strength of materials- bending stress in beam- hani aziz ameen

Strength of Materials Handout No.8

Bending Stress in Beam 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.8- Bending Stress in Beam- Dr. Hani Aziz Ameen

8 1 Introduction As mentioned before the lateral forces produce shearing force and bending moment , producing a stress which is called bending stress (flexure stress) ( tensile & compressive). Also shearing force represented by stress which is called shearing stresses. Generally in beam design two important factors must be considered. 1 strength (stresses) 2 rigidity (deflection) The beam shown in Fig(8-1) is subjected to bending , in each cross section a tensile and compressive force are induced in additional to

shearing force accompained . Fig(8-1) 8-2 Neutral Surface It is the region that separates the tensile forces from compressive force i.e. bending stress equal to zero. 8 2 1 Neutral Axis (N.A.) The intersection between any cross section with neutral surface is

(N.A.) The position of the neutral axis must pass through the centroid of the section. 8 3 Derivation of Flexural Formula To establish the bending stress formula several assumptions are used :- 1 The cross section of the beam is plane and must remain plane after bending. 2 3 The material must be free from any resistance force and from impurities, holes, or grooves. 4 The bending moment of elasticity in tension must be the same for compression. 5 The beam has constant cross section. 6 The beam is subjected to pure bending. Due to action of bending moment the beam will bend, let the radius of curvature of the neutral axis at a particular section be , Fig.(8-2)


Fig(8-2)


0

0L

LL

010 LLL L1= )1(LLL 000

Since , the deformation at y from the N.A. is ds(1+ ) hence, for a certain distance y, d will be

dsy)1(dsd

y y

y

Now , the forces applied on the cross section are

Applying the equilibrium equations on the cross section yields. 1-

)A(xx 0dA.0F

2- 000Fy 3- 000Fz 4- 000Mx 5-

)A(xy 0z.dA.0M

)A(

zxz 0My dA.0M


From these points only points (1) and (5) can be analyze 1-) 0dA.x

EyE

,y

x

)A(

0dA E y

)A(

0 dA yE

)A(

0dA y

It means that axis z passes through centorid of the cross section. 5-)

)A(x 0z .dA.

0dAEyz)A(

)A(

0dA z yE

)A(0dA z y

i.e. Iyz = 0, axes y , z are principal axes of inertia. It means from points (1) & (5) , that y & z are the centroidal principal axes of inertia.

6) )A(

zx 0My.dA.

)A( )A(

xz dA.EyydA.yM


)A(

2z dAyEM

zz IEM

z

zI

ME

we have Eyx

yE x

yIM x

z

z

z

zx I

y.M

The greatest bending stresses occur at the fibers most remote from z axis.( as shown in Fig(8-3) ) . If the distance to this extreme fiber is c (ymax.) , the max. bending stresses are.

)A(

A e

z

yIz

z

A

zzAintpomax

Z

MMy

I.M

ZMM

yIM

)B(e

z

yBIz

z

B

yyBintPo)max

where Ze is the elastic section modulus.

Fig(8-3)


8-4 Ze for Typical Cross-Section The section modulus ( Ze )is usually quoted for all standard sections and partically is of greater use than the second moment of area . strength of the beam sections depends mainly on the second modulus. The section modulii of several shapes are calculated below :-

a) Rectangular Section

Iz = bh3/12 , Ze = Iz/c

Ze = 2hzI = (bh3/2) / (2/h) = bh2/6

b) Square Section

Iz= a4/12 Ze = a3/6 c) Rhombus Section

Iz= a4/12 Ze= a3 2 /12


d) Polygon Section

Iz= (5 3 /16)R4

Ze= (5/8) R3

e) Polygon Section

Iz= (5 3 /16)R4

Ze= (5 3 /16)R3

f) Triangle Section

Iz= bh3/36 Ze1=bh2/24 Iy= hb3/48 Ze2= bh2/12


g) Circular section

Iz= D4/64

Ze= )2/D(

Iz = 2Iz/D = D3/32

h) Pipe (Hollow Circular Cross Section )

