beam is constructed from three boards glued together as shown in 
Fig. 7–17a. If it is subjected to a shear of V = 850 kN, determine the 
shear flow at B and B ′ that must be resisted by the glue.
SOLUTION
Section Properties. The neutral axis (centroid) will be located from 
the bottom of the beam, Fig. 7–17a. Working in units of meters, we have
 y =
Σy�A
ΣA
=
2[0.15 m](0.3 m)(0.01 m) + [0.305 m](0.250 m)(0.01 m)
2(0.3 m)(0.01 m) + (0.250 m)(0.01 m)
 = 0.1956 m
The moment of inertia of the cross section about the neutral axis is thus
 I = 2 c
1
12
 (0.01 m)(0.3 m)3 + (0.01 m)(0.3 m)(0.1956 m - 0.150 m)2 d
 + c
1
12
 (0.250 m)(0.01 m)3 + (0.250 m)(0.01 m)(0.305 m - 0.1956 m)2 d
 = 87.42(10-6) m4
The glue at both B and B ′ in Fig. 7–17a “holds” the top board to the 
beam. Here
 QB = y=BA=B = [0.305 m - 0.1956 m](0.250 m)(0.01 m)
 = 0.2735(10-3) m3
Shear Flow.
q =
VQB
I
=
850(103) N(0.2735(10-3) m3)
87.42(10-6) m4
= 2.66 MN>m
Since two seams are used to secure the board, the glue per meter 
length of beam at each seam must be strong enough to resist one-half 
of this shear flow. Thus,
 qB = qB′ =
q
2
= 1.33 MN>m Ans.
NOTE: If the board CC' is added to the beam, Fig. 7–17b, then y and I 
have to be recalculated, and the shear flow at C and C′ determined 
from q = V y′ C A′ C>I. Finally, this value is divided by one-half to obtain 
qC and qC′.
ExampLE 7.4
250 mm
10 mm
300 mm
125 mm 10 mm10 mm
AN
_
y
_
 y¿B
V � 850 kN
B B¿
(a)
A¿B
_ 
 y¿C
AN
C
(b)
C ¿
A¿C
Fig. 7–17
7.3 shear Flow in Built-up memBers 407
7
A box beam is constructed from four boards nailed together as shown 
in Fig. 7–18a. If each nail can support a shear force of 30 N, determine 
the maximum spacing s of the nails at B and at C to the nearest 5 mm 
so that the beam will support the force of 80 N.
SOLUTION
Internal Shear. If the beam is sectioned at an arbitrary point along 
its length, the internal shear required for equilibrium is always 
V = 80 N, and so the shear diagram is shown in Fig. 7–18b.
Section Properties. The moment of inertia of the cross-sectional 
area about the neutral axis can be determined by considering a 
75@mm * 75@mm square minus a 45@mm * 45@mm square.
I =
1
12
 (0.075 m)(0.075 m)3 -
1
12
 (0.045 m)(0.045 m)3 = 2.295(10-6) m4
The shear flow at B is determined using QB found from the darker 
shaded area shown in Fig. 7–18c. It is this “symmetric” portion of the 
beam that is to be “held” onto the rest of the beam by nails on the left 
side and by the fibers of the board on the right side.
Thus,
QB = y′A′ = (0.03m)(0.075m)(0.015m) = 33.75(10-6)m3
Likewise, the shear flow at C can be determined using the “symmetric” 
shaded area shown in Fig. 7–18d. We have
QC = y′A ′ = (0.03m)(0.045m)(0.015m) = 20.25(106)m3
Shear Flow.
 qB =
VQB
I
=
(80 N)[33.75(10-6) m3]
2.295(10-6) m4
= 1176.47 N>m
 qC =
VQC
I
=
(80 N)[20.25(10-6)m3]
2.295(10-6) m4
= 705.88 N>m
These values represent the shear force per unit length of the beam 
that must be resisted by the nails at B and the fibers at B′, Fig. 7–18c, 
and the nails at C and the fibers at C′, Fig. 7–18d, respectively. Since in 
each case the shear flow is resisted at two surfaces and each nail can 
resist 30 N, for B the spacing is
ExampLE 7.5
(a)
80 N
s
60 mm 15 mm
60 mm
15 mm
B
C
15 mm
(c)
0.075 m
B B9
AN
0.03 m
0.015 m
0.045 m
C ¿
AN
C
(d)
0.03 m
0.015 m
Fig. 7–18sB =
30 N
(1176.47>2) N>m
= 0.0510 m = 51.0 mm Use sB = 50 mm Ans.
And for C,
sC =
30 N
(705.88>2) N>m
= 0.0850 m = 85.0 mm Use sC = 85 mm Ans.
(b)
V (N)
x (m)
80
408 Chapter 7 transverse shear
7
Nails having a shear strength of 900 N are used in a beam that can be 
constructed either as in Case I or as in Case II, Fig. 7–19. If the nails 
are spaced at 250 mm, determine the largest vertical shear that can be 
supported in each case so that the fasteners will not fail.
