Strength of Materials- Combined Stresses- Hani Aziz Ameen

Strength of Materials Handout No.12

Combined Stresses Asst. Prof. Dr. Hani Aziz Ameen Technical College- Baghdad Ba ghdad Dies and Tools Eng. Dept. E-mail:[email protected] www.mediafire.com/haniazizameen

Strength of materials- Handout No.12- Combined Stresses - Dr. Hani Aziz Ameen

12-1 Introduction As should have become apparent from the preceding chapters , a description of the state of stress at a point of a stressed member can be found by using the convential formulas and may involve normal and shearing stresses. In this chapter , the method of redescribing the state of stress in iterms of the combined stressed will be applied to some particular cases of stressed members The curved members members , such as crane hooks and machine frames, are often subjected to axial loads and bending moments . Therefore the theory of curved beam will be also discussed in this chapter .

12-2

Stresses due to Combinations of o f Axial Loads and Bending Moments The simple simple stress stress formula for the member subjected to to axial axial P is ............( 12- 1) A as shown in Fig(12-1a)

load

Fig(12-1a) And for a member subjected to lateral load (bending) the stress is My 12- 2) I as shown in Fig(12-1 b ) .

Fig(12-1b) The stresses produced p roduced by axial loads and bending moments are such that stresses are normal to the cross-section cross -section of a member . Thus the resulting stresses at any point po int on the cross-section cross -section of the member are added algebraically .


The stresses given in Eq(12-1) & Eq(12-2) are both bot h normal to the cross section of the member , therefore the resultant stress acting on the section , due to the simultaneous action of the axial load P and the pure bending moment M , is the algebraic sum of the direct stress and the flexural stress. s tress. The equation of o f resultant stress is generally written in the form ; P

My

...................... (12-3)

A I The plus and minus signs indicated in Eq(12-3) are assigned

It can be seen that the resultant stress distribution due to the combined action of P & M is obtained by superposing the stress distribution distribution due to P & M . So , in this section it has been assumed that the eccentric load is applied to the member member on one of its principal axes, either either on the x-axis or on the y axis , thus the two cases cases are shown in Fig.(12-2). Fig.(12-2).


-a-

-bFig(12-2)

P

M yy. x

P

M xx .y

A

I yy

A

I xx

where : Myy=P.a Iyy

Mxx=P.b tia

Ixx

about y-axis

12-2-1

about x-axis

Stresses due to Loads not Applied to Principal Axis

Fig(12-3)


In order to position the load P (shown in Fig(12-3)) along the centroidal axis C of the member, we have h ave to move either directly from the point of application of the load to point C or we have to move through the distance ( a ) towards the Y axis and then through the distance (b) towards the X- axis . In order to obtain the equation equation of stress ( ) let us move the load through a distance ( a ) and then through thro ugh a distance ( b ). In moving through a distance dist ance (a) , we shall induce a bending movement Myy = P.a . plus the condition shown in Fig(12-2b ) .Thus , the stress acting at any point on the number of Fig.(12-3) is the algebraic sum of the M yy .x P M xx y and the as indicated below I yy A I xx P

M xx y

M yy .x

A

I xx

I yy

X & Y represent the coordinates of the point point at which which the stress in the the member of Fig(12-3) is required .

12-3 Direct Shear Combined with Torsion The shearing stress due to applied torque acting at any point on a shaft having a circular cross section is computed as T.r J and the shearing stress , due to a direct shearing force acting at any point on a beam of any section is V Ay Ib The resultant shearing stress at the point is given g iven by the algebraic sum as: T.r V A y .......... .......... ..... (12-4) J Ib the plus and minus signs in Eq.(12-4) are assigned to shearing stresses that act along the same line of action and have hav e respectively the same and opposite directions .

12.4 Combined Stresses due to Bending & Torsion A common application of combined stresses is that of o f a shaft subjected to bending & twisting and it is often convenient to express the resulting direct and shear stresses directly in terms of the applied moment &torque. If the bending moment moment is M and the torque T Fig(12-4 Fig(12 -4 a) a) , then the stresses acting on an element on the upper surface are as shown in the


plane view Fig(12- 4 b) ( those on the lower surface surface are the same same , except that x compressive).