Iz = (D4 d4) /64 Ze = ( /32)((D4 d4 )/D)

i) Ellipse cross-section

Iz= a3b/4 Ze = Iz/a = a2b/4


j- Ze of the following sections :-

12

bhBHI33

z

H6

bhBHZ33

e

k- Ze of the following sections :-

12

bhBHI33

z

H6

bhBHZ33

e


l- Ze of the following section can be found in appendix I -

8-5 Calculation of Bending Stress

Flexure strength calculations are made on base

e

max zmax Z

M

following the steps :-

1) Find Reactions 2) Plot B.M. diagram

3) Find e

max zmax Z

M

8-6 Efficient Use of Section in Bending

In the design of beam , the selected section must be strong in

resisting the max. bending moment as well as economical in weight per unit length . The condition of strength for beams in pure bending is satisfied provided that :-


e

max z)allowable( Z

M

This equation indicates that for a given allowable stress and a max. bending moment the section modulus Ze must not be less than the ratio

(.allow

.maxM ) . If the allowable stress of the material in tension is the same as

in compression the use of a section which is symmetrical about the neutral axis is preferred and the material from which the beam is made should be ductile. Structural steel is a good example of a ductile material. The next condition to be satisfied is the economy in weight of the beam . This is accomplished by distributing as much of the area of the cross- section as far as possible from the neutral axis of the section . For example for the same depths and areas the wide flange section of Fig8-4c , is capable of resisting more bending moment than the I- section of Fig 8-4b or the rectangular section of Fig 8-4a.

-a- -b- -c- Fig(8-4) Therefore , out of the three sections the wide flange section is considered to be the most efficient section this is because the value of section modulus Ze is largest If the section is not symmetrical about the neutral axis the compressive and tensile bending stresses at the extreme fibers of the material will not be equal . Beams having such sections should be made from brittle materials for example cast iron is a brittle material which is stronger in compression than in tension . The allowable bending stresses , according to I

My will be


e

maxallowable c Z

M

e

max allowable t Z

M

8-7 Stress Concentration in Bending

In pure bending the stresses vary linearly over the cross section of the beam , being zero at the neutral axis and max. at the extreme fibers of the material to either side of the neutral axis. Such stresses are computed by formula I

My if the cross-sectional area of the beam changes suddenly , the stress distribution over the cross-section of the beam is

I

MyK

where K is the stress- concentration factor


8-8 Examples The following examples explain the difference ideas of the

bending stresses problems. Example(8-1) A 250 mm ( depth) X 150 mm ( width) rectangular beam is subjected to maximum bending moment of 750 kNm : find

i) The maximum stress in the beam ii) The value of the longitudinal stress at a distance of 65 mm

from the top surface of the beam . Solution

i) Moment of inertia I= (bh3)/12 I = (0.15*0.253)/12 = 0.0001953 m4 Distance of N.A. from the top surface of the beam : y = h/2 = 0.25/2 = 0.125 m Using the relation = My/I =(750*103*0.125) / 0.0001953 = 4.8*108 N/m2

iii) using the relation , M / I = / y = 1 / y1

1=My1/I = ( 750*103*60*10 3*106) / 0.0001953 = 230.4 MN/m2 Example(8-2) A symmetrical section 200 mm, deep has a moment of inertia of 2.26*10 5 m4 about its N.A. . Find the longest span over which , when simply supported the beam would carry a uniformly distributed load 4 kN/m run without the stress due to bending exceeding 125 MN/m2

Solution From the relation = My/I , M = M = (125*106*2.26*10 5)/0.212 = 28.25 *103 Nm


Also the maximum bending moment due to uniformly distributed loads is = w L2/8 = 4*L2/8 = 0.5 L2

M = 0.5 L2 28.25 = 0.5 L2 L = 7.516 m

Example(8-3) Find the dimensions of beam a of a timber for span 8 m to carry a brick wall 200 mm thick and 5 m high . If the density of brick work is 1850 Kg/m3 and the maximum permissible stress is limited to 7.5 MN/m2 . given that the depth of beam is twice the width . Solution Total weight of the wall = W W = length of Span x thickness of wall x height of wall x density of brick work per m3 = 8 *0.2 * 5*(1850*9.81) = 145188 N = 0.145 kN R1+R2 = 0.145 R1*8 0.145 *4 = 0 R1 = 0.145/2 MN R2= 0.145/2 MN 0 x<4 Mx= (0.145/2) x At x= 0 Mx=0 At x=4 Mx= 0.29 MN.m 0 x<4 Mx=(0.145/2)*(4+x2) 0.145x At x= 0 Mx=0.29 MN.m At x=4 Mx= 0 Max. B.M = 0.29 MN.m Moment of resistance = M = ( * (bh3/12)) / (h/2) = h2 b/6) = (7.5*bh2)/ 6 Equating moment of resistance to B.M

