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 Dies and Tools Eng. Dept. E-mail:[email protected]

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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 , 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 Axial Loads

and Bending Moments The simple stress formula for the member subjected to axial load

is AP ............( 12- 1)

as shown in Fig(12-1a)

Fig(12-1a) And for a member subjected to lateral load (bending) the stress is

I

My 12- 2)

as shown in Fig(12-1 b ) .

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


The stresses given in Eq(12-1) & Eq(12-2) are both 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. The equation of resultant stress is generally written in the form ;

I

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

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 due to P & M . So , in this section it has been assumed that the eccentric load is applied to the member on one of its principal axes, either on the x-axis or on the y axis , thus the two cases are shown in Fig.(12-2).


-a- -b- Fig(12-2)

yy

.yy

IxM

AP

xx

xxI

y.MAP

where : Myy=P.a Mxx=P.b Iyy tia Ixx about y-axis about x-axis

12-2-1 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 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 of stress ( ) let us move the load through a distance ( a ) and then through a distance ( b ). In moving through a distance (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

yy

yy

Ix.M

and the xx

xxI

yMAP as indicated below

yy

yy

xx

xxI

x.MI

yMAP

X & Y represent the coordinates of the point at which the stress in 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

Jr.T

and the shearing stress , due to a direct shearing force acting at any point on a beam of any section is

yAIbV

The resultant shearing stress at the point is given by the algebraic sum as:

.........................yAIbV

Jr.T (12-4)

the plus and minus signs in Eq.(12-4) are assigned to shearing stresses that act along the same line of action and have respectively the same and opposite directions . 12.4 Combined Stresses due to Bending & Torsion A common application of combined stresses is that of 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 is M and the torque T Fig(12-4 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 are the same , except that x compressive).

-a- -b- Fig(12-4)

34 d32

M

d64

)2d(MI

My

3d16

TJr.T ,

Assuming solid shaft, the max . principal stress , max . is given by 22

xxmax 421

2

316

2

332

332

max dT4

dM

dM

21

i.e. 22max TMM

21 d

323

Me = 22 TMM21 - - - - equivalent bending moment

emax3 M d

32

e

e3e

max ZM

32d

M

and also 2

x2

Max 421

2

316

2

332

max dT4

dM

21


TM d16

22Max

3

Te = 22 TM .. equivalent torque

eMax3 T d

16

Jr.Te

dTe

316

Max

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

242d

tens dP4P

AP for axial load

and 3dT16

Jr.T for torque

222max ,

2

3dT16

2

2d2

p4max

228

Pd3max T

d16

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

23directbending dF4

dM32

and 3dT16

Jr.T and also,

Fig(12-6)

22

max 2 , 2

2

23max dT16

dF4

dM32


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 rigidly fixed at the base . It is acted upon by a horizontal wind pressure which is taken to be of a uniform intensity of 1kN/m2 of projected area for the lower 15m and to vary uniformly from 1 kN/m2 to 2 kN/m2 over the upper 15m. Find the maximum stress in the plates at the base , steel has a density of 7.8 Mg/m3 .

Fig(12-7) Solution The pressure distribution diagram is shown in Fig(12-7). The total wind force can be divided into P1, 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.

P1=1*103*30 = 30 kN

P2 = kN 5.715*1*210*1 3

.: moment about base =30*15+7.5*25=637.5 kN.m

I of cross- section = 44 98.0164

0 .003815 m4

.: maximum bending stress =I

M *y = 23

MN/m 6.835.0*03815.00

10* 5.637


Direct stress at base = specific weight * height

= 7.8 *103 * 9.81 * 30 = 2.3 MN/m2 .: total stress at base = 83.6 +2.3 =85.9 MN/m2 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 W/4 acting through a point on the 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 stress is not to exceed 80 MN/m2. What is the minimum stress in the section?

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

Ixx= 23

23

26.8*180*1012180*10 74.86*10*160

1210*160

+ 23

26.103*10*12012

10*120 = 29.85*106 mm4

Transferring the load ( W/4) to the axis 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 of the load From XX.