ExampLE 7.6
N
75 mm
25 mm
80 mm
10 mm
10 mmCase I
A N
250 mms
100 mm
10 mm
Case II
25 mm
250 mms
10 mm
AA
Fig. 7–19
SOLUTION
Since the cross section is the same in both cases, the moment of inertia 
about the neutral axis is
I = 
1
12
 (0.075 m)(0.1 m)3 - 
1
12
 (0.05 m)(0.08 m)3 = 4.1167(10-6) m4
Case I. For this design a single row of nails holds the top or bottom 
flange onto the web. For one of these flanges,
Q = y′A′ = (0.045 m)(0.075 m)(0.01 m) = 33.75(10-6) m3
so that
 q =
VQ
I
 
900 N
0.25 m
=
V[33.75(10-6) m3]
4.1167(10-6) m4
 
 V = 439.11 N = 439 N Ans.
Case II. Here a single row of nails holds one of the side boards onto 
the web. Thus
Q = y′A ′ = (0.045 m)(0.025 m)(0.01 m) = 11.25(10-6) m3
 q =
VQ
I
 
900 N
0.25 m
=
V[11.25(10-6) m3]
4.1167(10-6) m4
 
 V = 1.3173(103) N = 1.32 kN Ans.
7.3 shear Flow in Built-up memBers 409
7
F7–6. The two identical boards are bolted together to form 
the beam. Determine the maximum spacing s of the bolts to 
the nearest mm if each bolt has a shear strength of 15 kN. The 
beam is subjected to a shear force of V = 50 kN.
100 mm
100 mm
300 mm
V
s
s
Prob. F7–6
F7–7. Two identical 20-mm-thick plates are bolted to the 
top and bottom flange to form the built-up beam. If the beam 
is subjected to a shear force of V = 300 kN, determine the 
maximum spacing s of the bolts to the nearest mm if each 
bolt has a shear strength of 30 kN.
20 mm
10 mm
10 mm
200 mm
200 mm
10 mm
20 mm
300 mm
s
s
V
Prob. F7–7
F7–8. The boards are bolted together to form the built-up 
beam. If the beam is subjected to a shear force of V = 20 kN, 
determine the maximum spacing s of the bolts to the nearest 
mm if each bolt has a shear strength of 8 kN.
V
150 mm
150 mm
200 mm
s
s
50 mm
25 mm
25 mm
50 mm
Prob. F7–8
F7–9. The boards are bolted together to form the built-up 
beam. If the beam is subjected to a shear force of V = 75 kN,
determine the allowable maximum spacing of the bolts to 
the nearest multiples of 5 mm. Each bolt has a shear strength 
of 30 kN.
100 mm
75mm
75 mm
25 mm
25 mm
100 mm
25 mm
12 mm
12 mm
V
s
s
Prob. F7–9
FUNDamENTaL pROBLEmS
410 Chapter 7 transverse shear
7
*7–32. The double T-beam is fabricated by welding the 
three plates together as shown. Determine the shear stress in 
the weld necessary to support a shear force of V = 80 kN.
7–33. The double T-beam is fabricated by welding the 
three plates together as shown. If the weld can resist a shear 
stress tallow = 90 MPa, determine the maximum shear V 
that can be applied to the beam.
20 mm
50 mm50 mm 75 mm
20 mm20 mm 
V
150 mm
Probs. 7–32/33
7–34. The beam is constructed from two boards fastened 
together with three rows of nails spaced s = 50 mm apart. If 
each nail can support a 2.25-kN shear force, determine the 
maximum shear force V that can be applied to the beam.The 
allowable shear stress for the wood is tallow = 2.1 MPa.
7–35. The beam is constructed from two boards fastened 
together with three rows of nails. If the allowable shear stress 
for the wood is tallow = 1 MPa, determine the maximum shear 
force V that can be applied to the beam. Also, find the maximum 
spacing s of the nails if each nail can resist 3.25 kN in shear.
V
40 mm
s
s
150 mm
40 mm
Probs. 7–34/35
*7–36. The beam is constructed from four boards which 
are nailed together. If the nails are on both sides of the 
beam and each can resist a shear of 3 kN, determine the 
maximum load P that can be applied to the end of the beam.
P
2 m 2 m
3 kN
B CA
30 mm
30 mm
30 mm
100 mm
250 mm30 mm
150 mm
Prob. 7–36
7–37. The beam is fabricated from two equivalent structural 
tees and two plates. Each plate has a height of 150 mm and a 
thickness of 12 mm. If a shear of V = 250 kN is applied to 
the cross section, determine the maximum spacing of the 
bolts. Each bolt can resist a shear force of 75 kN.