0.29 = (7.5*bh2)/6 , ( h = 2b) b *(2b)2 = 0.232 b = ( 0.232/4)1/3 b = 0.387 m h = 2b = 0.774 m


Example(8- 4) A floor has to carry a load of 12 kN/m2 , the floor is supported on rectangular beams each 300mm * 100mm and 5 m long. Find the distance apart ( from the center to center) at which beams should be placed so that the maximum stress in the beams should not exceed 8 MN/m2 Solution I = bh3/12 = ( 0.1 * (0.3)3) / 12 = 2.25 * 10 4 m4 M = 4) /(0.3/2) = 0.012 MNm =12 kN m .. ( i) Let x be the spacing , in meter , of the beams from center to center . Each beam will share half of the load of the floor between two floor beams on either side . The length of each beam is 5 m therefore the total load on each beam ( if the spacing between two beams is x from center to center) = 12* x * 5 = 60 x kN Max. B.M. = wL/4 = 60x * 5 /4 ...........................(ii) From eq(i) and eq(ii) 12 = (60x * 5) / 4 x = (12*4) / ( 60*5) = 0.16 m Example(8-5) Fig(8-5) shows a cast iron water main 12 m long of 500 mm inside diameter are 25mm wall thickness runs full of water and is supported at its ends, find the maximum stress in the metal if density of cast iron is 7200 Kg/m3 and that of water is 1000 Kg/m3 .

Fig(8-5)


Solution Outside diameter of water main ; D = d+ 2t = 500 +2 * 25 = 0.55m Weight of water main per meter length = 0.04123 * 1 * 7200 * 9.81 = 2912.16 N Weight of water in one meter long main =

N19.192681.9*1000*1*)5.0(*4

2

Total weight of pipe when full of water = 2912.16 + 1926.19 = 4838.35N

It can be consider the beam with uniform distributed load

Bending moment , M = 8

12*35.48388

wL 22 = 87090.3 N.m

I = 344

10*1.442384 )5.055.0(64

m4

y = m275.0255.o

2D

= My/I = ( 87090.3 * 0.275 ) / ( 1.42384*10 3) = 16.82 MN/m2 Example(8-6) A hollow circular bar having outside diameter twice the inside diameter is used as a beam . From the bending moment diagram of the beam , it is found that the bar is subjected to a bending moment of 40 kNm . If the allowable bending stress in the beam is to be limited to 100 MN/m2 , find the inside diameter of the bar . Solution

)dD(64I 44 = 4

44 d6415

)d)d2((64

y =D/2 = (2d)/2 = d

= My/I = 100*106 = 46415 d*

d*1000*40

d3 = (40*1000*64)/(15 *1000*106) d = 0.0816 m


Example(8-7) Fig.(8-6) shows a cross section of two wooden planks 150mm * 50 mm each are connected to a 5 m T-section of a beam . If the amount of 3.4 kNm is applied around the horizontal N.A. inducing tension below the N.A., find the stresses at the extreme fibers of the cross section . Also calculate the total tensile force on the cross section .

Fig(8-6) Solution

Component 2) Centroidal distance from

the bottom force , y (mm)

A y (mm3)

Rectangular (1) 150*50=7500 175 1312500 Rectangular (2) 150*50 = 7500 75 562500 A = 1500 Distance of the N.A. xx from the bottom face

mm12515000

1875000AAyy

I = [ ((150*503)/12)+ 150*50*(75 25)2]+[(50*1503/12)+150*50*(125 75)2] = (1562500+18750000)+(14062500+18750000) = 5312.5 *104 mm4 Distance of c.g. from the upper extreme fiber yc = 200 125 = 75 mm , yt = 125 mm

M/I = / y , Tensile stress t = M yt / I Compression stress c= M yc / I


t = 8* 106 N/m2 c= 4.8 * 106 N/m2

We know that the section is subjected to compressive stresses above the xx axis (i.e.) N.A. and tensile stress below it as shown

Area in tension = 125*50*10 6 = 6.25 * 10 3 m2 Total tensile force = average tensile stress * area in tension = [ (8*106)/2] * 6.25= 25000 kN Example(8-8) Fig(8-7) shows a cross section of a beam simply supported at ends is loaded with a U.D.L , over whole of its span . If the beam is 8 m long , Find the U.D.L. if maximum permissible bending stress in tension is limited to 30 MN/m2 and in compression to 45 MN/m2 , find the actual maximum bending stresses setup in the section .

