strength of materials- axial force shear force bending moment diagrams- hani aziz ameen

Strength of Materials Handout No.7

Axial Force, Shear Force & Bending Moment Diagrams

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.7- Axial Force, Shear Force & Bending Moment Diagrams- Dr.Hani Aziz Ameen

7-1 Introduction The beam is a structural member either as a girder in a frame or

part in machine which is subjected to lateral forces that causes a tensile force in part of section and compressible force in another at the same section. The beam may be classified according to the type of support into the following :- 1 Simply Supported Beam , whose ends freely rest on walls or columns or knife edges

2 Over Hanging Beam Over hanging beam is one in which the supports are not situated at the ends.

3 Cantilever Beam , whose one end is fixed in a wall and the other end free.


4 Propped Beam , whose one end is fixed (built) and the other end freely rests on columns.

5 Fixed beam A beam whose both ends are rigidly fixed or built into supporting walls.

6 Continuous Beam A beam which has more than two supports

And according to the type of problem, the beam can be classified into :-

1 Statically determinate beams : The static equations are sufficient to determine the unknown reactions. The static Eqs. are 0Fy 0Fx 0M 2 Statically indeterminate beam : The static equations are not sufficient for solution so that we need another equation obtained from the deflection.


7 2 Types of Loading on Beam The beam may be subjected to the following types of static loading ( all of them or part of them).

a- Concentrated load b- Uniform distributed load c- Uniform variable load d- Non- uniform distributed load e- Direct moment or Torque

7 3 Bending Moment , Shear Force , Axial Force Diagrams 1 The diagrams of axial force, shear force & bending moment are closed

diagrams i.e. start from zero and end at zero. 2 In case of unloading , shear force diagram will be a horizontal line,

while the bending moment diagram will be inclined line. 3 The effect of concentrated load on the shear force diagram will be

vertical line in location of the effect of concentrated load, while for bending moment diagram will cause a hard changing.

4 In case of distributed load, shear force diagram will be an inclined line , while the bending moment diagram will be a curve of second order equation.

5 In case of the non distributed load (i.e. the load change from zero to a certain value) the shear force diagram will be a curve of second order equation , while the bending moment diagram will be curve of third order equation and so on .

6 The effect of coupling on the bending moment diagram is a vertical line, and there is no effect on the shear force diagram.


7 At any section in a beam carrying transverse loading the shearing force (S.F.) is defined as the algebraic sum of the forces taken on either side of the section.

Shear force sign convention forces upwards to left of a section or downwards to the right of the section are positive.

8 At any section of beam the bending moment is the algebraic sum of the moments of the forces about the section taken on either side.

Bending moment sign convection.

9 At any section in a beam the axial force is the algebraic sum of axial

forces taken on either side of section. Axial force sign convention

10 To construct the S.F. , B.M. & A.F. diagrams, select any section on

the beam at distance X from the concentrated load and before the other force and this part of the beam may be kept in equilibrium.


Let us take the following examples A Concentrated load If the beam carry a load at centre of beam R1+R2=P R1L= P L/2 R1=P/2 R2= P/2 0 2/Lx1 Fx1=R1=P/2 Mx1=R1 x1 x1=0 0Mx1 , Fx1= P/2 x1= L/2 2/LRMx 11 = (P/2)(L/2) = PL/4 , Fx1=P/2 0 2/Lx2 Fx2=R1 P = P/2 P Mx2 = R1(L/2 +x2) Px2 x2= 0 2Mx R1L/2 = PL/4 x2= L/2 2Mx R1(L/2 + L/2) PL/2 Mx2= (P/2 )(2L/2) (PL/2) = 0 If we have a beam with the loads as indicated in the Fig. Below .

First stage In plotting the B.M. , S.F. & A.F. diagrams first of all one must

find the reaction. Reaction 0R0F Axx 0PRPR0F 2B1Ay

RA+RB = P1+P2 0)2/a(Pa*Pa2*R0M 21AB


2*RA= 0.5*P2 P1 RA = 0.5 * P1 0.25*P2 RB+0.5P1 0.25P2 P1 P2 = 0 RB = 0.5 P1 + 1.25 P2 Second Stage

Make a section before and after each force .