266max mN/ 10*80

10*85.2910826.0*

406.0*W

0046.0*4W5

from which W=245 kN


Minimum stress

= 26

33MN/m 92.55

10*85.2908674.0*

406.0*10*245

0046.0*410*245*5

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

-a- -b- Fig(12-9) Solution Fig(12-9 b) shows the section of the pillar at a distance x m below the

top, the width of the section is 0.05+ m15x05.01.0*

5.1x so that the

load has an eccentricity of m30x with respect to centroid of the section.

Direct stress , x75.0

30}05.0{05.0

10*100

15x

3

d MN/m2

Bending stress, 22 )x75.0(x1205.22

)x75.0(90

x75.030 .. (i)

For to be a maximum m375.0x0dxd

Substituting in Eq.(i) yields

22 m/MN33.53

125.15.67


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 round the section the tensile stresses at these points are A , 90 MN/m2; B , 75MN/m2 ; C , 30MN/m2. Find :- a) The magnitude & location of the greatest & 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 AOB, At A . as in Fig(12-10) .

2d4

P +3d

32

cos*e*P21.0

4

P + cos*e*801P400

1.032

cos*e*P3

=90 MN/m2 ............... ( i ) Similarly , at B.

sin23cos

21*e*801P400120cos*e*801P400

=75 MN/m2 .............. ( ii )

At C , )120cos(*e*801P400

= 2m/MN30sin23cos

21(*e*801p400 ... (iii)

These equations simplify to

P409cos*e*801 .................. (iv)

P405.7)sin3cos(*e*401 ......... (v)

P403)sin3cos(*e*401 ............... (vi)

From Eq(v) and Eq(vi)


2 80 P405.10cos*e ..................... (vii)

From Eq(iv) and Eq(vi)

P = 120

5.19 = 0.51 MN

From Eq(v) and Eq(vi)

80 e3 * sinP40

5.4 = 0.693 ........ (viii)

And from Eq(iv)

80 386.01P40

9cos*e ........... (ix)

3386.0693.0tan =1.035 46

From Eq(viii) or Eq(ix) , 00695.0e m

)e801(P400max when 0

= 101)00695.0*801(51.0*400 MN/m2

)e801(P400min when 180°

= 9.28)00695.0*801(51.0*400 MN/m2

Fig( 12-10)


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 . If the yield stress is 300 MN/m2 and the long edges of the section are vertical, find the load when yielding first occurs . Solution

-a- -b- Fig(12-11)

Mmax = 4

wL (see Fig(12-11a)

max = e

maxZ

M = 4

wL * 2bh6

300*106= 208.0*04.0*46*4.1*w w = 36.57 kN

After yielding resisting moment =2 02.0*F035.0*F 21 ( see Fig(12-11b) )

i.e

44.1*w =

02.0*)04.0*03.0*10*150(035.0*)04.0*01.0*10*300(*2 66

w= 44.6 kN If yielding ceases at x m from the center

M = 2

10*6.44 3( 0.7 x ) = 300*106 *

608.0*04.0 2

x = 0.126m

Length over which yielding occurs 2x = 0.252 m


Example(12-6) Fig (12-12 )shows the section of a beam which is subject to a bending moment of such magnitude that yielding occurs at the lower part of the web over a depth of 50mm. The stress of 300MN/m 2 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

2m/MNh150

h300h150

300h

..................... (i)

Stress at underside of flange = h

20h

Therefore equating forces above and below the N.A, gives

*h20h20*120*

2h20h

20*)h150(*2

30020*50*30020*)20h(*2


From which :2000h200h

)hh250(3002

2

From Eq(i) and Eq(ii) h = 65.8 mm and 2m/MN5.234

Stress at underside of flange = MPa3.1635.234*8.65208.65

Assuming that the force on the flange acts its geometric center moment about N.A.

= m.N26650)01.00658.0(*02.0*12.0*2

10*)3.1635.234( 6

Moment about N.A of force on web above N.A.