Fig(8-7)


Solution The following table gives the calculations for location of horizontal N.A. Component 2 Centroidal distance

A y (mm3)

Top Flange 100*30 =3000 200 (30/2) = 185 mm 555000 Web 120*30 = 3600 50+(120/2) = 110 mm 396000 Bottom Flange 12*50 = 6000 50/2 = 25 mm 150000 Total

mm38.8712600

1101000AAyy

I = I top flange + I web + I bottom flange = [ ((100*303)/12) + 100*30*(112.62 15)2] + [ ((30*1203)/12)+30*120*(110 87.38)2] +[((120*503)/12)+120+50*(87.38 25)2] = 28813993 + 6161992 + 24597586 = 5957 * 104 mm4 Max. B.M. = wL2/8 = w * 82 / 8 = 8 w For tension side of the I section M/I = t / yt , M = I * t / yt , M = (5957*10 8*30*106)/(87.38*10 3) = 20452 Nm


for compression side of the I section M = I ( c / yc) , M = ( 5957*10 8*45*106)/ (112.62*10 3) = 23802.6Nm Moment of resistance = M = 20.452 kNm 8w = 20.452 or w = 2.556 kN/m Actual max. stress in the top- most fibers of the beam = (M/I) * yc = (20.452*103*112.62*10 3)/(5957*10 8) = 38.6*106 N/m2 actual max. stress in the bottom = 30 MN/m2 Example(8-9) Fig.(8-8) shows a cast iron bracket of cross section of I- form .Find :

i) Position of the N.A. and the moment of inertia of the section about the N.A.

ii) The maximum B.M. that should be imposed on this section if the tensile stress in the top flange does not to exceed 40 MN/m2 , what is then the value of the compressive stress in the bottom flange ?

Fig(8-8) Solution

i) Position of the N.A. The section may be split into three rectangular components thus ,

Component 2 Centroidal distance from bottom edge

A y(mm3)

Top flange 200*40=8000 260 mm 2080000 Web 200*40=8000 140 mm 1120000 Bottom flange 120*40=4800 20 mm 96000 Total 6000


Distance of the N.A. from the bottom edge

mm5.15820800

3296000AAyy

Hence , yc = 158.5 mm yt = (40+200+40) 158.5 = 121.5 mm

I = I top flange + I web + I bottom flange = [ ((200*403)/12) + 200*40*(121.5 20)2] + [ ((40*2003)/12)+40*200 * (158.5 140)2] +[((120*403)/12)+120+40*(158.3 20)2] = 83484667+29404667+92714800= 2.056*108 mm4

ii) M = I * t / yt , M = ( 40*106*2.056*10 4) / (121.5*10 3) = 67.7 kNm Compressive stress in the bottom flange c

c = M yc / I

c = ( 67.7*103*158.5*10 3) / ( 2.056*10 4) = 52.19*106 N.m


Example(8-10) Fig(8-9) shows the section of the horizontal beam is 4 m long and is simply supported at the ends . Find the maximum uniformly distributed load it can carry if the tensile and compressive stresses must not exceed 25 MN/m2 and 45 MN/m2 respectively .

Fig(8-9) Solution Divide the section into three rectangulars , the area of the individual components , their centroidal distance from the bottom edge and their moment about the bottom edge are tabulated below:

Component Area A mm2 Centroidal distance from

bottom edge y

A y (mm3)

Rectangular (1) 100* 20 =2000 50 100000 Rectangular (2) 100*20 = 2000 50 100000 Rectangular (3) 120*20=2400 10 24000 A = 640 A y =224000 Distance of the N.A. from the bottom edge

mm356400

224000AAyy

yt = 35 mm , yc = 100 35 = 65 mm I = I rect(1) + I rect(2) + I rect(3)

= [ (( 20*1003)/12) + 20*100*(50 35)2] + [((20*1003)/12) + 20*100*(50 35)2] + [((120*203)/12) + 120*20*(35 10)2] = 2 * 2116666+158000 = 5813332 mm4 M/I = t / yt , M/I = c / yc , c / t = yc / yt = 65/35 = 1.857 The tensile stress ( t ) according to the above ratio

45 * (1/1.857) = 24.23 MPa hence , the actual stress in the beam should be 45 MN/m2 compressive and 24.23 MN/m2 tensile and the U.D.L can be determined from these value .