Third Stage

Apply equilibrium equation to each section

ax0 1

Fy = 0 , Fx1 = RA 21Ax1 P* 0.25 - P * 0.5 RF

Mx1 = 0 , Mx1 = RA . x1 1211x x)P*25.0P*5.0(M Fx = 0 , RAx = 0

At x1 = 0 Fx1 = RA , Mx1 = 0 At x1 = a Fx1 = RA , Mx1 = RA . a = ( 0.5*P1 0.25*P2)*a

ax0 2 Fx2 = Fx1 P1 = RA P1 Mx2 = Mx1 + Fx1 x2 P1x2 = RA ( a+x2) P1x2

At x2 = 0 Fx2 = RA P1 Mx2 = RA . a


At x2 = a Fx2 = RA P1 Mx2 = RA ( a + a) P1 a = 2 a RA P1 a

2/ax0 3 Fx2 + RB Fx3 = 0 Fx3 = RA P1 + RB Mx2 RB x3 Fx2 x3 + Mx3 = 0

Mx3 = Mx2 + RB x3 + ( RA P1 ) x3 Mx3 = RA ( 2a + x3) P1 ( a + x3) +RB x3 At x3 = 0 Fx3 = RA P1 + RB Mx3 = 2 a RA a P1 At x3 = a/2 Fx3 = RA P1 + RB Mx3 = 2.5 a RA 1.5 a P1 + 0.5 a RB Fourth stage Located ( point ) the value of S.F , B.M & A.F . of each section and link each other

B Uniform distributed load Reaction

Fx = 0 , RBx = 0 Fy = 0 , RA q L + RB = 0

RA + RB = q L MB = 0 , RA L q L L/2 =0

RA = q L /2 RB = q L RA = q L /2


Lx0

Fx = RA q x = (q L /2) q x Mx = RA x q x (x/2) Mx = q (L /2) x q (x2/2) Ax = RBx = 0 At x = 0 Fx = qL/2 , Mx = 0 At x = L Fx = q(L/2) q L = q (L/2) Mx = q (L/2) L q (L2/2) = 0

To find the max. bending moment

02

qx22

qLdx

dM x

q (L/2) q x =0 , x = L / 2 Mmax = q (L/2) (L/2) (q/2)(L/2)2 = qL2/8


C- Uniform Variable load

Reaction Fy = 0 , RA + RB 0.5 qo L = 0 MB = 0 , RA L + 0.5 qo L (1/3) L = 0

Let = 0.5 qo L RA = /3 = qoL/6 RB = qoL/2 qoL/6 = qoL/3 RAx = 0

To find the relation for q(x) from similar triangular


xq

Lq )x(o

q(x) = qo ( x / L)

Lx0 Fx = RA 0.5 q(x) x Fx = RA 0.5 x qo ( x / L) = RA (qox2/2L) Fx = qoL/6 qo x2 / 2L Mx = RA x (qo/2L) x2 x = (qo L /6) x (qo/2L) x3 Ax = RAx = 0 At x = 0 Fx = qoL/6 , Mx =0 At x = L Fx = (qoL/6) (qo/2L)L2 = qoL/3 , Mx = 0

To find the max. bending

0xL2

q3Rdx

dM 2oA

x

x = L / 3 Mmax. = (qo L /3) (L/3) ( qo/2L) (L/3)3 Mmax = (qoL2/18) ( qo L2/54) = (qo L2) / 27


11- To find the max. bending moment , it can be known from the highest value of bending moment diagram but if the diagram has curve , it can be found by derivation of the moment equation for the curve and equalization to zero , from that can be found the value of x that M at it will be maximum or it can be known from the shear diagram , that the value of which SF= 0 , its point has the max. B.M. and also the value of B.M. can be found from the SFD and to prove that , let us take an element ds

where Axial force

0Fx

N cos(d /2) F sin (d /2)+ Pds + (N+dN) cos(d /2) + (F+dF) sin(d /2) = 8-1)

0Fy N sin(d /2) F cos (d /2)+ qds + (N+dN) sin(d /2)+ (F+dF) cos(d /2) = 0

8-2) 0Mo

M + (M+dM) F tan(d /2) ( F+dF) tan(d /2) + P ds [( /cos (d /2)) ] =0 . (8-3) For small d cos (d /2) 1 sin(d /2) d /2 tan(d /2) d /2 Hence , Eq.(8-1) will be

N *1 F *(d /2)+ Pds + N+dN (F dF)(d /2) = 0 F d + Pds + dN = 0 ( divided by ds )

F (d /ds)+ P + (dN/ds) = 0 ds = d 1/ = d /ds


( F/ ) + (dN/ds) + P = 0 and eq.(8-2) will be ( dF/ds) (N/ ) = q and eq. (8-3) will be (dM/ds) = F For straight beams we can take hence

