Mechanics of Materials Solutions Chapter09 Probs1 10

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9.1 For the following problems, a beam segment subjected to internal bending moments at sections Aand B is shown along with a sketch of the cross-sectional dimensions. For each problem:

(a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at

sections A and B. Indicate the magnitude of key bending stresses on the sketch.(b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and

show these resultant forces on the sketch.

(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine

the horizontal force required to satisfy equilibrium for the specified area and show the location and

direction of this force on the sketch.

Consider area (1) of the 400-mm-long beam segment, which is subjected to internal bending moments o M A = −6 kN-m and M B = −8 kN-m.

Fig. P9.1a Beam segment Fig. P9.1b Cross-sectional dimensions

Solution

Moment of inertia about the z axis:

Shape I C d = yi – d²A I C + d²A

(mm4) (mm) (mm

4) (mm

4)

top flange 1,333,333.3 60 36,000,000.0 37,333,333.3

left web 853,333.3 0 0.0 853,333.3

right web 853,333.3 0 0.0 853,333.3

bottom flange 1,333,333.3 –60 36,000,000.0 37,333,333.3

Moment of inertia about the z axis (mm4) = 76,373,333.3

(a) Bending stress distribution

(b) Resultant forces acting on area (1)

On section A, the resultant force on area (1) in the x direction is





1(6.2849 MPa 3.1425 MPa)(250 mm)(40 mm) 47,136.9 N 47.1 kN (T)

2 A F = + = = Ans.

and on section B, the horizontal resultant force on area (1) is

1(8.3799 MPa 4.1899 MPa)(250 mm)(40 mm) 62,849.2 N 62.8 kN (T)

2 B F = + = = Ans.

(c) Equilibrium of area (1)

47,136.9 N 62,849.2 N 15,712.3 N 015.71 kN

x

H

F F

Σ = − + = ≠

∴ = Ans.

The horizontal shear force is directed from section B toward section A at the interface between area (1)and the web elements.









(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determinethe horizontal force required to satisfy equilibrium for the specified area and show the location and


Consider area (1) of the 20-in.-long beam segment, which is subjected to internal bending moments o M A = 24 kip-ft and M B = 28 kip-ft.


Solution

Moment of inertia about the z axis:Shape I C d = yi – d²A I C + d²A

(in.4) (in.) (in.

4) (in.

4)

left web 864.000 0.000 0.000 864.000

top flange 12.505 10.250 1,287.016 1,299.521

bottom flange 12.505 –10.250 1,287.016 1,299.521

right web 864.000 0.000 0.000 864.000

Moment of inertia about the z axis (in.4) = 4,327.042








1(798.699 psi 565.745 psi)(3.5 in.)(3.5 in.) 8,357.227 lb 8.36 kips (C)

2 A F = + = = Ans.


1(931.816 psi 660.036 psi)(3.5 in.)(3.5 in.) 9,750.098 lb 9.75 kips (C)

2 B F = + = = Ans.


8,357.227 lb 9,750.098 lb 1,392.871 lb 0

1.393 kips

x

H

F

F

Σ = − = − ≠

∴ = Ans.

The horizontal shear force is directed from section A toward section B at the interface between area (1)and the web elements.











Consider area (1) of the 12-in.-long beam segment, which is subjected to internal bending moments o M A = 700 lb-ft and M B = 400 lb-ft.


Solution

Centroid location in y direction: (reference axis at bottom of tee shape)

Shape Width b Height h Area Ai

yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

top flange 4.5 1.0 4.50 6.50 29.25

stem 1.0 6.0 6.00 3.00 18.00

10.50 47.25

3

2

47.25 in.4.50 in.

10.50 in.

i i

i

y A y

A

Σ= = =

Σ (measured upward from bottom edge of shape)



(in.4) (in.) (in.

4) (in.

4)

top flange 0.375 2.000 18.000 18.375

stem 18.000 –1.500 13.500 31.500

Moment of inertia about the z axis (in.4) = 49.875








1(421.053 psi 252.632 psi)(4.5 in.)(1 in.) 1,515.792 lb 1,516 lb (C)

2 A F = + = = Ans.


1(240.602 psi 144.361 psi)(4.5 in.)(1 in.) 866.167 lb 866 lb (C)

2

B F = + = = Ans.


1,515.792 lb 866.167 lb 649.625 lb 0

650 lb

x

H

F

F

Σ = − = ≠

∴ = Ans.

The horizontal shear force is directed from section B toward section A at the interface between area (1)

and the stem.











Consider area (1) of the 500-mm-long beam segment, which is subjected to internal bending moments o M A = −5.8 kN-m and M B = −3.2 kN-m.