M = ( t/ yt )* I = ( c / yc )* I , sub t = 24.23 MPa t = 45 MN/m2 , yt = 35 mm , yc = 65 mm


M = ( t / yt) * I = 24.23 * 106 * [(581.33*10 8)/(35*10 3)] = 4024.5 Nm Let w be the uniformly distributed load in kN/m run Mmax = wL2/8 = (w*42)/8 = 4.0245

w = ( 4.0245*8) / 42 = 2.012 kN/m

Example(8-11) Fig.(8-10) shows the cross section of a cast iron beam when this beam is subjected to a bending moment, the tensile stress at the bottom edge is 30 MN/m2 , find

i) The value of the bending moment and ii) Stress induced at the top edge

Fig(8-10) Solution In order to locate the position of N.A. , divide the section into five rectangulars as shown in the figure The areas of the individual components , their centroidal distances from the bottom edge and their moments about the bottom edge are tabulated below :


Component Area A ( cm2)

Centroidal distance from bottom edge

y

A y (mm3)

Rectangular (1) 16*2 = 32 20 640 Rectangular (2) and (3)

2*15*2=60 11.5 690

Rectangular (4) & (5)

2* 9*4= 72 2 144

A = 164 A y =1474 Distance of the N.A. from the bottom edge (LL)

cm 9say 90.8164

1474AAyy

yt = 9 cm & yc = 12 cm I = I rect(1) + I rect(2)+(3) + I rect(4)+(5) Ixx = [ ((16*23)/12) + 16*2*(12 1)3]+[((12*153)/12) +2*15*(11.5 9)2]*2 + 2*[ ((9*43)/12) + 9*4*(9 2)2] = 3883 + 2*750+2*1812 = 9007 cm4

i) Bending moment , M ; M/I = t / yt , M = ( t / yt) * I , M = (30*106*9007*10 8)/ (9*10 2) =30023 Nm

ii) Stress at the top edge c c = M yc / I = (30*103*12*10 2) / ( 9007.6*10 8) = 40*106 Nm2


Example(8-12) Fig(8-11) shows a steel stanchion is built of a rolled steel beam of 1 section 45 cm x 20 cm united 1.5 cm thick and 30 cm wide plate fastened on each flange . the length of the stanchion is 5 m and is freely supported at both ends for the I- section; I = 35060 cm4 , find

i) Moment of inertia of the enlarged section ii) Greatest central point load the beam will carry if the

bending stress does not exceed 120 MN/m2 iii) Minimum length of 30 cm X 1.5 cm plates

Solution i) I = 35060 +2 [ ((350*1.53)/12) +30*1.5*((45/2)+(1.5/2))2] = 35060 + 2( 8.437+24325.313) = 83727.5 cm4

ii) Greatest central point load , w max. B.M. = wL / 4 = w*5/4 = 1.5 w B.M. M = I /y = (120*106*83727.5*10 8)/(24*10 2 ) = 418637.5 Nm 1.25 w = 418637.5 w = 334910 N Fig(8-11)

iii) Min. length of the plates Suppose the cover plates are absent for a distance of x m from each support , then at these point the bending moment must not exceed moment of resistance of I section alone i.e. I/y = 120*106*((35060*10 8 ) /( 24*10 2)) = 175300 Nm B.M. at x m from each support = (w/2) * x = 175300 = ( 334910/2)*x x = 1.05 Hence leaving 1.05 m from each support , for the middle 5 2.1 = 2.9 m the cover plates should be provided Example(8-13) Fig(8-12) shows a cross section of a simply supported beam .The beam carries a load W = 20 kN as shown . Its self weight is 7 kN/m , calculate the max. normal stress at 1-1


Fig(8-12) Solution To determine reaction at the supports taking moment about A, we get RB * 3.6 = 20*2.4+7*3.6*3.6/2 = 93.36 RB = 25.93 kN RA+RB = 20+7*3.6 = 45.2 RA=19.27 kN Bending moment at 1-1 M = 25.93 * 2.4 20*1.2 7*2.4*2.4/2 = 18.072 kNm Refer to cross section of the beam . the areas of the individual components , their centroidal distances from the bottom edge (LL) and their moments about the bottoms edge are tabulated below

Component Area A (cm2) Central distance from the bottom

edge (cm)