P = 0 , = and 1/ = 0 1) dN/dx = 0 2) dF/dx = q ..(8-4)

dF/dx = q ( This is called first principle of Szwedler) 3) dM/dx = F ( This is called second principle of Szwedler)

It can be seen that from the second principle of Szwedler , that if F = 0 ( i.e. shear force = 0 ) the term will be dM/dx = 0 and that the conditions of Mmax ( i.e. the Mmax occurs at F = 0 ) and also the B.M. can be formed from the SFD using 2nd Szwedler principle , that dM/dx = F

FdxdM

M = Fdx M = ( Area S.F)2 ( Area S.F)1 7-4 Examples The following examples explain the differences ideas of the drawing of the axial force , shear force & bending moment diagrams. Example(7-1) Fig.(7-1) shows the simply supported beam , with concentrated load indicated on the Figure . Draw the B.M. & S.F. & A.F. diagrams

Fig(7-1)


Solution Fy = 0 , R1 + R2 = 70 kN MA = 0 , 6 R2 = 50 * 2 + 20 * 7 , R2 = 40 kN

R1 =70 40 = 30 kN 0 < x1<2 Fx1 = R1 = 30 kN Mx1 = R1 x1 At x1 = 0 Fx1 = 30 , Mx1 = 0 At x1 = 2 Fx1 = 30 , Mx1 = 30*2 = 60 kN.m 0<x2<4 Fx2 = R1 50 = 30 50 = 20 kN Mx2 = R1(2+x2) 50 x2 At x2 = 0 Mx2 = 60 kNm , Fx2 = 20 kN At x2 = 4 Fx2 = 20 , Mx2 = 30(2+4) 50 *4 = 20 kN.m 0<x3<1 Fx3 = R1 50 + R2 = 30 50+40 = 20 kN Mx3 = R1(6+x3) 50(4+x3) + R2x3 At x3 = 0 Fx3 = 20 kN , Mx3 = 30 * 6 50*4 = 20 kN.m At x3 =1 Mx3 = 30* 7 50 * 5 + 40* 1 = 0 , Fx3 = 20 kN


Example(7-2) Fig(7-2) shows the simply supported beam with the load indicated in the figure . Draw the S.F. , B.M. and A.F diagrams .

Fig(7-2) Solution

Fy =0 R1 + R2 = 10 kN

MD = 0 40 10 * 7 + R1* 5 = 0 R1 = 6 kN R2 = 4 kN 0 x1<2 Fx1 = 10 kN Mx1 = 10 x1 At x1 = 0 Mx1 = 0 , Fx1 = 10 kN At x1 = 2 Mx1 = 20 , Fx1 = 10 kN 0 x2<3 Fx2 = 10+ R1 = 10+6 = 4 kN Mx2 = 10 ( 2+x2) + R1x2 At x2 = 0 Fx2 = 4 , Mx2 = 20 At x2 = 3 Fx2 = 4 , Mx2 = 32 0 x3<2 Fx3 = 10+6 = 4 kN Mx3 = 10(2+3+x3) + R1(3+x3)+40 At x3 = 0 Fx3 = 4 , Mx3 = 50 + 18 +40 = 8 At x3 = 2 Fx3 = 4 , Mx3 = 70 + 30 +40 = 0 Example(7-3) Fig(7-3) shows the simply supported beam with the distributed load as indicated in the figure . Draw the S.F. , B.M. and A.F diagrams .


Fig(7-3) Solution

Fy = 0 R1+R2 = 30 + 10 * 10 = 130

MA=0 10R2 = (10*10)*5 + 2*30 R2 = 56 kN R1 = 74 kN 0<x1<2 Fx1 = R1 10 x1 Mx1 = R1x1 10x1 .(x1/2) Mx1 = R1x1 5(x1)2 At x1 = 0 Fx1 = 74 kN , Mx1 = 0 At x1=2 Fx1 = 74 20 = 54 , Mx1 = 74*2 5*4 = 128 0<x2<8 Fx2 = R1 10(2) 30 10*x2 Mx2 = R1(2+x2) (10*2) * [ 0.5*(2) + x2] 30 x2 10 x2 (0.5 x2) At x2 =0 Fx2 = R1 20 30 = 74 20 30 = 24 kN Mx2 = 2 R1 20 = 2*74 20 = 128 kNm At x2 = 8 Fx2 = 74 50 10*8 = 24 8 = 56 kN Mx2 = 74 (10) 20(1+8) 30*8 10*8(4) = 0