Solution

Centroid location in y direction: (reference axis at bottom of shape)


yi (from bottom) yi Ai

(mm) (mm) (mm2

) (mm) (mm3

)top flange 160 30 4,800 285 1,368,000

left stem 20 270 5,400 135 729,000

right stem 20 270 5,400 135 729,000

15,600 2,826,000

3

2

2,826,000 mm181.154 mm

15,600 mm

i i

i

y A y

A

Σ= = =



Shape I C d = yi – d²A I C + d²A (mm4) (mm) (mm4) (mm4)

top flange 360,000 103.846 51,763,160.3 52,123,160.3

left stem 32,805,000 –46.154 11,503,035.3 44,308,035.3

right stem 32,805,000 –46.154 11,503,035.3 44,308,035.3

Moment of inertia about the z axis (mm4) = 140,739,231








1(4.898 MPa 3.661 MPa)(160 mm)(30 mm) 20,542 N 20.5 kN (T)

2 A F = + = = Ans.


1(2.702 MPa 2.020 MPa)(160 mm)(30 mm) 11,334 N 11.33 kN (T)

2 B F = + = = Ans.


20,542 N 11,334 N 9, 209 N 0

9.21 kN

x

H

F

F

Σ = − + = − ≠

∴ = Ans.

The horizontal shear force is directed from section A toward section B at the interface between area (1)

and the stems.











Consider area (1) of the 600-mm-long beam segment, which is subjected to internal bending moments o

M A = 1.6 kN-m and M B = 4.2 kN-m.


Solution




(mm) (mm) (mm2) (mm) (mm3)

left stem 40 250 10,000 125 1,250,000

bottom flange 170 40 6,800 20 136,000

right stem 40 250 10,000 125 1,250,000

26,800 2,636,000

3

2

2,636,000 mm98.358 mm

26,800 mm

i i

i

y A y

A

Σ= = =



Shape I C d = yi – d²A I C + d²A (mm

4) (mm) (mm

4) (mm

4)

left stem 52,083,333.3 26.6 7,097,850.3 59,181,183.6

bottom flange 906,666.7 –78.4 41,752,060.6 42,658,727.3

right stem 52,083,333.3 26.6 7,097,850.3 59,181,183.6









1(0.977 MPa 0.580 MPa)(170 mm)(40 mm) 5,294.6 N 5.29 kN (T)

2 A F = + = = Ans.


1(2.566 MPa 1.522 MPa)(170 mm)(40 mm) 13,898.2 N 13.90 kN (T)

2 B F = + = = Ans.


5,294.6 N 13,898.2 N 8,603.6 N 0

8.60 kN

x

H

F

F

Σ = − + = ≠

∴ = Ans.

The horizontal shear force is directed from section B toward section A at the interface between area (1)and the upright elements.











Consider area (1) of the 16-in.-long beam segment, which is subjected to internal bending moments o

M A = −3,300 lb-ft and M B = −4,700 lb-ft.


Solution




(in.) (in.) (in.2) (in.) (in.3)

left flange (1) 1.50 3.50 5.25 10.25 53.8125

right flange (2) 1.50 3.50 5.25 10.25 53.8125

central stem 1.50 12.00 18.00 6.00 108.0000

28.50 215.6250

3

2

215.625 in.7.5658 in.

28.50 in.

i i

i

y A y

A

Σ= = =



Shape I C d = yi – y d²A I C + d²A

(in.4) (in.) (in.

4) (in.

4)

left flange (1) 5.3594 2.6842 37.8262 43.1856

right flange (2) 5.3594 2.6842 37.8262 43.1856

central stem 216.0000 –1.5658 44.1305 260.1305









1(506.765 psi 106.767 psi)(1.5 in.)(3.5 in.) 1,610.522 lb 1,611 lb (T)

2 A F = + = = Ans.


1(721.756 psi 152.061 psi)(1.5 in.)(3.5 in.) 2,293.773 lb 2,290 lb (T)

2 B F = + = = Ans.


1,610.522 lb 2,293.773 lb 683.252 lb 0

683 lb

x

H

F

F

Σ = − + = ≠

∴ = Ans.


and the stem.











Consider area (1) of the 18-in.-long beam segment, which is subjected to internal bending moments o

M A = −42 kip-in. and M B = −36 kip-in.


Solution



yi

(from bottom) yi Ai

(in.) (in.) (in.

2

) (in.) (in.

3

)top flange (1) 6 2 12 11 132

bottom flange (2) 10 2 20 1 20

web 2 8 16 6 96

48 248

3

2

248 in.5.1667 in.

48 in.

i i

i

y A y

A

Σ= = =



Shape I C d = yi – y d²A I C + d²A (in.

4) (in.) (in.

4) (in.

4)

top flange (1) 4.0000 5.8333 408.3333 412.3333

bottom flange (2) 6.6667 –4.1667 347.2222 353.8889

web 85.3333 0.8333 11.1111 96.4444









1(332.690 psi 235.317 psi)(6 in.)(2 in.) 3,408.043 lb 3.41 kips (T)

2 A F = + = = Ans.


1(251.546 psi 154.173 psi)(6 in.)(2 in.) 2,921.180 lb 2.92 kips (T)

2 B F = + = = Ans.


3,408.043 lb 2,921.180 lb 486.863 lb 0

0.487 kips

x

H

F

F

Σ = − + = − ≠

∴ = Ans.

