SOLUTION Locate neutral axis

In document Mechanics of Materials 7th Edition Beer Johnson Chapter 6 (Page 47-98)

(2)(2)(10) (10)(4) 80 in2

 A  

(2)(2)(10)(5) (10)(4)(8) 520 in3

Ay   

1 (10)(4) (10)(4)(1.5) 566.67 in 12

3 0.110447 in

4 4 8

100 mm

PROBLEM 6.44

A beam consists of three planks connected as shown by steel bolts with a longitudinal spacing of 225 mm. Knowing that the shear in the beam is vertical and equal to 6 kN and that the allowable average shearing stress in each bolt is 60 MPa, determine the smallest permissible bolt diameter that can be used.

bolt bolt bolt

131.25 10 mm 131.25 10 m (7500)(50) 375 10 mm

375 10 m

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939 6 in.

1 in.

1 in.

PROBLEM 6.45

A beam consists of five planks of 1.5 6-in. cross section connected by steel bolts with a longitudinal spacing of 9 in. Knowing that the shear in the beam is vertical and equal to 2000 lb and that the allowable average shearing stress in each bolt is 7500 psi, determine the smallest permissible bolt diameter that can be used.

(89.888)(9) 809 lb

 

  

q

F qs

bolt bolt 2

bolt bolt

bolt bolt

2 bolt

bolt bolt bolt

809 0.1079 in 7500

12 mm 400 mm

PROBLEM 6.46

Four L102  102  9.5 steel angle shapes and a 12  400-mm plate are bolted together to form a beam with the cross section shown. The bolts are of 22-mm diameter and are spaced longitudinally every 120 mm. Knowing that the beam is subjected to a vertical shear of 240 kN, determine the average shearing stress in each bolt.

(263.78 kN/m)(0.120 m) 31.65 kN

  

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PROBLEM 6.47

A plate of 14-in. thickness is corrugated as shown and then used as a beam. For a vertical shear of 1.2 kips, determine (a) the maximum shearing stress in the section, (b) the shearing stress at point B. Also, sketch the shear flow in the cross section.

SOLUTION

2 2 2

(1.2) (1.6) 2.0 in. (0.25)(2.0) 0.5 in

BD BD

L    A  

Locate neutral axis and compute moment of inertia.

Part A(in )2 y(in.) Ay(in )3 (in.)d Ad2(in )4 I(in )4

(2)(0.25)(0.4) 0.2 in (0.5)(0.25)(0.2) 0.025 in 0.225 in

a d

A plate of 2-mm thickness is bent as shown and then used as a beam. For a vertical shear of 5 kN, determine the shearing stress at the five points indicated and sketch the shear flow in the cross section.

SOLUTION

2 (2)(48) (2)(52) (20)(2) (20)(2)(25)

12 12 12

133.75 10 mm 133.75 10 mm (2)(24)(12) 576 mm 576 10 mm 0

(12)(2)(25) 600 mm 600 10 m (2)(24)(12) 1.176 10 mm 1.176 1

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PROBLEM 6.49

An extruded beam has the cross section shown and a uniform wall thickness of 3 mm. For a vertical shear of 10 kN, determine (a) the shearing stress at point A, (b) the maximum shearing stress in the beam.

Also, sketch the shear flow in the cross section.

SOLUTION

tan 16 28.07

 30   

(2)(3 sec )(11.932) 1 11.932 2

2.14776 10 484.06 2.6318 10 mm 2.6318 10 m

PROBLEM 6.49 (Continued)

3 3

3 3

(28)(3)(18.077) 1.51847 10 mm 1.51847 10

(50.9) 2.14776 10 36.0 MPa

B

B B A

A

Q Q

Q

  

  

Multiply shearing stresses by (3 mmt  0.003 m) to get shear flow.

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PROBLEM 6.50

A plate of thickness t is bent as shown and then used as a beam. For a vertical shear of 600 lb, determine (a) the thickness t for which the maximum shearing stress is 300 psi, (b) the corresponding shearing stress at point E. Also, sketch the shear flow in the cross section.