A y ( cm3)

Rectangular (1) 30*20 = 600 15 9000 Circular hole (2) ( /4)*152=176.7 20 3534 Total A = 423.3 A y = 5466 Distance of the N.A. ( xx axis) from the bottom edge is

cm9.12yAAy

y

yt = 12.9 cm & yc = 30 12.9 = 17.1 cm I = I rec I hole = [ (20*303/12)+20*30*(15 12.9)2] [ ( /64)*154+( /4)*152*(17.1 10)2] = 47646 11393 = 36253 cm4 Maximum normal stress at 1-1 M = I /y

c= Myc/I = (18.072*103*17.1*10 2)/(36253*10 8) = 8.52*106 N/m2


Example (8-14) i) Find the dimension of the strongest section that can be cut

out of a circular log of wood 25 cm in diameter. ii) Also specify the safe maximum span for the beam of

rectangular section when it is to carry a U.D.L of 2.5 kN/m and the bending stresses are limited to 10 MN/m2 .

iii) What will be the longest span if the circular log itself is used as a beam? Solution i) The diagonal of the section = diameter of circular log = 25 cm :. b2+h2 =252 or h2=252 b2 Ze=I/y = [(bh3/12)/(h/2)] = bh2/6 = b(252 b2)/6 Ze=(625b b3)/6 For Ze to be max. dZe/db should be equal to zero .

cm43.14b0b3625b362561

dbdZ 22e

h2=252 14.432=416.78 h=20.41 ii) Safe max. span : Ze=(bh2)/6 Ze=[14.43*(20.41)2]/6 = 1001.85 cm3 M = * Ze M = 10*106*1001.85*10 6 = 10 kN.m ....(i) We know that , in case of beam (length L) carrying a U.D.L of 2.5 kN/m Mmax. = wL2/8 = 2.5 L2/8 = 0.3125 L2 ...(ii) Equating (i) and (ii) , we get 0.3125L2 = 10 or L = 5.66m

iii) Diameter of log = 25 cm

Ze=I/y = 363332

m10*18.1534cm18.15343225*

32d

2/d64/d

M= * Ze=10*106*1534.18*10 6=15341.8 N.m= 15.34 kN/m ...(iii) Mmax. = wL2/8 = (2.5*L2)/8 = 0.3125 L2 ...(iv) 0.3125 L2 = 15.34 L = 7m

Example (8-15) Fig.(8-13) shows a cast iron beam is simply support at its end and carries a load of 18 kN at mid. span. Find the maximum allowable span if


the stress due to bending is not to exceed 30 MN/m2 tension. Neglect the weight of the beam, find also the maximum compressive stress?

Fig(8-13) Solution Divide the section into two triangles and place 1/3 of the area of each triangle at the mid point of its sides . Take moments about the base 600*120+4800*60+1800*0 = 7200 y y = 50 mm :. I = 600*702+4800*102+1800*502=7 920 000 mm4 Mmax. = wL/4 = (18*103*L)/4 = 4500 L

m056.1L10*7920000

05.0*L450010*30I

My12

6

The maximum compressive stress is given by 2m/MN4230*

05.007.0

Example (8-16) Fig (8-14) shows a steel tube 40 mm outside diameter and 30 mm inside diameter is used as a simply supported beam on a span of 1 m and it is found that the maximum safe load it can carry at mid span is 1.2 kN. Four of these tubes are placed parallel to one another and firmly fixed together to form in effect a single beam, the centres of the tubes forming a square of 40 mm side with one pair of centres, vertically over the other pair. Find the maximum central tube which this beam can carry if the maximum stress does not to exceed that of the single tube above. Solution The load W M Ze for a given value of For a single tube,


444 mm85900304064

I Fig(8-14)

:. Ze=85900/20 = 4295 mm3 For four tubes,

4222 mm122360020*30404

859004I

Ze = 1223600/40 = 30950 mm3 Therefore the load which can be carried by the four tubes = (30950/4292)*1.2=8.65 kN Example (8-17) Fig. (8-15) shows the section of a steel beam in the shape of an inverted semicircular channel with flanges. Find the position of the N.A ((XX)) and the second moment of area of the section about XX. Find also the moment of resistance of the beam in Nm if the maximum stress due to bending is 125 MN/m2. For a semicircular area of radius (r) , the distance of the centroid from the diameter is 0.4244 r.

