Page422, Concept Problem 6.1. Solve this problem completely,butwith the following changes. (1) Continue toassume that the upper plank is 100 mm wide. (2) Assumenowthat the lower plank isreduced to80 mm wide. (3)Assume that each nail shouldwithstand ashearingforce of no more than 100 N.The questions are: (a) What is therequired spacing of the nails in the upper plank? (b) What is the required spacing of the nails in the lower plank?Please keep in mind that thesechangeswill result in many other changes across the board. Discuss all practicalissues in your findings. 0.020 m20 mm100 mm0.100 mN.A.C'(a)100 mm20 mm20 mm↑-0.100 my = 0.060 mN.A.Concept Application 6.1A beam is made of three planks, 20 by 100 mm in cross section, and nailedtogether (Fig. 6.8a). Knowing that the spacing between nails is 25 mm andthe vertical shear in the beam is V = 500 N, determine the shearing force ineach nail.Determine the horizontal force per unit length q exerted on the lower faceof the upper plank. Use Eq. (6.5), where Q represents the first moment withrespect to the neutral axis of the shaded area A shown in Fig. 6.8b, and I isthe moment of inertia about the same axis of the entire cross-sectional area(Fig. 6.8c). Recalling that the first moment of an area with respect to a givenaxis is equal to the product of the area and of the distance from its centroidto the axis,*0.100 m(b)(c)Fig. 6.8 (a) Beam made of three boardsnailed together. (b) Cross section forcomputing Q. (c) Cross section forcomputing moment of inertia.0.020 mQ = Ay = (0.020 mx 0.100 m) (0.060 m)= 120 x 10-6 m³I=/(0.020 m) (0.100 m)³+2[(0.100 m) (0.020 m)³+(0.020 mx 0.100 m) (0.060 m)²]= 1.667 × 10-6 + 2(0.0667 +7.2)10-6= 16.20 x 10-6 m²Substituting into Eq. (6.5),q=VQ(500 N) (120 x 10-6 m³)16.20 x 10-6 m²= 3704 N/mSince the spacing between the nails is 25 mm, the shearing force in eachnail isF = (0.025 m)q = (0.025 m) (3704 N/m) = 92.6 N

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Page422, Concept Problem 6.1. Solve this problem completely,butwith the following changes. (1) Continue toassume that the upper plank is 100 mm wide. (2) Assumenowthat the lower plank isreduced to80 mm wide. (3)Assume that each nail shouldwithstand ashearingforce of no more than 100 N.The questions are: (a) What is therequired spacing of the nails in the upper plank? (b) What is the required spacing of the nails in the lower plank?Please keep in mind that thesechangeswill result in many other changes across the board. Discuss all practicalissues in your findings.

0.020 m
20 mm
100 mm
0.100 m
N.A.
C'
(a)
100 mm
20 mm
20 mm
↑
-0.100 m
y = 0.060 m
N.A.
Concept Application 6.1
A beam is made of three planks, 20 by 100 mm in cross section, and nailed
together (Fig. 6.8a). Knowing that the spacing between nails is 25 mm and
the vertical shear in the beam is V = 500 N, determine the shearing force in
each nail.
Determine the horizontal force per unit length q exerted on the lower face
of the upper plank. Use Eq. (6.5), where Q represents the first moment with
respect to the neutral axis of the shaded area A shown in Fig. 6.8b, and I is
the moment of inertia about the same axis of the entire cross-sectional area
(Fig. 6.8c). Recalling that the first moment of an area with respect to a given
axis is equal to the product of the area and of the distance from its centroid
to the axis,*
0.100 m
(b)
(c)
Fig. 6.8 (a) Beam made of three boards
nailed together. (b) Cross section for
computing Q. (c) Cross section for
computing moment of inertia.
0.020 m
Q = Ay = (0.020 mx 0.100 m) (0.060 m)
= 120 x 10-6 m³
I=/(0.020 m) (0.100 m)³
+2[(0.100 m) (0.020 m)³
+(0.020 mx 0.100 m) (0.060 m)²]
= 1.667 × 10-6 + 2(0.0667 +7.2)10-6
= 16.20 x 10-6 m²
Substituting into Eq. (6.5),
q=
VQ
(500 N) (120 x 10-6 m³)
16.20 x 10-6 m²
= 3704 N/m
Since the spacing between the nails is 25 mm, the shearing force in each
nail is
F = (0.025 m)q = (0.025 m) (3704 N/m) = 92.6 N

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