Example(7-4) Fig(7-4) shows the simply supported beam with the load as indicated in the figure . Draw the S.F. , B.M. and A.F diagrams

Fig(7-4) Solution

Fy = 0 R1+R2 = 20*6 + 20 +40 = 180 kN

MA = 0 20 *6*3+40*4 R1*2 R2*6 =0 solving the above equations gives : R1 = 140 kN , R2 = 40 kN 0<x1<2 Fx1= 20 20 x1 Mx1 = 20 x1 20 x1 ( x1/2 ) At x1 = 0 Fx1 = 20 kN , Mx1 = 0 At x1 = 2 Fx1 = 60 kN , Mx1 = 80 kN.m 0<x2<2 Fx2 = 20 + 140 20*(x2+2) Mx2= 20 *(2+x2) + 140 x2 20* (x2+2)( (x2+2)/2) At x2 = 0 Fx2 = 80 kN , Mx2 = 80 kN.m At x2 = 2 Fx2 = 20 +140 20*4 = 40 kN Mx2 = 20 *4 + 140* 2 20 (2+2)2/2= 40 kN.m 0<x3<2 Fx3= 20+140 40 20(4+x3) Mx3= 20(4+x3)+140(2+x3) 40x3 20(4+x3) * (4+x3)/2 At x3 = 0 Fx3 = 20 + 140 40 20*4 =0 Mx3 = 20*4+140*2 (20/2) (4*4) = 40 kN.m At x3 = 2 Fx3 = 20 ( 4+2)+ 140(2+2) 40*2 20* (4+2)2/2 =0


Example(7-5) Fig(7-5) shows the simply supported beam with the load as indicated in the figure . Draw the S.F. , B.M. and A.F diagrams

Fig(7-5) Solution

Fy = 0 R1+R2 = 60 kN MA = 0 5R2 = 60*3 R2 =36 kN , R1 = 24 kN

0 x1<2 Fx1 = R1 = 24 kN Mx1 = R1 x1 At x1 = 0 Fx1 = 24 , Mx1 = 0


At x1 = 2 Fx1 =24 , Mx1 = 48 kN.m 0 x2<2 Fx2 = R1 30 x2 Mx2 = R1(2+x2) 30 x2 ( 0.5 x2) = R1 (2+x2) 15* (x2)2 At x2 = 0 Fx2 =24 , Mx2 = 48 kN.m At x2 = 2 Fx2 = 24 60 = 36 , Mx2 = 96 60 = 36 0<x3<1 Fx3 = R1 60 = 24 60 = 36 kN Mx3 = R1(2+2+x3) 60 (1+x3) At x3 = 0 Fx3 = 36 , Mx3 = 96 60 = 36 At x3 = 1 Fx3 = 36 , Mx3 = 24(5) 60(2) = 0


Example(7-6) Fig(7-6) shows the simply supported beam with the variable load as indicated in the figure . Draw the S.F. , B.M. and A.F diagrams

Fig(7-6) Solution

Fy = 0 R1+R2 = 0.5 * 3 * 20 = 30 , (y/x1) = (20/3) , y = (20/3) x1

MA=0 5 R2 30*(2/3)*3 = 0 , R2 = 12 kN , R1 = 18 kN 0 x1<3 Fx1 = R1 0.5 x1 y Fx1 = 18 0.5 x1 ( 20x1/3) = 18 (10(x1)2)/3 Mx1 =R1 x1 (10/9)(x1)3 At x1 = 0 Fx1 = 18 kN , Mx1 =0 At x1 = 3 Fx1 = 12 kN , Mx1 = 18 * 3 (10/9) (3*9) = 24 kN.m 0 x2<2 Fx2 = R1 30 = 18 30 = 12 kN Mx2 = R1 (3+x2) 30(1+x2) At x2 = 0 Fx2 = 12 kN , Mx2 = 18*3 30 = 24 kN.m At x2 =2 Fx2 = 12 kN , Mx2 = 90 90 = 0


Example(7-7) Fig(7-7) shows the simply supported beam with the load as indicated in

the figure . Draw the S.F. , B.M. and A.F diagrams Fig(7-7) Solution

Fx= 0 RAx = 3 + 14 = 17 kN Fy =0 RA + 4 + RD = 0 MA = 0 7*RD 28+6+8 =0 RD = 2 kN MD = 0 7 RA + 28 6 + 4* 5 = 0 RA = 6 kN

