Mechanics of Materials
Mechanics of Materials
10th Edition
ISBN: 9780134321158
Author: HIBBELER
Publisher: PEARSON
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Textbook Question
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Chapter 7, Problem 7.1RP

The beam is fabricated from four boards nailed together as shown. Determine the shear force each nail along the sides C and the top D must resist if the nails are uniformly spaced at s=3 in. The beam is subjected to a shear of V=4.5 kip.

Chapter 7, Problem 7.1RP, The beam is fabricated from four boards nailed together as shown. Determine the shear force each

Expert Solution & Answer
Check Mark
To determine

The shear force (FC) each nails along the side C.

The shear force (FD) each nails along the top D.

Answer to Problem 7.1RP

The shear force (FC) each nails along the side C is 197lb_.

The shear force (FD) each nails along the top D is 1.38kip_.

Explanation of Solution

Given information:

The shear force is 4.5kip.

The uniform nail spacing is 3 in.

Calculation:

Sketch the diagram of the T section as shown in Figure 1.

Mechanics of Materials, Chapter 7, Problem 7.1RP , additional homework tip  1

Refer Figure 1,

The area of the beam is the sum of area of three rectangles 1, 2, and 3.

The dimensions of rectangle 1 as width b1=10in. and depth d1=1in.

The dimensions of rectangle 2 as width b2=4in. and depth d2=2in.

The dimensions of rectangle 2 as width b3=12in. and depth d2=1in.

Find the value of area section 1 as shown below:

a1=b1d1 (1)

Substitute 10 in. for b1 and 1 in. for d1 in Equation (1).

a1=10×1=10in2

Find the value of area section 2 as shown below:

a2=b2d2 (2)

Substitute 4 in. for b2 and 2 in. for d2 in Equation (2).

a2=4×2=8in2

Find the value of area section 3 as shown below:

a3=b3d3 (3)

Substitute 12 in. for b3 and 1 in. for d3 in Equation (3).

a3=12×1=12in2

Calculate the centroid of y¯ as shown in below:

y¯=a1y1+a2y2+a3y3a1+a2+a3 (4)

Here, a1,a2, and a3 area of section 1,2, and 3 and y1,y2 y3 is the centroid of section 1,2, and 3.

Substitute 10in2 for a1, 8in2 for a2, 12in2 for a3, 0.5in. for y1, 2in. for y2, and 7in. for y3 in Equation (4).

y¯=0.5(10)+2×8+7×1210+8+12=5+16+8430=3.5in

Sketch the diagram of y¯ as shown in Figure 2.

Mechanics of Materials, Chapter 7, Problem 7.1RP , additional homework tip  2

Calculate the moment of inertia of the beam (I) as follows:

INA=I1+I2+I3=[b1d1312+a(y¯y1)+b2d2312+a(y¯y2)+b1d1312+a(y3y¯)] (5)

Refer to Figure 2:

The value of y1 is 0.5in.

The value of y2 is 2in.

The value of y3 is 7in.

Substitute 10 in. for b1, 1 in. for d1, 4 in. for b2, 2in for d2, 12 in. for b3, and 1 in. for d2 in Equation (5).

INA=[(10)(13)12+(10)(1)(3.500.5)2+2×4312+2×4(3.502)2+1×12312+1×12(73.50)2]=90.8333+28.6667+291=410.5in4

Calculate the first moment area (QC) as shown in below:

QC=y¯1A (6)

Here, y¯ is the distance between the neutral axis and centroid of the area under consideration and A is the area of cross section.

Refer to Figure 2.

The value of y¯1 is 1.5in..

Substitute 1.5in. for y¯1 and (4×1)in2 for A in Equation (6).

QC=1.5×4×1=6.00in3

Calculate the first moment area (QD) as shown in below:

QD=y¯2A (7)

Here, y¯ is the distance between the neutral axis and centroid of the area under consideration and A is the area of cross section.

Refer to Figure 2.

The value of y¯2 is 3.50in..

Substitute 3.50in. for y¯2 and (12×1)in2 for A in Equation (7).

QC=3.5×12×1=42.0in3

Show the formula for shear flow (qC) as follows:

qC=VQCI (8)

Here, V is the shear force, I is the moment of inertia, and y¯ is the distance between the neutral axis and centroid of the area under consideration.

Substitute 4.5kip for V, 6.00in3 for QC, and 410.5in4 for I in Equation (8).

qC=4.5×103×6.00410.5=27,000410.5=65.773lb/in.

Show the formula for shear flow (qD) as follows:

qD=VQDI (9)

Here, V is the shear force, I is the moment of inertia, and y¯ is the distance between the neutral axis and centroid of the area under consideration.

Substitute 4.5kip for V, 42.0in3 for QD, and 410.5in4 for I in Equation (9).

qD=4.5×103×42.0410.5=189,000410.5=460.414lb/in.

Calculate the shear force (FC) using the relation:

FC=qCs (10)

Here, s is the spacing and qC is the shear flow at point C.

Substitute 65.773lb/in. for qC and 3 in. for s in Equation (10).

FC=65.773×3=197lb

Hence, the shear force (FC) resisted nail at C is 197lb_.

Calculate the shear force (FD) using the relation:

FD=qDs (11)

Here, s is the spacing and qD is the shear flow at point D.

Substitute 460.41lb/in. for qD and 3 in. for s in Equation (11).

FD=460.41lb/in.×3=1.38kip

Hence, the shear force (FD) resisted nail at D is 1.38kip_.

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Chapter 7 Solutions

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