SOLUTION

2 2

4.8 2 5.2 in.

BD EF

LL   

Neutral axis lies at 2.4 in. above AB.

Calculate I.

2

0 (3)(0.23168)(2.4) 1.668 in3

E EF FG

2 in.

The design of a beam calls for connecting two vertical rectangular 38 -in. plates 4 by welding them to horizontal 12 -in. plates as shown. For a vertical shear V, 2 determine the dimension a for which the shear flow through the welded surface is maximum.

4.041667 2 in (2) 1 in

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947 a a

PROBLEM 6.52

The cross section of an extruded beam is a hollow square of side a  3 in. and thickness t  0.25 in. For a vertical shear of 15 kips, determine the maximum shearing stress in the beam and sketch the shear flow in the cross section.

SOLUTION

4 4

1 (3.25 2.75 )

  12 

u v

I I

4.53125 in4

Since products of inertia  0,

  

x y u v

I I I I

4.53125 in4 x

I

15 kips

V

2[(3 in. 0.25 in.)(1.0607 in.)] 1.59105 in3

  

QNA

3 4

(15 kips)(1.59105 in ) (2 ) (4.53125 in )(2)(0.25 in.)

m NA NA

VQ

  I t

 

m 10.53 ksi ◄

(a) a

a

(b)

PROBLEM 6.53

An extruded beam has a uniform wall thickness t. Denoting by V the vertical shear and by A the cross-sectional area of the beam, express the maximum shearing stress as maxk V A( / ) and determine the constant k for each of the two orientations shown.

SOLUTION

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(a) Determine the shearing stress at point P of a thin-walled pipe of the cross section shown caused by a vertical shear V. (b) Show that the maximum shearing stress occurs for

  and is equal to 2V/A, where A is the cross-sectional area of the pipe. 90

PROBLEM 6.55

For a beam made of two or more materials with different moduli of elasticity, show that Eq. (6.6)

ave

VQ

  It

remains valid provided that both Q and I are computed by using the transformed section of the beam (see Sec. 4.4), and provided further that t is the actual width of the beam where ave is computed.

SOLUTION

Let E be a reference modulus of elasticity. ref

1 2

1 2

ref ref

, , etc.

E E

n n

E E

 

Widths b of actual section are multiplied by n’s to obtain the transformed section. The bending stress distribution in the cross section is given by

x

nMy

   I

where I is the moment of inertia of the transformed cross section and y is measured from the centroid of the transformed section.

The horizontal shearing force over length x is

( ) ( ) ( )

( x) n M y M Q M

H dA dA ny dA

I I I

   

Q

ny dA first moment of transformed section.

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PROBLEM 6.56

A composite beam is made by attaching the timber and steel portions shown with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of 4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress at the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION

Let steel be the reference material.

10 GPa

(a) Shearing stress in the bolts.

For the upper wooden portion, Qw (90)(140)(4245)1.0962 10 mm 6 3 For the transformed wooden portion,

6 3 3 6 3

(0.05)(1.0962 10 ) 54.81 10 mm 54.81 10 m

Qn Qw w      

Shear flow on upper wooden portion:

6 6

(4000)(54.81 10 )

7616 N/m

bolt (7616)(0.200) 1523.2 N

Fqs  

2 2 2 6 2

bolt bolt (12) 113.1 mm 113.1 10 m

4 4

A d

Double shear: bolt bolt 6

bolt

PROBLEM 6.56 (Continued)

(b) Shearing stress at the center of the cross section.

For two steel plates, Qs  (2)(6)(9042)(9042) 76.032 10 mm 3 3 76.032 10 m 6 3 For the neutral axis, Q 54.81 10 6 76.032 10 6 130.842 10 m 6 3

Shear flow across the neutral axis:

6 3

6

(4000)(130.842 10 )

18.181 10 N/m 28.787 10

q VQ I

    

Double thickness: 2t 12 mm 0.012 m Shearing stress:

3 6

18.181 10

1.515 10 Pa

2 0.012

  q  

t

1.515 MPa

  

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PROBLEM 6.57

A composite beam is made by attaching the timber and steel portions shown with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of 4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress at the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION

Widths of transformed section:

3 2 3

2 (150)(12) (150)(12)(125 6) (7.5)(250)

12 12

2[0.0216 10 30.890 10 ] 9.766 10 71.589 10 mm 71.589 10 m

bolt bolt 6

bolt

(4 10 )(235.8 10 )

13.175 10 N/m 71.589 10

2 in.