-a-

-bFig(12-4)

My

M(d 2)

I 64 T

T.r J

M

d4

32

d3

,

d3

16 Assuming solid shaft, the max . principal stress , 1 max

2 x

2

2 max

i.e.

32

Me = 32

1

M

M

2

3 d 32

3 d 32

1

d3 1 2

max

M2

M

d3

max

2

T2

2

4

M2

M

Me

Me

d3

Ze

and also Max

2

2

x

4

2

2 max

1

M

2

3 d 32

2

4

3 d 16

- - - - equivalent equivalent bending moment moment

32 1

T

T2

Me

max

. is given by

2

4

x

max

T 3 d 16


16

d3

M2

Te = 16

M2

Max

d3

T2

Te 16

.. equivalent torque

Te

Max

Max

T2

Te.r

d3

J

12.5 Combined Torsion &

Axial Load

If the axial load P and torque T are applied to the rod as shown in Fig(12-5)

Fig(12-5) Then tens

P

P

4P

A

d2

d

4

T.r

and

16T

J

d 2

max

max

Pd 2 8

d3

4p

,

2

16

for torque

3

2

for axial load

2

max

2

2 16T d3

2 d2

T2

12.6 Combined Axial Load , Bending & Torsion If the axial load F, bending moment M and Torsion T are applied to the rod as shown in Fig(12-6 ) . Then 32 M 4 F bending

direct

T.r

and

J

d3

16T d

d2

and also,

3

Fig(12-6) 2 2 max

2

,

max

32M

4F

d3

d2

2

16T d2


12- 7 Examples The following examples explain the different concepts of the combined stresses problems . Example(12-1) Fig.(12-7) shows a steel chimney is 30m high , 1 m external diameter and 10 mm thick . It is is rigidly fixed at the base . It is acted acted upon by a horizontal wind pressure which is taken to be of a uniform intensity of 2 1kN/m of projected area for the lower 15m and to vary uniformly from 2 2 1 kN/m to 2 kN/m over the upper upper 15m. Find the maximum stress in the 3 . plates at the base base , steel has a density of of 7.8 Mg/m

Fig(12-7) Solution The pressure distribution diagram is shown in Fig(12-7). The total wind force can be divided into P 1, the force due to a uniform of 1kN/m2 over the whole height and P2 , the force due to the additional pressure over the upper 15 m. 3 P1=1*10 *30 = 30 kN

P2 =

1 * 10 3

* 1 * 15 7.5 kN 2 .: moment moment about about bas basee =30*15+ =30*15+7.5* 7.5*25= 25=637. 637.5 5 kN.m I of cross- section = .:

64

14

0.98 4

0 .003815 m

M

637.5 * 103

maximum bending stress =

I

4

*y =

00.03815

* 0.5

83.6 MN/m 2


Direct stress stress at at base = specific specific weight weight * height 3 = 7.8 *10 * 9.81 * 30 = 2.3 MN/m2 2 .: total stress at base = 83.6 +2.3 =85.9 MN/m Example(12-2) Fig (12-8) shows a short column of I-section 200mm x160mm . A vertical load W acts through the centroid of the section together with a parallel load load W/4 acting through a point on the center center line of the web , distance 60 mm from the centroid measured towards the longer flange . Find the greatest allowable value of W if the maximum compressive 2 stress is not to exceed 80 MN/m . What is the minimum stress in the section?

Fig(12-8) Solution Taking moments about about the top edge edge 160*10*5+180*10*100+120*10*195 160*10*5+180*10*100+12 0*10*195 =4600 y1 .: y1=91.74mm and y2=108.26mm

Ixx=

160 * 103 12 +

120 * 103 12

160 * 10 * 86.74

10 * 1803

2

120 * 10 * 103.26

12 2

10 * 180 * 8.26 2 6

4

= 29.85*10 mm

Transferring the load ( W/4) to the axis XX, XX, there is then a total direct load of ( 5 W/4 ) ,together with a bending moment about XX of magnitude ( We / 4 ) where ( e ) is the eccentricity eccentricity of the load From XX. 5W W * 0.06 0.10826 * 80 * 10 6 N/ m 2 max 6 4 * 0.0046 4 29.85 * 10 from which W=245 kN


Minimum stress =

5 * 245 * 103

245 * 103 * 0.06

0.08674

*

55.92 MN/m 2

6 4 * 0.0046 4 29.85 * 10 Example(12-3) Fig. (12-9 ) shows a pillar 1.5 m high is of rectangular section 50 mm think and tapers longitudinally longitud inally from a width of 150 mm at the base to 50mm at the top . A compressive load of 100 kN acts through throug h the centroid at the top end and parallel to the vertical edge. edge. Find the magnitude of the maximum maximum compressive stress and the crosssection at which it will occur.

