Mechanics Of Materials 7th edition Beer Johnson chapter 5 PDF

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CHAPTER 5

PROBLEM 5.1

w A

B

L

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves.

SOLUTION Reactions:

M B  0:  AL  wL 

L  0 2

A

wL 2

M A  0: BL  wL 

L  0 2

B 

wL 2

Free body diagram for determining reactions: Over whole beam,

0 x  L

Place section at x. Replace distributed load by equivalent concentrated load. Fy  0:

wL  wx  V  0 2 L  V  w  x   2  

M J  0:  M 

wL x x  wx  M  0 2 2

w ( Lx  x 2) 2 M 

Maximum bending moment occurs at x 

w x (L  x )  2

L . 2 Mmax 

wL2  8

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PROBLEM 5.2

P A

B

C

a

b

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves.

L

SOLUTION Reactions:



M C  0: LA  bP  0

A

Pb L

M A  0: LC  aP  0

C 

Pa L

0 x  a

From A to B,

Fy  0:

Pb V  0 L



Pb  L

V   M J  0: M 

Pb x0 L M 

Pbx  L

a x L

From B to C,

Fy  0: V 

Pa 0 L V 

 M K  0:  M 

Pa ( L  x)  0 L M 

Pa (L  x )  L M 

At section B,

Pa  L

Pab  L2

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PROBLEM 5.3

w0

A

B L

For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the equations of the shear and bendingmoment curves.

SOLUTION Free body diagram for determining reactions.

Reactions:  F y  0: R A 

w0 L 0 2

RA 

w0 L 2

 w L  2L  M A  0: M A   0   0  2  3  M

A



w0L2 w L2  0 3 3

Use portion to left of the section as the free body.

Replace distributed load with equivalent concentrated load. Fy  0:

w0 L 1 w0 x   x V  0 2 2 L V 

w 0L w 0x 2   2 2L

M J  0: w0L2  w0L  3  2

  1 w0x  x   x    M  0  ( x)    2 L  3  M 

w0 L2 w0 Lx w0 x 3    3 2 6L

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PROBLEM 5.4

w B

A L

For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the equations of the shear and bendingmoment curves.

SOLUTION Free body diagram for determining reactions.

Reactions: Fy  0: RA  wL  0

R A  wL  L M A  0: M A  (wL)    0 2

M

A



w0L2 2

Use portion to the right of the section as the free body.

Replace distributed load by equivalent concentrated load.

Fy  0: V  w( L  x)  0 V  w( L  x)   L x M J  0:  M  w(L  x )  0  2 

M 

w (L  x )2  2

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P

PROBLEM 5.5

P B

C

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves.

A a

a

SOLUTION 0x a

From A to B:

 Fy  0 :

 P V  0 V   P 



M J  0 :

Px  M  0 M   Px 

a  x  2a

From B to C:

Fy  0 :

 P  P V  0 V   2 P 

M J  0 :

Px  P (x  a )  M  0 M  2 Px  Pa 



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w

PROBLEM 5.6

w B

C

D

A a

a

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves.

L

SOLUTION Reactions:

A  D  wa

From A to B,

0 x a

Fy  0:

wa  wx  V  0

V  w(a  x)  MJ  0: wax  ( wx)

x M 0 2  x2  M  w ax    2  

a  x  La

From B to C, Fy  0:

wa  wa  V  0 V 0  a  wax  wa  x    M  0 2 

MJ  0: From C to D, Fy  0:

M 

1 2 wa  2

La  x  L V  w(L  x )  wa  0

V  w( L  x  a)  L  x   M  w( L  x )    wa( L  x )  0  2 

M J  0:

1   M  w  a( L  x)  ( L  x) 2   2  

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3 kN A

C

0.3 m

5 kN

2 kN

PROBLEM 5.7

E

B

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

2 kN D

0.3 m

0.3 m

0.4 m

SOLUTION Origin at A:



Reaction at A: Fy  0: RA  3  2  5  2  0

M

A

 0: M

A

RA  2 kN

 (3 kN)(0.3 m)  (2 kN)(0.6 m)  (5 kN)(0.9 m)  (2 kN)(1.3 m)  0 M A  0.2 kN  m





From A to C:  Fy  0:

V  2 kN

M1  0:

0.2 kN  m  (2 kN)x  M  0 M  0.2  2 x

From C to D: Fy  0: 2  3  V  0

V  1 kN M 2  0:

 0.2 kN  m  (2 kN)x  (3 kN)( x  0.3)  M  0 M  0.7  x

From D to E: Fy  0: V  5  2  0

M 3  0:

V  3 kN

 M  5(0.9  x )  (2)(1.3  x )  0 M   1.9  3x

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PROBLEM 5.7 (Continued)

From E to B: Fy  0: V  2 kN

M 4  0:

 M  2(1.3  x)  0 M   2.6  2x

 

(a) (b) M

V max

max

 3.00 kN 

 0.800 kN  m 

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100 lb

250 lb C

100 lb

D

E

A

B

15 in.

20 in.

25 in.

PROBLEM 5.8 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

10 in.

SOLUTION Reactions:

MC  0:

RE (45 in.)  100 lb(15 in.)  250 lb(20 in.)  100 lb(55 in.)  0 R E  200 lb

F y  0:

RC  200 lb  100 lb  250 lb  100 lb  0

RC  250 lb At any point, V is the sum of the loads and reactions to the left (assuming + ) and M the sum of their moments about that point (assuming ).

(a) Vmax  150.0 lb 

(b)

M max  1500 lb  in. 

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PROBLEM 5.8 (Continued)

Detailed computations of moments: M

A

 0

M C   (100 lb)(15 in.)   1500 lb  in. M D  (100 lb)(35 in.)  (250 lb)(20 in.)   1500 lb  in. M E  (100 lb)(60 in.)  (250 lb)(45 in.)  (250 lb)(25 in.)  1000 lb  in. M B   (100 lb)(70 in.)  (250 lb)(55 in.)  (250 lb)(35 in.)  (200 lb)(10 in.)  0 (Checks)

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PROBLEM 5.9

25 kN/m C

D

A

B 40 kN

0.6 m

40 kN

1.8 m

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

0.6 m

SOLUTION The distributed load is replaced with an equivalent concentrated load of 45 kN to compute the reactions. (25 kN/m)(1.8 m)  45 kN M A  0:  (40 kN)(0.6 m)  45 kN(1.5 m)  40 kN(2.4 m)  RB (3.0 m)  0 RB  62.5 kN  Fy  0:

RA  62.5 kN  40 kN  45 kN  40 kN  0

RA  62.5 kN

At C: F y  0: V  62.5 kN

M 1  0: M  (62.5 kN)(0.6 m)  37.5 kN  m

At centerline of the beam: Fy  0:

62.5 kN  40 kN  (25 kN/m)(0.9 m)  V  0

V 0 M 2  0: M  (62.5 kN)(1.5 m)  (40 kN)(0.9 m)  (25 kN/m)(0.9 m)(0.45 m)  0 M  47.625 kN  m

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PROBLEM 5.9 (Continued)

Shear and bending-moment diagrams:

(a)

(b)

M

V

max

max

 62.5 kN 

 47.6 kN  m 

From A to C and D to B, V is uniform; therefore M is linear. From C to D, V is linear; therefore M is parabolic.

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2.5 kips/ft

PROBLEM 5.10

15 kips C

D

A

B

6 ft

3 ft

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

6 ft

SOLUTION M B  0: 15R A  (12)(6)(2.5)  (6)(15)  0 R A  18 kips M A  0: 15RB  (3)(6)(2.5)  (9)(15)  0 RB  12 kips Shear: VA  18 kips VC  18  (6)(2.5)  3 kips C to D:

V  3 kips

D to B :

V  3  15   12 kips

Areas under shear diagram:  1

A to C :

 V dx   2  (6)(18  3)  63 kip  ft  

C to D:

 V dx  (3)(3)  9 kip  ft

D to B :

 V dx  (6)( 12)  72 kip  ft MA  0

Bending moments:

MC  0  63  63 kip  ft M D  63  9  72 kip  ft M B  72  72  0 V 



M

max

max

 18.00 kips 

 72.0 kip  ft 

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3 kN

3 kN

PROBLEM 5.11

E

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

450 N ? m A

B C

D

300 mm

300 mm 200 mm

SOLUTION

M B  0: (700)(3)  450  (300)(3)  1000 A  0 A  2.55 kN M A  0:  (300)(3)  450  (700)(3)  1000 B  0 B  3.45 kN At A:

V  2.55 kN

A to C:

V  2.55 kN

M 0

 MC  0:

At C:

(300)(2.55)  M  0 M  765 N  m C to E:

V  0.45 N  m  MD  0:

At D:

(500)(2.55)  (200)(3)  M  0 M  675 N  m  MD  0:

At D:

(500)(2.55)  (200)(3)  450  M  0 M  1125 N  m E to B:

V  3.45 kN  ME  0:

At E:

M  (300)(3.45)  0 M  1035 N  m At B:

V  3.45 kN,

M 0 (a) (b)

M

V

max

max

 3.45 kN 

 1125 N  m 

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400 lb

1600 lb

PROBLEM 5.12

400 lb G

D

E

8 in.

F

A

B

8 in.

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

C

12 in.

12 in.

12 in.

12 in.

SOLUTION M G  0:  16C  (36)(400)  (12)(1600)  (12)(400)  0 Fx  0:  C  Gx  0

G x  1800 lb

F y  0: 400  1600  G y  400  0

G y  2400 lb

A to E:

V  400 lb

E to F:

V  2000 lb

F to B:

V  400 lb

At A and B,



C  1800 lb

M 0

At D ,

M D  0: (12)(400)  M  0

At D +,

M D  0: (12)(400)  (8)(1800)  M  0

M  9600 lb  in.

At E,

M E  0: (24)(400)  (8)(1800)  M  0

M  4800 lb  in.



M  4800 lb  in.

At F ,

M F  0: M  (8)(1800)  (12)(400)  0 M  19,200 lb  in.

At F ,+

M F  0: M  (12)(400)  0

(a) (b)

M  4800 lb  in.

Maximum |V |  2000 lb  Maximum | M|  19, 200 lb  in. 

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1.5 kN

1.5 kN

C

D

A

PROBLEM 5.13 B

0.3 m

0.9 m

0.3 m

Assuming that the reaction of the ground is uniformly distributed, draw the shear and bending-moment diagrams for the beam AB and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION Over the whole beam, Fy  0: 1.5w  1.5  1.5  0

w  2 kN/m

0  x  0.3 m

A to C:

Fy  0: 2 x  V  0

V  (2 x) kN

 x M J  0: (2 x)   M  0  2

At C ,

M  ( x2 ) kN  m

x  0.3 m V  0.6 kN, M  0.090 kN  m  90 N m

C to D:

0.3 m  x  1.2 m Fy  0: 2x  1.5  V  0

V  (2 x  1.5) kN

x  M J  0:  (2 x)    (1.5)( x  0.3)  M  0 2 

M  ( x2  1.5 x  0.45) kN  m At the center of the beam, x  0.75 m V 0

M  0.1125 kN  m   112.5 N m

At C +,

x  0.3 m,

V  0.9 kN (a) Maximum |V |  0.9 kN  900 N  (b)

Maximum |M |  112.5 N  m 

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24 kips

2 kips/ft C

D

PROBLEM 5.14

2 kips/ft E

A

B

3 ft

3 ft

3 ft

Assuming that the reaction of the ground is uniformly distributed, draw the shear and bending-moment diagrams for the beam AB and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

3 ft

SOLUTION Over the whole beam, F y  0: 12w  (3)(2)  24  (3)(2)  0 A to C:

w  3 kips/ft

(0  x  3 ft)

Fy  0: 3x  2x  V  0 MJ  0:  (3x ) At C,

V  ( x) kips

x x  (2x )  M  0 2 2

M  (0.5 x 2) kip  ft

x  3 ft V  3 kips, M  4.5 kip  ft

C to D:

(3 ft  x  6 ft) Fy  0: 3x  (2)(3)  V  0  MK  0: (3 x) 

V  (3x  6) kips