1 in.

1.5 in.

Aluminum

Steel

PROBLEM 6.58

A steel bar and an aluminum bar are bonded together as shown to form a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the modulus of elasticity is 29 10 6 psi for the steel and 10.6 10 6 psi for the aluminum, determine (a) the average stress at the bonded surface, (b) the maximum shearing stress in the beam. (Hint: Use the method indicated in Prob. 6.55.)

Steel 4.1038 0.5 2.0519 0.6335 1.6469 0.3420

Σ 7.1038 8.0519 3.8994 1.3420

2 4

8.0519 1.1335 in.

7.1038

(1.5)(1.8665) 2.6129 in 2

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PROBLEM 6.59

A steel bar and an aluminum bar are bonded together as shown to form a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the modulus of elasticity is 29 10 6psi for the steel and 10.6 10 6 psi for the aluminum, determine (a) the average stress at the bonded surface, (b) the maximum shearing stress in the beam. (Hint: Use the method indicated in Prob. 6.55.)

Steel 8.2074 2.0 16.4148 0.2318 0.4410 2.7358

Alum. 1.5 0.5 0.75 1.2682 2.4125 0.1250

(2.7358)(1.5)(1.2318) 3.1133 in

Q  2 

A J

x

C E K B Plastic

Neutral axis P

C'E' y

yY

PROBLEM 6.60

Consider the cantilever beam AB discussed in Sec. 6.5 and the portion ACKJ of the beam that is located to the left of the transverse section CC and above the horizontal plane JK, where K is a point at a distance yyY above the neutral axis. (See figure.) (a) Recalling that x Y between C and E and x (Y/ )y yY between E and K, show that the magnitude of the horizontal shearing force H exerted on the lower face of the portion of beam ACKJ is

1 2

2 Y 2 Y Y

H b c y y

y

    

(b) Observing that the shearing stress at K is

0 0

1 1

lim lim

xy A x

H H H

A b x b x

 

   

 

  

 

and recalling that y is a function of x defined by Eq. (6.14), derive Y Eq. (6.15).

SOLUTION

Point K is located a distance y above the neutral axis.

The stress distribution is given by

for 0 and for .

Y Y Y Y

Y

y y y y y c

  y      

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957

PROBLEM 6.60 (Continued)

For equilibrium of horizontal forces acting on ACKJ,

2 2

Differentiating, 3 2 2

2 3

e

PROBLEM 6.61

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

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PROBLEM 6.62

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

PROBLEM 6.62 (Continued)

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PROBLEM 6.63

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

A

PROBLEM 6.64

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

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PROBLEM 6.65

An extruded beam has the cross section shown. Determine (a) the location of the shear center O, (b) the distribution of the shearing stresses caused by the vertical shearing force V shown applied at O.

SOLUTION

2

3 3

6 4 6 4

1 192 1

2 (72)(6) (72)(6) (12)(192)

12 2 12

15.0431 10 mm 15.0431 10 m I

(576)(72) 41.472 10 mm 41.472 10 m 6 mm 0.006 m

(110 10 )(41.472 10 ) (15.0431 10 )(0.006)

B

PROBLEM 6.65 (Continued)

Part BD:

Point B: 3 3 6 3

3 6

6

96 mm 41.472 10 mm 41.472 10 m 12 mm 0.012 m

(110 10 )(41.472 10 ) (15.0431 10 )(0.012)

B

y Q

t VQ

It

    

 

 

 

 

25.271 10 Pa6

  25.3 B  MPa 

Point C:

3 3 3 6 3

0, 0.012 m

41.472 10 (12)(96) 96 96.768 10 mm 96.768 10 m 2

y t

Q

 

 

       

3 6

6 6

(110 10 )(96.768 10 )

58.967 10 Pa (15.0431 10 )(0.012)

VQ It

 

   

  59.0 C  MPa 

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PROBLEM 6.66

An extruded beam has the cross section shown. Determine (a) the location of the shear center O, (b) the distribution of the shearing stresses caused by the vertical shearing force V shown applied at O.