-a-

-bFig(12-9)

Solution Fig(12-9 b) shows the the section section of the pillar pillar at a distance distance x m below the the x x * 0.1 0.05 m so that the top, the width of the section is 0.05+ 1.5 15 x m with respect to centroid of the section. load has an eccentric e ccentricity ity of 30

Direct stress , Bending stress, For

100 * 103 d

0.05{0.05

30 x } 15

30 0.75

to be a maximum

0.75 90

x d

(0.75 0

dx Substituting in Eq.(i) yields 67.5 2 53 . 33 MN / m 2 1.125

x

x

MN/m2

22.5 120x x)

2

(0.75

0.375m

x)

2

.. (i)


Example (12- 4) A steel bar of circular section , 100 mm diameter , carries a longitudinal pull whose line of action is parallel to the axis of the bar . At a certain transverse section the longitudinal stresses are measured at the surface of the bar at three points A , B & C . These points being equally spaced 2 round the section the tensile stresses at these points are A , 90 MN/m ; B , 2 2 75MN/m ; C , 30MN/m . Find :a) The magnitude & location of the greatest g reatest & least stresses at the section b) The magnitude & eccentricity of the applied pull . Make a diagram showing the stresses & their positions relative to the points A, B, & C . Solution It will be evident that the line of action of P lies with in the sector sector AOB, At A . as in Fig(12-10) .

P d

+

P * e * cos

2

P

400P

P * e * cos

2

3

32 ( i)

2

.............. 400P

At C , =

400p

1 80 * e * cos

0.1

400P

1 80 * e * cos 120

=75 MN/m

400P

3

0.1 4 .... .......... .........

d 32 2 =90 MN/m Similarly , at B. 4

+

1 80 * e *

1 2

1

1 80 * e * (

sin

2

) 3

cos

2

sin

30MN / m

These equations simplify to 9 40P

1 40 * e * ( cos

3 sin )

1 40 * e * ( cos

3 sin )

From Eq(v) and Eq(vi)

2

( ii )

1 80 * e * cos(120

1 80 * e * cos

3

cos

.................. (iv) 7.5 40P 3 40P

......... (v)

............... (vi)

2

... (iii)


80 e * cos

2

10.5 40P

..................... (vii)

From Eq(iv) and Eq(vi) 19.5 P= = 0.51 MN 120 From Eq(v) and Eq(vi) 4.5

80 3e * si sin

40P

= 0.693 ........ (viii)

And from Eq(iv) 9

80 e * cos tan

0.693 0.386 3

=1.035

40P

1 0.386 ........... (ix)

46

From Eq(viii) or Eq(ix) , e 0.00695m 400P (1 80e) when 0 max =

400 * 0.51

400P min min

=

2

(1 80 * 0.00695) 101 MN/m

(1 80e)

400 * 0.51

when

(1 80 * 0.00695)

180

°

28.9 MN/m

Fig( 12-10)

2


Example(12-5) A steel bar of rectangular section 80mm x 40mm is used as a simply supported beam on a span of 1.4m and loaded at mid-span mid-span . If the the yield yield 2 stress is 300 MN/m and the long edges of the section are vertical, find the load when yielding first occurs . Solution

-a-

-bFig(12-11)

Mmax = max

= 6

wL 4

M max

300*10 =

(see Fig(12-11a)

Ze

=

wL 4

*

6 bh

2

w * 1.4 * 6

w = 36.57 kN

2

4 * 0.04 * 0.08 After yielding resisting moment =2 F1 * 0.035 F2 * 0.02 ( see Fig(12-11b) ) i.e w * 1.4 = 4 6 6 2 * (300 * 10 * 0.01 * 0.04) * 0.035 (150 * 10 * 0.03 * 0.04) * 0.02 w= 44.6 kN If yielding ceases at x m from the center M=