Fig(8-15) Solution Centroid of outer semi circle Form base = 0.4244*55 = 23.35 mm Centriod of inner semi circle from base = 0.4244 *50 = 21.20 mm Taking moments about the base, 60*5*2.5+ y50*

255*

25*602.21*50*

235.23*55*

22222

:. y =25.3 mm

2224423

95.2*55*2

35.23*55*2

55*8

8.22*5*6012

5*60I

422224 mm4136001.4*50*2

2.21*50*2

50*8

M= Nm174010*125*0253.0055.0

10*413600MyI 6

12


Note I of semi circle about N.A. of section = I about the diameter area * ( distance from diameter to centroid of semi circle ) 2 + area * ( distance of centroid of semi circle to N.A. of section)2 Example (8-18) Fig. (8-16) shows the square cross-section .Prove that the moment of resistance of a beam of square section with its diagonal in the plane of bending is increased by flattening the top and bottom corners and that the moment of resistance of resistance is a maximum when y=(8/9) Y

Fig(8-16) Solution The section may be divided into a rectangle and a square which is being bent about its diagonal , I = [(2(Y y) * (2y)3)/12]+ y4/3= (4/3) Y y3 y4

4334 yYy

y*M For to be minimum, d /dy = 0

i.e. (4/3) *Y y3 y4 = y(4Yy2 4y3) from which y = (8/9)Y Note The area of each half of the square is y2 . Placing y2/3 at the mid point of each of the sloping sides and (2/3) y2 at the mid point of the diagonal I = 4 * (y2/3) * (y/2)2 = y4/3 The area (2/3) y2 has no second moment about (xx)


Example(8-19) Fig(8-17) shows a horizontal cantilever 3m long is of rectangular cross section 60 mm wide throughout its length, the depth varying uniformly from 60 mm at the free end to 180 mm at the fixed end . A load of 4 kN acts at the free end . Find the position of the most highly stressed section, and find the value of the maximum bending stress induced . Neglect the weight of the cantilever itself .

Fig(8-17) Solution At a section x m from the load M =4*103 x d = 60 + (x/3)*120 = 60+40 x mm Ze = (bh2)/6 = (60/6) * (60+40x)2 = 4000 * (3+2x)2 mm2 =M/Ze = (4*103 x) / (4000*(3+2 x)2 * 10 9) = 109 x / (3+2 x)2 ....(i) for to be a maximum , d /dx = 0 ( 3+ 2 x)2 = 4x ( 3+2 x) x = 1.5 m Substituting in eq.(i) gives max = (109* 1.5) /(3+3)2 = 41.6 MN/m2 Example(8-20) Fig(8-18) shows a timber beam 80 mm wide by 160 mm deep is to be reinforced with two steel plates 5 mm thick. Compare the moments of resistance for the same value of the maximum bending stress in the timber when the plates are alternate :

a) 80 mm wide and fixed to the top and bottom surface of the beam b) 160 mm deep and fixed to the vertical sides of the beam ,

E steel = 20*E timber Solution a) It = (80*1603)/12 = 27.3*106 mm4

Is = (80/12)*(1703 1603) = 5.447*106 mm4 M = Ms+Mt (Ms/Mt) = (EsIs)/(EtIt) = 20* (5.447/27.3) =3.98 - a- - b- M = 3.98 Mt + Mt = 4.98 M Fig(8-18) It = 27.3 * 106 mm4


(b) Is = (10 * 1603) / 12 = 3.413*106 mm4 Ms/Mt = 20*(3.413/2.73) = 2.5 M = 2.5 Mt + Mt = 3.5 Mt Therefore ratio of moments of resistance in cases (a) and (b) for the same value of Mt ( i.e. for the same value of t) = 4.98/3.5 = 1.425 Example(8-21) Fig(8-19) shows a compound beam is formed by joining two bars rigidly together one of steel and the other of brass , each bar being 40 mm wide . The bars are of thickness t1 and t2 respectively , so that the total depth is ( t1+t2) . If E for the steel is twice that for the brass , find the ratio of t1 and t2 so that the N.A. of the section is at the dividing line of the two bars If the total depth of the section is to be 20 mm and the stresses in the steel and brass are not to exceed 110 and 40 MN/m2 respectively . Find the maximum moment of resistance of the beam .