SOLUTION

1(0.125)(3) 1.125 in 3

1 (4)(0.125) (4)(0.125)(3) 4.50065 in 12

PROBLEM 6.66 (Continued)

Part EF: By symmetry with part BD, EF 42 Vt

FI

Its moment about H is 3 EF 126Vt

FI

Part FG: By symmetry with part AB, FG 4.5VT

FI

Its moment about H is 4 FG 18 Vt

FI

Moment about H of force in part DE is zero.

(18 126 0 126 18) 144 144 (2.88)(0.125)

13.50

H

Vt Vt

Ve M

I I

e t I

       

  e2.67 in. 

(b) QAQG 0 AG   0

4.5QBQFt

(2.75)(4.5 ) 13.50

B F B

VQ t

It t

  

  0.917 B F  ksi

QDQE 16.5t

0 (2.75)(16.5 )

13.50

D E

VQ t

It t

  

   D E 3.36 ksi 

At H (neutral axis), QHQDt(3)(1.5) 21 t (2.75)(21 )

13.50

H H

VQ t

It t

 

 4.28 H  ksi 

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PROBLEM 6.67

An extruded beam has the cross section shown. Determine (a) the location of the shear center O, (b) the distribution of the shearing stresses caused by the vertical shearing force V shown applied at O.

SOLUTION

PROBLEM 6.67 (Continued)

(b) At A D F, , , and ,G Q 0 ADF G   0 Just above B: Q1QAB (2 )(4) 8tt

1 1

(2.75)(8 ) (19.417)

VQ t

It t

 

 11.133 ksi 

Just to the right of B: Q2QBD (3) (4) 12tt

2 2

(2.75)(12 ) (19.417)

VQ t

It t

   2 1.700 ksi 

Just below B: Q3Q1Q2 20t

3 3

(2.75)(20 ) (19.417)

VQ t

It t

   3 2.83 ksi 

At H (neutral axis), QHQ3QBH 20t t (3)(1.5) 24.5 t (2.75)(24.5 )

(19.417)

H H

VQ t

It t

 

 3.47 H  ksi 

By symmetry, 432.83 ksi 

 5 2 1.700 ksi 

6 1 1.133 ksi

   

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PROBLEM 6.68

An extruded beam has the cross section shown. Determine (a) the location of the shear center O, (b) the distribution of the shearing stresses caused by the vertical shearing force V shown applied at O.

SOLUTION

3 2 6 4

3 2 6 4

1 (30)(6) (30)(6)(45) 0.365 10 mm 12

1 (30)(4) (30)(4)(15) 0.02716 10 mm 12

1.14882 10 mm IAH (2)(1.14882 10 )

AB (2)(1.14882 10 )

DE

PROBLEM 6.68 (Continued)

(b) Calculation of shearing stresses.

3 6 4

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PROBLEM 6.69

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

(1.25)(3) 3.75 in

3 3

(2)(3.75) (2)(8.1667) 23.833 in4

  

E

PROBLEM 6.70

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

(420)(35) 171.5 10 mm

3 3

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PROBLEM 6.71

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

20 3.125 23.125

2 2

2 (23.125)(2.5) 0.62061

6 177.917

10[1 0.62061]

e 3  1.265 e in. 

Note that the lines of action of F and AB F pass through point K. Thus, these forces have zero moment DE about point K.

60 mm

PROBLEM 6.72

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

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975 e

O a t A

B

PROBLEM 6.73

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

For whole cross section, A2at

2 3 1 3

2 2

JAa  a t IJ a t Use polar coordinate  for partial cross section.

2

t

PROBLEM 6.74

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

For a thin-walled hollow circular cross section, A2at

2 3 1 3

2 2

Ja A a t IJ a t For the half-pipe section, 3

I 2a t Use polar coordinate  for partial cross section.