44.6 * 103

6

0.04 * 0.082

( 0.7 0.7 x ) = 300 300*1 *10 0 * 2 6 Leng Length th over over whic which h yie yield ldin ing g occu occurs rs

x = 0.126m

2x = 0.25 0.252 2m


Example(12-6) Fig (12-12 )shows the section of a beam which is subject to a bending moment of such magnitude that that yielding occurs at the lower part of the 2 web over a depth of 50mm. The stress of 300MN/m may be assumed constant over the yield area , while over the remainder of the section the stress is proportional to the distance from the N.A Find : a) The position of the N.A . b) The stress at the top of the section . c) The moment of resistance of the section

-a-

-b Fig(12-12)

Solution The stress distribution diagram is shown in Fig(12-12 b) . If the N.A after yielding is at a depth h mm below the top , then 300 300h MN / m 2 ..................... (i) h 150 h 150 h h 20 Stress at underside of flange = h Therefore equating forces above and below the N.A, gives h 20 h

2 2

* (h

*120 * 20 20) * 20

h

20 h

*

300 * 50 * 20

300 2

* (150

h ) * 20


h2)

300(250h

From which :

h2

200h

2000

From Eq(i) and Eq(ii)

h = 65.8 mm and

Stress at underside of flange =

234.5MN / m2

65.8 20

* 234.5 163.3MPa 65.8 Assuming that the force on the flange acts its geometric center moment about N.A. (234.5 163.3) * 10 6

* 0.12 * 0.02 * (0.0658 0.01) 2 Moment about about N.A N.A of force on web above N.A. =

=

163.3 * 106

26650N.m

* (0.0658 0.02) * 0.02 * 23 (0.0658 0.02)

2 Moment about N.A of force on web below N.A. =

300 * 10

2285N.m

6

* (0.15

0.0658) * 0.02 * 23 (0.15

0.0658) 14170N.m

2 Moment about N.A of force on plastic part of web 6 =300*10 *0.05*0.02*(0.175 0.0685) = 31950 N.m Total moment of resistance =75055 N.m

Example(12-6) Fig(12-13) shows a steel bar of rectangular section (38 mm)wide by 101.6 mm deep is subjected to compressive force of 53.37 kN acting as shown in Fig. Find the th e maximum tensile and compressive stresses normal to the cross-section of the bar and sketch stress distribution over the cross section of the bar .

Fig(12-13) Solution The forces can be represented as


The compressive stress My bendng

I=

bh 3 12

P c

53.37 6

A

38 * 101.6 * 10 53.37 * 10 (0.0508 12.7 * 10 3 ) y

=

3

I 38 * (101.4)3

I 3.33 * 10 6

mm 4

12 103 * 53.37 * (0.0508 12.7 *10 3 ) * 0.0508

bending

3.33 *10

P

12.79 MPa

6

31.027MPa

My

A I 12.74 31.07 = 43.817 Mpa (comp.) ( top fiber) (bottem fiber) 12.47 31.07 18.23MPa (tension ) Example(12-7) Fig(12-14) shows a steel beam with an overhang is simply supported at A & B . If the beam is rectangular in section 50.8mm wide by 152.4mm deep . Find the largest stress acting normal to the cross-section of the beam neglect the weight of the beam .

Fig(12.14)


Solution Resolving the inclined force of 53.37 kN

3

Fx =

10 1

Fy =

* 53.37 * 10

2

1 Fx

3

2

3

50.707kN

* 53.37 * 10

3

16.902kN

o

H Ax

Fx

Fy

o

MB

50.707kN RA

0

RB

40 *103 16.902 *103

0.2032 * R A

40 *103 *152.4 *10

0 3

16.902 *103 * 0.0508 0

RA=25.798 kN RB= 31.136 kN

50.707 * 103

P A

50.8 * 10

Bending stress (

0

x1

3

* 152.4 * 10

bending

3

6.55MPa

)

50.8

Mx1=25.798x1 At x1 = 0 Mx1=0 3 3 At x1=50.8 mm Mx1=25.798*10 *50.8*10 =15.725 kN.m