The horizontal shear force is directed from section A toward section B at the interface between area (1)

and the web.











Consider area (2) of the beam segment shown in Problem 9.7.


Solution




(in.) (in.) (in.2) (in.) (in.

3)

top flange (1) 6 2 12 11 132 bottom flange (2) 10 2 20 1 20

web 2 8 16 6 96

48 248

3

2

248 in.5.1667 in.

48 in.

i i

i

y A y

A

Σ= = =



Shape I C d = yi – y d²A I C + d²A

(in.4

) (in.) (in.4

) (in.4

)top flange (1) 4.0000 5.8333 408.3333 412.3333

bottom flange (2) 6.6667 –4.1667 347.2222 353.8889

web 85.3333 0.8333 11.1111 96.4444









1(251.546 psi 154.173 psi)(10 in.)(2 in.) 4,057.195 lb 4.06 kips (C)

2 A F = + = = Ans.


1(215.611 psi 132.149 psi)(10 in.)(2 in.) 3,477.595 lb 3.48 kips (C)

2 B F = + = = Ans.


4,057.195 lb 3,477.595 lb 579.599 lb 0

0.580 kips

x

H

F

F

Σ = − = ≠

∴ = Ans.


and the web.











Consider area (1) of the 300-mm-long beam segment, which is subjected to internal bending moments o M A = 7.5 kN-m and M B = 8.0 kN-m.


Solution



yi

(from bottom) yi Ai

(mm) (mm) (mm2) (mm) (mm

3)

left stiff (1) 40 90 3,600 275 990,000

flange (2) 150 40 6,000 300 1,800,000

right stiff (3) 40 90 3,600 275 990,000

stem 40 280 11,200 140 1,568,000

24,400 5,348,000

3

2

5,348,000 mm219.180 mm

24,400 mm

i i

i

y A y

A

Σ= = =








(mm4) (mm) (mm

4) (mm

4)

left stiff (1) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87

flange (2) 800,000.00 80.8197 39,190,916.42 39,990,916.42

right stiff (3) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87

stem 73,173,333.33 –79.1803 70,218,672.40 143,392,005.73

Moment of inertia about the z axis (mm4

) = 210,676,939.89




1(3.589 MPa 0.385 MPa)(40 mm)(90 mm) 7,153.755 N 7.15 kN (C)

2 A F = + = = Ans.


1(3.828 MPa 0.411 MPa)(40 mm)(90 mm) 7,630.672 N 7.63 kN (C)

2 B F = + = = Ans.


7,153.755 N 7, 630.672 N 476.917 N 0

0.477 kN

x

H

F

F

Σ = − = − ≠

∴ = Ans.

The horizontal shear force is directed from section A toward section B at the interface between area (1)and area (2).











Combine areas (1), (2), and (3) of the beam segment shown in Problem 9.9.


Solution

(a) Centroid location in y direction: (reference axis at bottom of shape)


yi

(from bottom) yi Ai

(mm) (mm) (mm2) (mm) (mm3)

left stiff (1) 40 90 3,600 275 990,000

flange (2) 150 40 6,000 300 1,800,000

right stiff (3) 40 90 3,600 275 990,000

stem 40 280 11,200 140 1,568,000

24,400 5,348,000

3

2

5,348,000 mm219.180 mm

24,400 mm

i i

i

y A y

A

Σ= = =




(mm4) (mm) (mm

4) (mm

4)

left stiff (1) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87

flange (2) 800,000.00 80.8197 39,190,916.42 39,990,916.42

right stiff (3) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87

stem 73,173,333.33 -79.1803 70,218,672.40 143,392,005.73




E f hi k b d d b i f di ib i f fi b i f i i i l l




1(3.589 MPa 0.385 MPa)(40 mm)(90 mm) 7,153.755 N

2 A F = + =


1

(3.828 MPa 0.411 MPa)(40 mm)(90 mm) 7,630.672 N2 B F = + =

Resultant forces acting on area (3)

The forces acting on area (3) are identical to those acting on area (1).

Resultant forces acting on area (2)


1(3.589 MPa 2.165 MPa)(150 mm)(40 mm) 17,262.855 N

2 A F = + =


1(3.828 MPa 2.309 MPa)(150 mm)(40 mm) 18,413.697 N

2 B F = + =

Resultant forces acting on combined areas (1), (2), and (3)

On section A, the resultant force on combined areas (1), (2), and (3) is

2(7,153.755 N) 17,262.855 N 31,570.363 N 31.6 kN (C) A F = + = = Ans.


2(7,630.672 N) 18,413.697 N 33,675.054 N 33.7 kN (C) B F = + = = Ans.

(c) Equilibrium of combined areas (1), (2), and (3)

31,570.363 N 33,675.054 N 2,104.691 N 0

2.10 kN

x

H

F

F

Σ = − = − ≠

∴ = Ans.

The horizontal shear force is directed from section A toward section B at the interface between area (2)and the stem of the tee.

Mechanics of Materials Solutions Chapter09 Probs1 10

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