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Solucionario mecanica de materiales, Apuntes de Mecánica de Materiales

solucionario completo sobre problemas del libro de mecanica de materiales capitulo 4 y 6

Tipo: Apuntes

2022/2023

Subido el 31/03/2025

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Chapter 3
3-1
0
o
M
18 6(100) 0
B
R
33.3 lbf .
B
R
Ans
0
y
F
100 0
oB
RR
66.7 lbf .
o
R
Ans
33.3 lbf .
CB
R
RA ns
______________________________________________________________________________
3-2
Body AB:
0
x
F
A
xBx
R
R
0
y
F
A
yBy
R
R
0
B
M (10) (10) 0
Ay Ax
RR
A
xAy
R
R
Body OAC:
0
O
M (10) 100(30) 0
Ay
R
300 lbf .
Ay
R
Ans
0
x
F 300 lbf .
Ox Ax
R
RA  ns
0
y
F 100 0
Oy Ay
RR
200 lbf .
Oy
R
Ans
______________________________________________________________________________
Chapter 3 - Rev. A, Page 1/100
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Chapter 3

M (^) o  0 18 RB  6(100)  0 RB  33.3 lbf Ans.

F (^) y  0 RoRB  100  0 Ro  66.7 lbf Ans. RCRB  33.3 lbf Ans.

______________________________________________________________________________

3- Body AB :  F (^) x  0 RA (^) xRBxF (^) y  0 RA (^) yRByM (^) B  0 RAy (10)  RAx (10)  0 R (^) AxRAy

Body OAC :  M (^) O  0 RAy (10) 100(30)  0 R (^) Ay  300 lbf Ans.  F (^) x  0 ROx   R (^) Ax  300 lbf Ans.  F (^) y  0 ROyRAy  100  0 ROy   200 lbf Ans. ______________________________________________________________________________

0.8 (^) 1.39 kN. O tan 30 R  (^)  Ans 0.8 (^) 1.6 kN. A sin 30 R  (^)  Ans

______________________________________________________________________________

Step 1: Find RA & RE 4.5 (^) 7.794 m tan 30 0 9 7.794(400 cos 30 ) 4.5(400sin 30 ) 0 400 N.

A E

E

h M R

R Ans

 

2 2

0 400cos30 0 346.4 N 0 400 400sin 30 0 200 N 346.4 200 400 N.

x Ax Ax y Ay Ay A

F R
R
F R
R

R Ans

Step 2: Find components of RC on link 4 and RD

4

305.4 N.
0 305.4 N
0 ( ) 400 N

C D D x Cx y Cy

M
R

R Ans F R F R

 

 M C  0
 1500 R 1  300(5)  1200(9) 0 

R 1 (^) 8.2 kN Ans.  F (^) y  0 8.2  9  5  R 2  0 R 2 (^) 5.8 kN Ans.

M (^) 1  8.2(300)  2460 N m Ans. M (^) 2  2460  0.8(900)  1740 N m Ans. M (^) 3  1740  5.8(300)  0 checks! _____________________________________________________________________________

3-

F (^) y  0 R 0 (^)  500  40(6) 740 lbf Ans.  M 0  0 M (^) 0  500(8)  40(6)(17)  8080 lbf in Ans.

M (^) 1   8080  740(8)  2160 lbf in  Ans. M (^) 2   2160  240(6)  720 lbf in  Ans.

3 720 1 (240)(6) 0 checks! 2

M    
______________________________________________________________________________
 M B  0
2.2 R 1  1(2) 1(4)  0

R 1 (^)  0.91 kN Ans.  F (^) y  0 0.91  2  R 2  4  0 R 2 (^) 6.91 kN Ans.

M 1 (^)  0.91(1.2)  1.09 kN m  Ans.

M (^) 2  1.09  2.91(1)  4 kN m  Ans. M (^) 3   4  4(1)  0 checks!

______________________________________________________________________________

Break at the hinge at B

Beam OB : From symmetry, R 1 (^)  VB 200 lbf Ans.

Beam BD :  M (^) D  0 200(12)  R 2 (10)  40(10)(5) 0  R 2 (^) 440 lbf Ans.

F (^) y  0  200  440  40(10)  R 3  0 R 3 (^) 160 lbf Ans.

1 2 1 0 0 0 0 1 0 1 1 0 0 1 2 2 0 0

at 20 in, 0, Eqs. (1) and (2) give

q R x M x x x x V R M x x x x M R x M x x x x V M R

   

0 ^  0

0 0 2 0

500 40 20 14 0 740 lbf. (20) 500(20 8) 20(20 14) 0 8080 lbf in.