0

x2

152.4

Mx2=25.798*(0.0508+x2) 40 * x2 At x1=0 Mx2=15.725 kN.m At x2=152.4 Mx2=25.798*0. =25.798*0.203 203

40*0.1524 40*0.1524 = 10.302 10.302 kN.m

0 x3<50.8 Mx3=25.798* (0.203+x3) 40*(0.1524+x 3) +31.136 x3 AT x3=0 Mx3=25.798*0 =25.798*0.203 .203 40*0.1524= 40*0.1524= 10.302 10.302 kN.m kN.m At x3= 50.8 Mx3= 25.798 *0.254 40 * 0.203 + 31.136 *0.0508 = 0 Mmax = 15.725 kN.m IN.A. =

bh 3

50.8 *152.4 3

12

12

My I

14.98 mm 4

15.725 * 10 3 * 76.2 * 10 6

14.98 * 10 the largest stress 6.55 79.98 86.53MPa

3

79.98 MPa

Example(12-8) Fig(12-15) shows the cast iron clamp clamp having the dimensioned cross-section cross-s ection shown . Find the max load P that can be applied . The allowable stresses are 20.68 MPa in tension and 62.05 MPa in compression .

Fig(12-15)


Solution

Ai yi

y

Ai

9.525 * 50.8 * 4.76 6.35 * (63.5 9.525) * ( (9.525 * 50.8 6.35 * (635 9.525) y

50.8 * (17.907) 3

I N .A

3

(50.8

63.5 9.525 2 17.907mm

y

6.35)(17.907

9.525)

9.525) 3

6.35(63.5 17.907)3 / 3

0.288 * 10 6 mm 4 The force acting on the clamp clamp of Fig(12-16) is equivalent to a force P acting acting along the centroidal axis of the section and to a bending moment of the magnitude M=P(0.1143+ y ) P(0.1143 0.017907) as shown in Fig(12-16)

0.1322 * P

Fig(12-16)

P A

direct stress where A P

9.525 * 50.8

6.35 * (63.5

1210.9 P and and bending stress is

9.525)

My

I 825.8 * 10 6 The flexure stress at each of the point A & B of Fig(12-16) 0.1322 * P * 17.907 * 10 A

0.288 * 10 6 0.1322 * (63.5 * 10 3

B

0.288 * 10

3

8.219 * 10 3 P tension

19.05 * 10 3 )

20.4 * 103 P comp

6

P

Allowable tensile stress at point A 6

A

20.68*10 = 121 1210.9 0.9 * P 2040 20400 0 *P Allowable compressive stress at point B 6

62.05*10 = 1210.9 P 20400* P Max. safe load is = 2.192 kN

My I P 2.192kN P My

A I P = 3.144 kN

825.8mm

2


Example(12-9) Fig(12-16) shows a concrete dam has the dimensioned dimensioned cross section 3 3 shown, if concrete weighs 23.55 kN/m and water weighs 9.796 kN/m . Find the max. height (h) of the water that can be maintained behind the dam without causing tension on the foundation of the dam .

Fig(12-16) Solution Firstly to find the centroid of the the section section to the back of of the dam :

y

A i yi Ai

thus , 1.8288*13.716*1.8288+0.5(4.5 0.5(4.572 72 1.8288) 1.8288)*13. *13.716 716* * [(4.572 1.8288) 1.8288) /3+1.8288] /3+1.8288] = [1.8288* [1.8288*13.71 13.716+0.5 6+0.5*(4.57 *(4.572 2 1.8288)*13.716] * y y 1.697m The weight of concrete acting at the foundation is W=volume*23550 W=volume*23550 =0.5(1.8288+4.572)*13.716*0.3048*2355 =0.5(1.8288+4.572)*13.716*0.3048*23550 0 =315.25 kN This load acts through the centroid of the dam as shown in Fig(12-17).