- a- - b Fig(8-19) Solution Fig(8-19a) shows the cross section of the composite bar and fig(8-19b) shows the equivalent brass section . For the N.A. to be on the dividing line , the moments of the parts on each side of this line must be equal

i.e. 80* t1 *( t1/2) = 40 * t2 *(t2/2) , 2

1tt

2

1

t1 + t2 = 20 mm and 2

1tt

2

1

t1 = 8.28 mm and t2 = 11.72 mm If the maximum permissible stress in the steel is 110 MN/m2 , the corresponding stress at the top edge of equivalent brass section is 110/2 = 55 MN/m2 since 12 t2t , it is evident that the stress at the bottom edge will reach 40 MN/m2 before that at the top edge reaches 55 MN/m2 so that the moment of resistance is that which will produce the fomer stress , I = (80*8.283)/3 + (40*11.723)/ 3 = 36000 mm4 M = I / y = (36630 *10 12*40*106) / 0.01172 = 125 Nm


Example(8-22) Find the following cross section dimensions

a) rectangle h x b if h/b = k = 3 b) circular d c) hollow cross section d / D = =2/3

if max = 100 MPa , Mmax=100 kN.m Solution. a) Ze = bh2/6 , = Mmax/Ze max

max6

bhmax

2

M , b = h / k

max2kh

max

hM*

max3max

hM*k*

m0564.0100

10*3*6

Mk 6h 33

3max

max

b = h / 3 = 5.64/3 = 1.88 cm = 2 cm

b) Ze = D3/32

max32Dmax

3

M

m0467.0100*10*32M32D 3

33

max

max

c) Ze = ( /32) * ( D4 d4)/D =( /32) (D4/D) [1 (d/D)4] = ( /32)D2(1 4)

max4332

max

)1(DM

cm03.5D

100*])(1[10*32

)1(M32D 3 4

32

33

max4max

d = * D = (2/3) * 5.03 = 3.35 cm


Example(8- 23) Fig(8-19) shows a beam with hinge at 2m from the left span . Find the suitable standard cross section for the beam depicted in fig .

Fig(8-20) Solution

Fx = 0 , RGx = 0 Fy = 0 , RA +5 q + P RC RG = 0 MA = 0 , Mo 5 q (5/2) + RC * 4 P * 5 +RG * 7 = 0

Bending moment on the hinge = 0

MB = 0 RA * 2 + Mo q*2*(2/2) = 0 Solving the above equations we get RA = 3* 104 N RC = 13.166 * 104 N RG = 3.1667 * 104 N To find the max. bending moment , we draw the B.M. Diagram 0 x1<4 Fx1 = RA q x1 Mx1 = RA x1 + Mo ( q x1

2/2) At x1 = 0 Fx1 = RA = 3*104 N Mx1 = Mo = 8*104 N m At x1 = 2 Fx1 = RA 104*2 = 3*104 104*2 = 5*104 N At x1 = 4 Fx1 = 3*104 104*4 = 7*104 N Mx1 = 3*104 * 4 + 8*104 (42*104/2) = 12*104 N.m 0 x2<1 Fx1 = RA q*4 + Rc q x2 Mx1 = RA (4+x2) q*4(2+x2)+Rc x2 q x2 x2/2 +Mo At x2 = 0 Fx1 = 3*104 104 * 4 + 13.16*104 104 *0 Fx1 = 7*104 + 13.1666* 104 = 6.1666*104 N


Mx1 = 3*104*4 104*4*2+8*104 Mx1 = 12*104 Nm At x2 = 1 Fx1 = 3 * 104 104*4+13.166*104 104*1=5.1666*104 N Mx1 = 3*104*5 104*4*3+13.166*104*1 104*0.5+8*104 = 6.33 * 104 N.m 0 x3<2 Fx3 = RA 104*5+RC 2 *104 = 30*104 104*5+13.166*104 2*104 = 3.166 * 104 N Mx3 = RA(5+x3) + Mo+ RC(1+x3) q*5* (2.5+x3) P x3 = 3*104(5+x3)+8*104+13.166*104(1+x3) 5*104* (2.5+x3) 2*104 x3 At x3 =0 Fx3 = 3.166*104 N Mx3 = 3*104*5+8*104+13.166*104 5*104*2.5 0 = 6.33*104 N.m At x3 = 2 Fx3 = 3.166*104 N Mx3 = 0

we

maxZ

M

6

4

we 10*160

10*12MZ

Ze = 0.00075 m3

strength of materials- bending stress in beam- hani aziz ameen

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