= m.N2285)02.00658.0(*02.0*)02.00658.0(*2

10*3.16332

6

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

= m.N14170)0658.015.0(*02.0*)0658.015.0(*2

10*30032

6

Moment about N.A of force on plastic part of web =300*106*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 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 6c 10*6.101*3837.53

AP = 12.79 MPa

Iy)10*7.120508.0(10*37.53

IMy 33

bendng

I= 4633

mm 10*33.312

)4.101(*3812bh

MPa027.3110*33.3

0508.0*)10*7.120508.0(*37.53*106

33

bending

IMy

AP

fiber) top( 12.74 31.07 = 43.817 Mpa (comp.) )(tension MPa23.1807.3147.12fiber) bottem(

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

Fx = kN707.5010*37.53*103 3

Fy = kN902.1610*37.53*31

1 322

oFx

kN707.50FH xAx

010*902.1610*40RRoF 33BAy

00508.0*10*902.1610*4.152*10*40R*2032.00M 333AB

RA=25.798 kN RB= 31.136 kN

MPa55.610*4.152*10*8.50

10*707.50AP

33

3

Bending stress ( bending )

8.50x0 1 Mx1=25.798x1 At x1 = 0 Mx1=0 At x1=50.8 mm Mx1=25.798*103*50.8*10 3=15.725 kN.m


4.152x0 2

Mx2=25.798*(0.0508+x2) 40 * x2 At x1=0 Mx2=15.725 kN.m At x2=152.4 Mx2=25.798*0.203 40*0.1524 = 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*0.203 40*0.1524= 10.302 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. = 12bh3

98.1412

4.152*8.50 3 mm 4

6

33

10*98.1410*2.76*10*725.15

IMy 79.98 MPa

the largest stress MPa53.8698.7955.6

Example(12-8) Fig(12-15) shows the cast iron clamp having the dimensioned cross-section 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

i

ii

AyA

y

17.907mmy y )525.9635(*35.68.50*525.9(

)525.92

525.95.63(*)525.95.63(*35.676.4*8.50*525.9

46

333

A.N

mm 10*288.0

3/)907.175.63(35.6)525.9907.17)(35.68.50(3

)907.17(*8.50I

The force acting on the clamp of Fig(12-16) is equivalent to a force P acting along the centroidal axis of the section and to a bending moment of the magnitude M=P(0.1143+ P*1322.0)017907.01143.0(P)y as shown in Fig(12-16) Fig(12-16)

2mm8.825)525.95.63(*35.68.50*525.9A wherestress directAP

IMy is stress bending and P 9.1210

10*8.825P

6

The flexure stress at each of the point A & B of Fig(12-16)

tension P 10*219.810*288.0

10*907.17*P*1322.0 36

3

A

comp P 10*4.2010*288.0

)10*05.1910*5.63(*1322.0 36

33

B

Allowable tensile stress at point AI

MyAP

20.68*106 = 1210.9 * P 20400 *P kN192.2P

Allowable compressive stress at point BI

MyAP

62.05*106 = 1210.9 P 20400* P P = 3.144 kN Max. safe load is = 2.192 kN


Example(12-9) Fig(12-16) shows a concrete dam has the dimensioned cross section shown, if concrete weighs 23.55 kN/m3 and water weighs 9.796 kN/m3 . 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 section to the back of the dam :

i

iiA

yAy

thus , 1.8288*13.716*1.8288+0.5(4.572 1.8288)*13.716* [(4.572 1.8288) /3+1.8288] = [1.8288*13.716+0.5*(4.572 1.8288)*13.716] * y

m697.1y The weight of concrete acting at the foundation is W=volume*23550 =0.5(1.8288+4.572)*13.716*0.3048*23550 =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

F = average pressure *Area = 2Wh211*h*hW0

21

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

concrete is AW , where A is the area of foundation

Fig(12-19) 2

3m/kN831.226

3898.110*25.315

Now applying the flexure formula I

My

M=M2 M1 M! ..... bending 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 m.kN45.185)697.1286.2(*10*25.315y572.4 3

M2= 333

2 h*16326

h*97966

wh3h.wh

21

3Fh

M=1632 h3 185.45 The moment of inertia around the N.A of Fig(12-19)