Fig(12-17)


Force due to average pressure is 1 1 F = average pressure *Area = 0 W h * h * 1 Wh 2 2 2 This force acts through the centroid of the pressure distribution at a distance of ( h/3 ) from the foundation of the dam. The force acting on Fig(12-17) is equivalent to that shown in Fig(12-18)

Fig(12-18) From this figure the direct compressive stress due to the weight of W concrete is , where A is the t he area of foundation A

Fig(12-19) 315.25 * 103 1.3898

226.831kN / m 2

Now applying the flexure formula

My I

M=M2 M1 M! ..... bending moment moment due to weight W of concrete M2 ..... .due to pressures force F Taken about the of area of Fig(12- 19) (the foundation)


M1=Wa =W 4.572 M2=

3

y

315.25 * 10 * ( 2.286 1.697) 185.45kN.m

Fh

1

2

wh .

h

wh

3

9796 * h

3

1632 * h 3

3 2 3 6 6 3 M=1632 h 185.45 The moment of inertia around the N.A of Fig(12-19) IN.A=

bh 3

0.3048 * 4.5723

12 (1632 * h 3

185.45 m 4

12 185.45) * 2.286 2.427

Therefore the resultant stress at point B must be zero hence (1632 * h

3

B

185.45) * 2.286 2.427

0

h=9.509 m 226.185kN / m 2 (comp) ( dired ) flexual )

A

226.185kN / m 2 ten .at B , comp. at A

226.185

226.185

213.021kN / m 2 (comp) &

B

0

Example(12-10) Fig(12-20 ) shows a short short member member having having a rectangu rectangular lar section section 8 x10 is subjected to a compressive force of P = 180 kN . Find the stresses normal to the cross- section section A , B , C and D locate the neutral axis of the cross-section ABCD .


Fig(12-20) Solution

P

180 * 10

3

3.5MPa comp .at point A, B, C, D A 0.203 * 0.254 Bending moment about the X-axis is Mxx=P*0.025=180*0.025= 4.572 kN.m this Bending moment cause tension at A & B and comp at C&D 254 * 203.2 3

I XX

My

177.6 * 10 6 mm 4 & y 101.6mm

12 4.572 * 10 3

6

I

2.586MPa

177.6 * 10 tensile at A & B and comp. at C & D Moment of inertia of the entire section about the y-axis is 3 6 4 Iyy=(203.2*10 )/12 = 277.48*10 mm & X=127 mm M yy .X 6778.2 * 0.127 3.102MPa 6 I yy 277.48 * 10 tensile at B & C compression at A& D P

M xx

M yy

A

I xx

I yy

A=

3.5+ 3.5+2. 2.58 586 6 3.10 3.102 2 = 4.65 4.65 Mpa Mpa (com (comp) p)

B=

3.5+2.586+ 3.5+2.586+3.102 3.102 = 2.24 Mpa (ten.) (ten.) 3.5 2.58 2.586+ 6+3. 3.10 102 2 = 2.93 2.93 Mpa( Mpa(co com mp) C = 3.5 D=

3.5 3.5 2.5 2.586 3.1 3.102 = 9.1 9.13 Mpa Mpa (com (comp p)

The stress distribution is as shown in Fig(12-21).

Fig(12-21) Fig(12-21)

Strength of materials- Handout No. 12- Combined Stresses- Dr. Hani Aziz Ameen

Example(12-11) Fig(12- 22) shows a short column of T - section 25cm x 20cm has a 2 cross-sectional area 52 cm and max radius of gyration 10.7cm 1 0.7cm a vertical load W kN acts through the centroid of the section together with a parallel load of ( W/4 W/4 ) kN acting through a point on the center line of the web distant 6 cm cm from the centroid . Find ( a ) the greatest greatest allowable 2 value of W if the max stress is not to exceed exceed 65 MN/m (b) what is then the min stress? Solution 2 2 4 (a) I = A K = 5 (10.7)=5953.48 cm

(W ( dinect )

W ) 4 4

0.24W MPa

52 * 10 M (W / 4) * 0.06

( bending )

Ze

Ze

0.015W ( bed )

Ze

0.25

y

* 10

3

* 10

* 10

0.015W (I / y)

0.015 W -8

(5953.48 * 10 / 0.125)

3

* 10

3

0.031 W MPa

( bend ) d

* 10

0.125m

2 ( bend )

3

65

b

0.24 W

0.0315 W

65 Fig(12-22)

W= (b)

65 0.24 min

0.0315 d

239.4 kN b

=W * (0.24 0.0315) = 234.4(0.24 0.015)*10 3 = 49.9 MPa


Example (12-12) Fig(12-12 ) shows a rectangular plate with a hole drilled in it .Find the greatest and least intensities of stress at the critical cross-section of the plate when subjected to an axial pull of of 90 kN.