2

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PROBLEM 6.75

A thin-walled beam has the cross section shown. Determine the location of the shear center O of the cross section.

SOLUTION

e

A thin-walled beam has the cross section shown. Determine the location of the shear center O of the cross section.

SOLUTION

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979

PROBLEM 6.76 (Continued)

Likewise, for Part DE, 2 3 233 3

1 2 3

  

F h V

h h h

and for Part FG,

33

3 3 3 3

1 2 3

  

F h V

h h h

MH  MH: 2 3 3 23 3 333

1 2 3

2 2

  

 

bh bh

Ve bF bF V

h h h

3 3 3 3

2 3

3 3 3 3 3 3

1 2 3

2 (60) (2)(40) (80) (60) (40) (50)

 

 

   

h h

e b

h h h

21.7 mm

e 21.7 mm ◄

b

A thin-walled beam of uniform thickness has the cross section shown. Determine the dimension b for which the shear center O of the cross section is located at the point indicated.

F is not required, since its moment about O is zero. DE

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981

PROBLEM 6.77 (Continued)

2 3

2

2 3 2 2 2

0 : ( ) 0

2 2 0

2 2 1 0

2 6 2

2 1 1 2 1 1

2 6 2 2 0

O AB FG B BD F EF

AB B BD

A AB AB

B B B

A AB AB A AB AB B B

M b F F y F y F

b F y F

y l l

Vt V

b y Q b ty b

I I

Vt Vt

y l l b y l l y b y b

I I

     

 

   

      

        

   

   

Dividing by 2Vt

I and substituting numerical data,

2 3 2 2 2

1 1 1 1

(90)(60) (60) (90)(60) (60) (30) (30) 0

2 6 b 2 b 2 b

       

   

   

3 3 2

126 10 b108 10 b450b  0

3 2

18 10 b450b  0 and0 bb40.0 mm 

b

A thin-walled beam of uniform thickness has the cross section shown.

Determine the dimension b for which the shear center O of the cross section is located at the point indicated.

SOLUTION

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.

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983

PROBLEM 6.79

For the angle shape and loading of Sample Prob. 6.6, check that

q dz0 along the horizontal leg of the angle and

q dy P along its vertical leg.

SOLUTION

Refer to Sample Prob. 6.6.

Along horizontal leg: 3 ( )(3 3 ) 3 3 2 2

PROBLEM 6.80

For the angle shape and loading of Sample Prob. 6.6, (a) determine the points where the shearing stress is maximum and the corresponding values of the stress, (b) verify that the points obtained are located on the neutral axis corresponding to the given loading.

SOLUTION

Refer to Sample Prob. 6.6.

(a) Along vertical leg: 3 ( )(3 5 ) 3 3 2 2

tan tan 1 14.036

4

45 14.036 30.964

     

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985

PROBLEM 6.80 (Continued)

Neutral axis intersects vertical leg at

tan 30.964 y y z

1 1

tan 30.964 0.400

4 4 a a

 

   

2 y5a  Neutral axis intersects horizontal leg at

tan (45 + ) 1 1

tan 59.036 0.667 4 4

z z y

a a

  

 

   

2 z3a 

2a

PROBLEM 6.81

Determine the distribution of the shearing stresses along line D B  in the horizontal leg of the angle shape for the loading shown. The x and y axes are the principal centroidal axes of the cross section.

SOLUTION

15.8 cos sin

( ) (2 ) 1(2 ), 0

Coordinate transformation. 2 1

cos sin

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987

PROBLEM 6.81 (Continued)

x y

y x

V Ax V Ay I t I t

 

   

3

3

3

( cos )(2 )( )(0.13614 0.06961 ) (0.1557 )( )

( sin )(2 )(0.48111 0.36612 ) (1.428 )( )

(2 )(0.750 0.500 )

P a y t y a

ta t

P a y y a

a t t

P a y y a

ta

ta

In document Mechanics of Materials 7th Edition Beer Johnson Chapter 6 (Page 47-98)