Solution of strength of materials problems PDF

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Strength of Materials Solutions Problem #1

 x  10500 psi, Tensile  y  5500 psi  xy  4000 psi 3 0

Principal stresses:

1,  2 

x y 2

  y     xy2   x 2   2

Substitute values from above yields:  1  11444 psi  2   6444 psi

The maximum shear stress is determined by these two principal stresses as:

Max( max,12 ,  max,13 ,  max,23 )

 max,12   max

1  2

 max,1,3 

1   3

2 2 11444  6444   8944 psi 2

 max,23 

2 3 2

Note that the other maximum shear stresses are less than this value.

Problem #2 The total strain is:

total   t 

L 0.05   0.0005 L 100

This total strain is equal to:

 t   M  T  Substituting:

F  T EA

 t  0.0005 E  30 * 106 A1

  6.5 * 10 6 T  100

and solving for F we get: F=-4500 lbs The stress is 4500 psi compressive. Problem #3 2000 lb 6 ft Q 20000 lb 4.5 ft

4.5 ft

The 2000 lbs creates a bending stress at Q that is tensile and is equal to:

x 

Mzy Iz

Where M z  2000* 4.5 *12 d 2 2 d 4 Iz  64 y

Substituting into the bending formula, we get

 x  17188 psi

The stress due to the axial load is compressive and is equal to:  x' 

F 20000   1591 psi A  (2) 2

The total stress is:   171881591 15600

Problem #4

a

D

c 7/8

b y

3/8 1.5 Location of centeroid: Ay  2 Aa y a  Ab y b 1.125y  2(.375)(.75)  .375(.5) y  0.667 The area moment of inertia: I z  2 Ia  Ib 1 Ia (.25)(1.5)3  .375(.75 .667)2 12 1 I b  (1.5)(.25) 3  .375(.667  .375  .125) 2 12

The answer is: I z  0.158 in

4

The stress at D is:



M zc 10000* (0.833)   57200 0.158 Iz

note that c=1.5-0.667=0.833 Problem #5: Curved beam The neutral axis radius is:

rn 

ro  ri 4 2  2.8854  ln( ro / ri ) ln( 4 ) 2

psi

The stress at inner radius (critical point) is:

i 

M (rn  ri ) 30000(2 .8854  2)   57945 psi eAri (3  2 .8854)(2)(2)

There is also an axial stress of 5000 psi acting on the cross-section making the total stress become

 total  62945 psi Problem #6: Torsional stresses The maximum torsional shear stress is:

 

Tr 16T 16(2000* 6 *12)  3   11460 psi 3  (4) J d

Form Problem # 3, the normal stress on the surface is 15600 psi. The state of stress is shown below: xy x

The principal stresses are calculated as before using:

x

2

  2 1 ,  2    x    xy  21662 and 2  2  The maximum shear stress at point Q is:

 max  Problem #7

21662  6062  13860 psi 2

 6062

psi

30”

20”

The forces in the upper portion (Fu) and lower portion (FL) are:

FU 

KU F KU  K L

FL 

KL F KU  K L

Where KU 

EA 30

and K L 

EA 20

Substituting into the force expressions:

FU 

1 30

F 

1 1  30 20 3 FL  F  480 5

2 (800)  320 lbs 5

The maximum stress is (480/0.5)=960 psi Problem #8

4 ft

6 ft

The torque is divided according to torsional stiffnesses. In this case the left supports picks us (6/10)=0.6 of the torque and the right support takes 0.4 of the torque. Problem #9 P

y-bar Cross-section yc

The stress is

 

VQ IZ b

Finding the centroid is as before: y

2 * 4 * (2  4  1)  3 .5 2(2 * 4)

The area moment of inertia is: I

1 1 (2)(4) 3  (2)(4)(3.5  2) 2  (4)(2)3  (2)(4)(5  3.5)2  49.3 12 12

Q is Q  ( 2)(4)(5  3.5)  12

and



VQ V (12)   11  V  90.4 I Z b 49.3( 2)

Problem #10

lbs

Y

Z



VQ 250 * (1 * 6)(4.5)   11.8 IZ b 285.6(2)

psi

Problem #11

For this thin-walled tube:

 

T 200 *1000   34.6 Mpa 2 At 2(38 * 38)2

The angle of rotation is:



200 *1000* (4 * 38) * 50 *10 TSL   0.011 rad 4(38 * 38) 2 * 79 *103 * 2 4A 2Gt

  0.66 deg . Problem #12 The critical point is the inner radius. The tangential stress is:

P P Pi ri2  Po ro2  ri2 ro2  o 2 i   r  t  2 2 ro  ri Setting r=ri and Pi=0 we get

2ro2 2(0.875)2  t   Po 2 2  11200*  45733 psi ro  ri 0.8752  0.6252 The state of stress is simple – just this tangential stress which is also the principal stress. From theory, we know that there are no shear stresses on these surfaces when the stress element in oriented with radial edges.

x

The factor of safety is:

SF 

57000  1.25 45733

Problem #S13

The critical point is the inner radius. Using the formula:

ri 2ro2 1  3 2 3 2 2 r ) )(ri  ro  2   t   ( r 8 3 3  .24 1  3(.24) )(12.5 2  752  752  (12.5)2 )(106 )  t  (3320)(216.7) 2 ( 8 3  .24  t  714963 pa 2

Problem #S14 The interface pressure is:

E P r R

 (ro2  R 2 )(R2  ri 2 )    2 2 2  2R (ro  ri ) 

The radial interference is 0.013 mm. Substituting all the numbers:

E  207 *103

R  20 ri  0 ro  40

The answer is P=50.4 Mpa. Problem #S15 and #S16

6” 12” 6” 1000

# S15

1000

# S16

Problem #S15: Using the impact formula and simplifying for h>>:

 2hk  2 * 1 * 2.5 * 10 6     (1000)  70.7 * 103 Fe  W  1000  W  AE 1 * 30 * 106 K   2.5 * 106 12 L

lbs

The stress is 70.7 ksi.

Problem #S16: The two bars form a pair of two springs in series. The equivalent spring is: Ke 

K1 K2 AA E 1(2) 30 * 10 6 )  33.3 * 10 6 (  1 2 ( ) 6 K1  K 2 A1  A 2 L1 1  2

 2hk  W  Fe    W    max  81.6 ksi

lb / in

2 * 1 * 3.33 * 106 (1000)  81.6 * 103 lbs 1000

Problem #17 The area, moment of inertia, and radius of gyration: A  r 2  1.767 and and

k

I

d 4 64



 (1.5) 4 64

 .2485

I  .375 A

The slenderness ratio is: l 60   160 k .375 The limit for the use of Euler versus Johnson formula is: l     k 1

2 2CE  Sy

2 2 (1)(30 * 10 6 )  93.3 68000

Since the slenderness ratio is larger than the limit, the Euler formula applies:

Pcr  C

 2 EI 2

L



 2 (30 *106 )(.2485) 60

2

The factor of Safety is: SF=20438/5655 = 3.6 b) For this case: A  0.60 I  0 .0288 k  0 .219

The slenderness ratio is:

 20438

l  k

18 0.0288 .60

 82

We have to use the Johnson Formula: 2   Sy L  1  1 68000*18 2    Pcr  AS y     0.6 68000  ( )   24975 lbs 6 30 *10 2  (.219)     2 k  CE  24975  4.42 SF  5655

Problem #S18: Based on DET: Tr 16T 16 * 20000  3   12732.4 J d  (2)3 .58S y .58 * 50000   2.27 FS   12732.4



FS 

.5S y



psi

2

Based on MST: Problem #S19 For this cast iron : Sut  30000   0.211   .26 lbs /in 3

The critical point is the inner radius

3   2 2 ri2 ro2 1  3 2 )( ri  ro  2  r )  t   ( 8 3  r .26 3  .211 1  3(.211) 2 ( ) 2 ( )(2 * 52  32  (3) ) t  386 8 3  .211  t  .0147 2 2

Since this is a principal stress and the other principal stress is zero (radial stress is zero on the inner radius), we equate this stress to S ut.

0.0147 2  30000    1428 rad / sec  13600 rpm Problem #S21 The question in this problem is the factor of safety against eventual fatigue failure. First we calculate the maximum nominal shear stress:

Tr 56(103 )(10)    35.6  4 J (20) 32

Mpa

We would apply the fatigue stress concentration factor to the nominal stress to get the actual stress

  1.48(35.6)  52.8 Mpa

The need to find the VonMises stress and compare it to strength

 v ,a  3 2  91.4 Mpa On the strength side, the estimate of the endurance limit of the rotating bending fatigue specimen is half of the tensile strength for steels:

Se'  (0.5)518  259 Mpa Applying the correction factors to estimate the endurance limit of this part:

Se  (0.9)(0.78)(256)  182 Mpa The factor of safety is:

n

182 2 91...