Fig(12-23) Solution Axial load =90 kN

The location of centroid y2 =

20 * 20 * 10

80 * 20 * 160

20 * 20 80 y1=200-130=70 mm The moment moment of inertia inertia about about the XX axis. Ixx =

20 * 803

20 * 80 * 70

12 6

40

2

130mm

20

20 * 203

20 * 20 * 130 10

12

2

4

Ixx = 8.066*10 m Eccentricity, e = 100 y1= 100 100 70 = 30 mm mm or e = y2 100= 130 100 = 30 mm P 90 3 * 10 45 MPa (tens) d A 2 * 10 3 My P*e *y b

I

max .atuu

45

I P * e * y2

45

90 * 30 * 10

I xx 2 2 = 88.5 kN/m = 88.5 kN/m (tens)

min min at vv

45

P * e * y1

45

3

8.066 * 10

90 * 30 * 10

3

I xx 8.066 * 10 = 21.57 MPa. =21.57 MPa (tens.)

* 0.07 6

* 0.13 6

* 10

3

* 10

3


Example(12-13) Fig.(12-24 ) shows a short hollow cast iron column having an external diameter of 600 mm and diameter 400 mm was cast in a factory. On inspection it was found that that the bore is eccentric as shown in the figure. If the column carries a load of 1600 kN along the axis of the bore, find the extreme intensities of stress induced in the section.

Fig(12-24) Solution External diameter = 600 mm = 0.6 m Internal diameter = 400 mm = 0.4 m

Net area =

0.6

2

0.4

2

0.157m

2

4 Load , W =1600 kN To find out the c.g. of the section OG = x

4

* 0.6 2 * OG1 4

* 0.6 2

4 4

* 0.4 2 * OG 2 * 0.4 2

where OG1=0.3 m , OG2=0.34 m x 268mm So GG1 = 0.3 0.268 = 0.032 m GG2 = 0.34 0.268 = 0.072 m Eccentricity = OG2 x =0.34 0.268 = 0.072 m =72 mm Moment M = 1600*0.072 =115.2 kN.m. Iyy = Iyy =

4 64

0.64 0.6 4

4

4 0.4 4

* 0.6 2 * GG1

4

2

0.6 2 * 0.032 2

64

* 0.4 4

4

* 0.4 2

0.4 2 * 0.072 2

GG 2

4.743 * 10

2

3

m4


Bending stress (tensile)

bt

bt

MX1 I yy

115.2 * 0.268 9.743 * 10

3

* 10

3

6.51 MPa.

Bending stress (comp.) bc 8.06 MPa. W 1600 10 3 10.19 MPa.(comp.) direct A 0.157 Max. comp. Stress = d bc 10.19 8.06 18.25 MPa Min. comp. Stress = d 10.19 6.51 3.66 MPa bt Example(12-14) Fig(12-25) shows a steel rod 40 mm diameter passes through a copper tube of 60 mm internal diameter and 80 mm external diameter. Rigid cover plates are provided at each end of the tube and steel rod passes through these cover plates also. Nuts are screwed on the projecting ends of the rod so that the cover plates put pressure on the ends of the tube. If the center of the rod is 10 mm out of the center of the tube, find the max. stress in the copper tube when one of the nuts is tightened to produce a linear strain of 0.002 0.002 in the rod. 2 2 Take E steel steel =210 GN/m and E copper copper =105 GN/m

Fig(12-25) Solution Diameter of steel rod , d s=40 mm External diameter of the copper tube , D c=80 mm Internal diameter of the copper tube , d c= 60 mm

Area of of the cross section section of of the rod , As = * 40 2 4 Area of the cross section section of of the rod , As = * 80 2 4 Strain in the steel rod , s 0.002

1256.6 mm 2 60 2

2199.1mm 2

Strength of Materials- Combined Stresses- Hani Aziz Ameen

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