IN.A= 433

m 45.18512

572.4*3048.012bh

427.2286.2*)45.185h*1632( 3

Therefore the resultant stress at point B must be zero hence

0427.2

286.2*)45.185h*1632( 3

B

h=9.509 m )comp(m/kN185.226 2

)dired(

B.at ten m/kN185.226 2)flexual , comp. at A

0&)comp(m/kN021.213185.226185.226A B

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


Fig(12-20) Solution

DC,B,A,point .at comp MPa5.3254.0*203.0

10*180AP 3

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

mm6.011y & mm10*6.17712

2.203*254I 463

XX

D&Cat comp. and B&Aat tensile

MPa586.210*6.17710*572.4

IMy

6

3

Moment of inertia of the entire section about the y-axis is Iyy=(203.2*103)/12 = 277.48*106 mm4 & X=127 mm

MPa102.310*48.277

127.0*2.6778I

X.M6

yy

yy

tensile at B & C compression at A& D

yy

yy

xx

xxI

MI

MAP

A = 3.5+2.586 3.102 = 4.65 Mpa (comp) B= 3.5+2.586+3.102 = 2.24 Mpa (ten.) Fig(12-21) C= 3.5 2.586+3.102 = 2.93 Mpa(comp) D = 3.5 2.586 3.102 = 9.13 Mpa (comp)

The stress distribution is as shown in 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 cross-sectional area 52 cm2 and max radius of gyration 10.7cm a vertical load W kN acts through the centroid of the section together with a parallel load of ( W/4 ) kN acting through a point on the center line of the web distant 6 cm from the centroid . Find ( a ) the greatest allowable value of W if the max stress is not to exceed 65 MN/m2 (b) what is then the min stress? Solution

(a) I = A K2 = 52 (10.7)=5953.48 cm4

MPa W24.010*10*52

)W(4

4W

)dinect(

65 W0.0315 W24.065

MPa W 031.0

10*)125.0/10*(5953.48

W0.015

m125.0225.0y

10*)y/I(W015.010*

ZW015.0

10*Z

06.0*)4/W(ZM

bd

)bend(

38 -)bend(

33

e)bed(

3

ee)bending(

Fig(12-22)

W = kN 4.2390315.024.0

65

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

Strength of materials- Handout No. 12- Combined Stresses- Dr. Hani Aziz Ameen 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 90 kN.

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

The location of centroid y2 = mm130208020*20

160*20*8010*20*20

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

Ixx = 23

23

10130*20*2012

20*204070*80*2012

80*20

Ixx = 8.066*10 6 m4 Eccentricity, e = 100 y1= 100 70 = 30 mm or e = y2 100= 130 100 = 30 mm

(tens) MPa 4510*10*290

AP 3

3d

Iy*e*P

IMy

b

36

3

xx

2atuu.max 10*

10*066.813.0*10*30*9045

Iy*e*P45

= 88.5 kN/m2 = 88.5 kN/m2 (tens) 3

6

3

xx

1vv atmin 10*

10*066.807.0*10*30*9045

Iy*e*P45

= 21.57 MPa. =21.57 MPa (tens.)

Strength of materials- Handout No. 12- Combined Stresses- Dr. Hani Aziz Ameen 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 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 = 222 m157.04.06.04

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

OG = 24

24

22

412

4

4.0*6.0*

OG*4.0*OG*6.0*x

where OG1=0.3 m , OG2=0.34 m mm268x

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 = 22

2421

24 GG4.0*4

4.0*64

GG*6.0*4

64.04

Iyy = 43222244 m10*743.4072.0*4.0032.0*6.04

4.06.064

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

Bending stress (tensile) MPa. 51.6

10*10*743.9

268.0*2.115I

MX

bt

33

yy

1bt

Bending stress (comp.) MPa. 06.8bc

.)comp.(MPa 19.1010157.0

1600AW 3

direct

Max. comp. Stress = MPa 25.1806.819.10bcd Min. comp. Stress = MPa 66.351.619.10btd 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 in the rod. Take Esteel=210 GN/m2 and Ecopper=105 GN/m2

Fig(12-25) Solution Diameter of steel rod , ds=40 mm External diameter of the copper tube , Dc=80 mm Internal diameter of the copper tube , dc= 60 mm

Area of the cross section of the rod , As = 22 mm6.125640*4

Area of the cross section of the rod , As = 222 mm1.21996080*4

Strain in the steel rod , 002.0s

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Strength of Materials- Combined Stresses- Hani Aziz Ameen