R Ans R M M

         Ans

0 8 : 740 lbf, 740 8080 lbf in 8 14 : 740 500 240 lbf 740 8080 500( 8) 240 4080 lbf in 14 20 : 740 500 40( 14) 40 800 lbf 740 8080

x V M x x V M x x x x V x x M x

   500( x  8)  20( x  14) 2   20 x^2  800 x  8000 lbf in Plots of V and M are the same as in Prob. 3-6. ______________________________________________________________________________

3- 1 1 1 1 1 2 0 0 0 1 2 1 1 1 1 2

q R x x R x x V R x R x x M R x x R x x

 ^   ^   ^   

at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2),

Solving Eqs. (3) and (4) simultaneously,

1 2 1 2 1 2 1 2

R R R R
R R R R

R 1 = -0.91 kN, R 2 = 6.91 kN Ans. 0 1.2 : 0.91 kN, 0.91 kN m 1.2 2.2 : 0.91 2 2.91 kN 0.91 2( 1.2) 2.91 2.4 kN m 2.2 3.2 : 0.91 2 6.91 4 kN 0.91 2(

x V M x x V M x x x x V M x x

   1.2)  6.91( x  2.2)  4 x  12.8 kN m Plots of V and M are the same as in Prob. 3-7. ______________________________________________________________________________

1 1 1 0 0 1 1 2 3 0 0 1 1 0 1 2 3 1 1 2 2 1 1 2 3 1

0 at 8 in 8 400(

q R x x R x x x R x V R x R x x x R x M R x x R x x x R x M x R

 ^   ^   ^       

    8  4)  0  R 1 200 lbf Ans. at x = 20+, V = M = 0. Applying Eqs. (1) and (2), 2 3 2 3 2 2 2

200(20) 400(16) (10) 20(10) 0 440 lbf.

R R R R
R R A

3 600 440 160 lbf^.

ns R    Ans

0 4 : 200 lbf, 200 lbf in 4 10 : 200 400 200 lbf, 200 400( 4) 200 1600 lbf in 10 20 : 200 400 440 40( 10) 640 40 lbf 200 400( 4)

x V M x x V M x x x x V x x M x x

    440( x 10)  20  x  10  2   20 x^2  640 x  4800 lbf in

Plots of V and M are the same as in Prob. 3-8. ______________________________________________________________________________

3-13 Solution depends upon the beam selected. ______________________________________________________________________________

3- (a) Moment at center,

2

c

c

x l^ a

M l^ l a l^ l a

 ^ ^  ^ 

w wl

At reaction, Mrwa^2 a = 2.25, l = 10 in, w = 100 lbf/in

 2 

100(10) (^10) 2.25 125 lbf in 2 4 100 2. 253 lbf in. 2

c

r

M

M Ans

 ^  

(b) Optimal occurs when M (^) cMr

(c)

2 2 1 2

(^24 10) 17 kpsi 2 (^24 10) 7 kpsi 2 7 6 9.22 kpsi 17 9.22 26.22 kpsi 17 9.22 7.78 kpsi

C
CD
R

 

(^1 90) tan 1 7 69.7 ccw p 2 6

  ^  ^ 

 ^ 

 

1 9.22 kpsi s 69.7^45 24.7^ ccw

 R

 ^  ^  

(d)

2 2 1 2

(^12 22) 5 kpsi 2 (^12 22) 17 kpsi 2 17 12 20.81 kpsi 5 20.81 25.81 kpsi 5 20.81 15.81 kpsi

C
CD
R
 ^  

(^1 90) tan 1 17 72.39 cw p 2 12

  ^  ^ 

 ^ 

 

1 20.81 kpsi s 72.39^45 27.39^ cw

 R

______________________________________________________________________________

(c)

2 2 1 2

(^12 4) 4 MPa 2 (^12 4) 8 MPa 2 8 7 10.63 MPa 4 10.63 14.63 MPa 4 10.63 6.63 MPa

C
CD
R

(^1 90) tan 1 8 69.4 ccw p 2 7

  ^  ^ 

 ^ 

 

1 10.63 MPa s 69.4^45 24.4^ ccw

 R

 ^  ^  

(d)

2 2 1 2

(^6 5) 0.5 MPa 2 (^6 5) 5.5 MPa 2 5.5 8 9.71 MPa 0.5 9.71 10.21 MPa 0.5 9.71 9.21 MPa

C
CD
R

(^1) tan 1 8 27.75 ccw p 2 5.