7–38. The beam is fabricated from two equivalent structural 
tees and two plates. Each plate has a height of 150 mm and a 
thickness of 12 mm. If the bolts are spaced at s = 200 mm
determine the maximum shear force V that can be applied to
the cross section. Each bolt can resist a shear force of 75 kN.
75 mm
75 mm
A
V
12 mm
25 mm
12 mm
150 mm
s
N
Probs. 7–37/38
pROBLEmS
7.3 shear Flow in Built-up memBers 411
7
7–39. The double-web girder is constructed from two 
plywood sheets that are secured to wood members at its top 
and bottom. If each fastener can support 3 kN in single 
shear, determine the required spacing s of the fasteners 
needed to support the loading P = 15 kN. Assume A is 
pinned and B is a roller.
*7–40. The double-web girder is constructed from two 
plywood sheets that are secured to wood members at its top 
and bottom. The allowable bending stress for the wood is 
sallow = 56 MPa and the allowable shear stress is 
tallow = 21 MPa. If the fasteners are spaced s = 150 mm
and each fastener can support 3 kN in single shear, 
determine the maximum load P that can be applied to the 
beam.
7–42. The simply supported beam is built up from three boards 
by nailing them together as shown. The wood has an allowable 
shear stress of tallow = 1.5 MPa, and an allowable bending 
stress of sallow = 9 MPa. The nails are spaced at s = 75 mm,
and each has a shear strength of 1.5 kN. Determine the 
maximum allowable force P that can be applied to the beam.
7–43. The simply supported beam is built up from three 
boards by nailing them together as shown. If P = 12 kN,
determine the maximum allowable spacing s of the nails to 
support that load, if each nail can resist a shear force of 1.5 kN.
P
B
s
A
50 mm
50 mm
250 mm
150 mm
12 mm 12 mm
50 mm
50 mm
1.2 m 1.2 m
Probs. 7–39/40
7–41. A beam is constructed from three boards bolted 
together as shown. Determine the shear force in each bolt if 
the bolts are spaced s = 250 mm apart and the shear is 
V = 35 kN.
s � 250 mm
250 mm100 mm
25 mm
25 mm
25 mm
350 mm
V
Prob. 7–41
1 m 1 m
s
P
100 mm
200 mm
25 mm
25 mm
25 mm
A B
Probs. 7–42/43
*7–44. The T-beam is nailed together as shown. If the nails 
can each support a shear force of 4.5 kN, determine the 
maximum shear force V that the beam can support and the 
corresponding maximum nail spacing s to the nearest 
multiples of 5 mm. The allowable shear stress for the wood 
is tallow = 3 MPa.
300 mm
300 mm50 mm
50 mm
V
s
s
Prob. 7–44
412 Chapter 7 transverse shear
7
7–45. The nails are on both sides of the beam and each can 
resist a shear of 2 kN. In addition to the distributed loading, 
determine the maximum load P that can be applied to the 
end of the beam. The nails are spaced 100 mm apart and the 
allowable shear stress for the wood is tallow = 3 MPa.
P
1.5 m 1.5 m
B CA
40 mm
20 mm
20 mm
100 mm
200 mm
200 mm
2 kN/m
Prob. 7–45
7–46. Determine the average shear stress developed in the 
nails within region AB of the beam. The nails are located on 
each side of the beam and are spaced 100 mm apart. Each 
nail has a diameter of 4 mm. Take P = 2 kN.
P
1.5 m 1.5 m
B CA
40 mm
20 mm
20 mm
100 mm
200 mm
200 mm
2 kN/m
Prob. 7–46
7–47. The beam is made from four boards nailed together 
as shown. If the nails can each support a shear force of 
500 N, determine their required spacing s′ and s to the 
nearest mm if the beam is subjected to a shear of V = 3.5 kN.
P
P
1
—
4 P
1
—
4
0.8 m 0.8 m1 m 1 m
A B
20 mm
40 mm
30 mm
60 mm
40 mm
Prob. 7–48
*7–48. The beam is made from three polystyrene strips 
that are glued together as shown. If the glue has a shear 
strength of 80 kPa, determine the maximum load P that can 
be applied without causing the glue to lose its bond.
7–49. The timber T-beam is subjected to a load consisting 
of n concentrated forces, Pn. If the allowable shear Vnail for 
each of the nails is known, write a computer program that 
will specify the nail spacing between each load. Show an 
application of the program using the values L = 5 m, 
a1 = 1.5 m, P1 = 3 kN, a2 = 3 m, P2 = 6 kN, b1 = 40 mm,h1 = 200 mm, b2 = 200 mm, h2 = 25 mm, and 
Vnail = 900 N.
40 mm
250 mm
50 mm
B
V
25 mm
250 mm
25 mm
A
25 mm 
C
s
s
D
s¿
s¿
Prob. 7–47
b1
h2
h1
b2
P1 P2 Pn
a1
L
a2
an
s1 s2
s3 sn
A B
Prob. 7–49