  ^ ^ 

1 9.71 MPa s^45 27.75^ 17.25^ cw

 R

 ^  ^  
______________________________________________________________________________

(a)

2 2 1 2

(^12 6) 9 kpsi 2 (^12 6) 3 kpsi 2 3 4 5 kpsi 5 9 14 kpsi 9 5 4 kpsi

C
CD
R

(^1) tan 1 4 26.6 ccw p 2 3

  ^ ^ ^ 

1 5 kpsi s^45 26.6^ 18.4^ ccw

 R

 ^  ^  

(b)

2 2 1 2

(^30 10) 10 kpsi 2 (^30 10) 20 kpsi 2 20 10 22.36 kpsi 10 22.36 32.36 kpsi 10 22.36 12.36 kpsi

C
CD
R

(^1) tan 1 10 13.28 ccw p 2 20

  ^ ^ 

1 22.36 kpsi s^45 13.28^ 31.72^ cw

 R

 ^  ^  

(a)

2 2 1 2 3

(^80 30) 55 MPa 2 (^80 30) 25 MPa 2 25 20 32.02 MPa 0 MPa 55 32.02 22.98 23.0 MPa 55 32.0 87.0 MPa

C
CD
R
 ^   

1 2 2 3 1 3

(^23) 11.5 MPa, 32.0 MPa, 87 43.5 MPa 2 2

(b)

2 2 1 2 3

(^30 60) 15 MPa 2 (^60 30) 45 MPa 2 45 30 54.1 MPa 15 54.1 39.1 MPa 0 MPa 15 54.1 69.1 MPa

C
CD
R

1 3

1 2

2 3

39.1 69.1 (^) 54.1 MPa 2 39.1 (^) 19.6 MPa 2 69.1 (^) 34.6 MPa 2

(c)

2 2 1 2 3

(^40 0) 20 MPa 2 (^40 0) 20 MPa 2 20 20 28.3 MPa 20 28.3 48.3 MPa 20 28.3 8.3 MPa z 30 MPa

C
CD
R

1 3 1 2 2 3

48.3 (^30) 39.1 MPa, 28.3 MPa, 30 8.3 10.9 MPa 2 2

  ^       

(d)

2 2 1 2 3

(^50) 25 MPa 2 (^50) 25 MPa 2 25 30 39.1 MPa 25 39.1 64.1 MPa 25 39.1 14.1 MPa z 20 MPa

C
CD
R

1 3 1 2 2 3

64.1 (^20) 42.1 MPa, 39.1 MPa, 20 14.1 2.95 MPa 2 2

  ^       

______________________________________________________________________________

(a) Since there are no shear stresses on the stress element, the stress element already represents principal stresses. 1 2 3

10 kpsi 0 kpsi 4 kpsi

x

y

1 3

1 2

2 3

10 ( 4) (^) 7 kpsi 2 (^10) 5 kpsi 2 0 ( 4) (^) 2 kpsi 2

3-20 From Eq. (3-15),

3 2 2 2 2 2 2 3

 ^2 

     ^           

Roots are: 21.04, 5.67, –26.71 kpsi Ans. 1 2

2 3

max 1 3

21.04 5.67 (^) 7.69 kpsi 2 5.67 26.71 (^) 16.19 kpsi 2 21.04 26.71 (^) 23.88 kpsi. 2 Ans

_____________________________________________________________________________

From Eq. (3-15)

3 2 2 2

(^2 2 )

3 2

 ^2 

    ^       
 ^       

Roots are: 60, 20, –40 kpsi Ans.

1 2

2 3

max 1 3

(^60 20) 20 kpsi 2 (^20 40) 30 kpsi 2 (^60 40) 50 kpsi. 2

Ans

_____________________________________________________________________________

From Eq. (3-15)

3 2 2   2  ^2

2 2 2 3 2

    ^        
 ^         

Roots are: 90, 0, 0 MPa Ans.

2 3 1 2 1 3 max

(^90) 45 MPa. 2 Ans

_____________________________________________________________________________

  

 

 

2

6

1 6

(^15000) 33 950 psi 34.0 kpsi. 4 0.

33 950 60 0.0679 in. 30 10 0.0679 (^) 1130 10 1130. 60

F (^) Ans A FL L (^) Ans AE E

Ans L

From Table A-5, v = 0.

   

2 1 6 6 2

330 10 (0.75) 248 10 in.

v A d d An

ns s

 

_____________________________________________________________________________

  

 

 

2

6

6 1

(^3000) 6790 psi 6.79 kpsi. 4 0.

6790 60 0.0392 in. 10.4 10 0.0392 (^) 653 10 653. 60

F (^) Ans A FL L (^) Ans AE E

Ans L

From Table A-5, v = 0.

   

2 1 2 6 6

217 10 (0.75) 163 10 in.

v Ans d d An s

 