0 x1<2 Fx1 = 6 kN , Ax1 = Rax = 17 kN Mx1 = 6 x1 At x1 = 0 Fx1 = 6 kN , Mx1 = 0 , Ax1 = 17 kN At x1 = 2 Fx1 = 6 kN , Mx1 = 12 kNm , Ax1 = 17 kN 0 x2<2 Fx2 = 6+4 = 2 kN , Ax2 = RAx 3 = 17 3 =14 kN Mx2 = 6 (2+x2) 6 + 4 x2 At x2 = 0 Fx2 = 2 kN , Mx2 = 18 kN.m , Ax2 = 14 kN At x2 = 2 Fx2 = 2 kN , Mx2 = 22 kN.m , Ax2 =14 kN 0 x3<3 Fx3 = 6+4 = 2 kN , Ax3 = RAx 3 14 = 0 Mx3 = 6(2+2+x3) + 4(2+x3) 6+28 At x3 = 0 Fx3 = 2 kN , Mx3 = 6 kN.m , Ax3 = 0


At x3 = 3 Fx3 = 2 kN , Mx3 = 0 , Ax3 = 0

Example(7-8) Fig(7-8) shows a crank shaft , contract the S.F. , B.M. & A.F. diagrams for the crank shaft .Takes a= 10* 10 2 m , L = 20*10 2 m and P = 5 kN .

Fig(7-8) Solution


Fx = 0 RAx = 0 Fy = 0 RA+RG = P MA=0 P * 2a + RG * 4a = 0 RG = P/2

RA = P (P/2) =P/2 RA = RG = 2.5 kN First interval 0 x1 a Nx1 = 0 Fx1 = RA = P/2 = 2.5 kN Mx1 = RA . x At x1 = 0 Mx1 = 0 At x1 = a Mx1 = 2.5*10*10 2 = 25 kN.m Second interval 0 y2 L Ny2 = RA = 2.5 kN Fy2 =0 My2 = RA a = a (P/2) = 25 kN.m Third interval 0 x3 a Nx3 = 0 Fx3 = RA = 2.5 kN Mx3 = RA (a+x3) At x3 = 0 Mx3 = RA . a = 25 kN.m At x3 = a Mx3 = 2.5 (a+a) = 50 kN m Fourth Interval 0 x4<a Nx4 = 0 Fx4 = RA P = 2.5 5 = 2.5 kN Mx4 = RA(2a+x4) Px4 = 50 2.5x4 At x4 = 0 Mx4 = 50 At x4 = a Mx4 = 50 2.5*10 =25 kN.m Fifth Interval 0 y5 L Ny5 = RA P = 2.5 5 = 2.5 kN Fy5 = 0 My5 = RA * 3 a P a = (P/2)*3a Pa = (P/2) a = 0.5*5*10=25 kNm Sixth Interval 0<x6<a Nx6 = 0 Fx6 = RA P = 2.5 5 = 2.5 kN Mx6 = RA(3a+ x6) P(a+x6) = 25 2.5 x6 At x6 = 0 Mx6 = 25


At x6 = a Mx6 = 25 25 = 0

B.M. S.F. A.F. Example(7-9) Fig(7-9) shows a beam ABC is continuous over two spans (i.e. two beams are connected by a hinge) , being supported as shown in the figure , find the reaction .

Fig(7-9)

Solution

0Mhinge 1.524 R3 12200.7 8.826*103*0.6096 =0 R3 = 11.564 kN


For the whole structure

0Fy R1+R2=26.68 17.792+8.896 11.564 R1+R2= 6.22 (i)

0MB 26.268 ( (0.6096/2)+ 1.8288+0.6096) + R1 * (1.8288+0.6096) +17.792 * 0.6096 + 8.896 (0.6096+ 0.6096) + 12.2007 11.564 ( 0.9144 +0.6096 +0.6096) = 0

73.09 + 2.438*R1 + 10.846 + 10.85 +12.2 24.67 = 0 R1= 26.21 kN R2 = 20 kN


7-5 Problems

7-1) Fig(7-10) shows a simply supported beam loaded as indicated in the Fig. Draw to scale the shear force and bending moment diagrams and find how much may be added to the load at mid span without increasing the maximum bending moment on the beam .

Fig(7-10)

7-2) Fig(7-11) shows the beam ABC is continuous over two spans , being supported as shown in Fig. A hinge , capable of transmitting shearing force but not bending moment , is placed at the center of span AB . The loading consists of distributed load , of total weight 20 kN , spread over the span AB , and a concentrated load of 30 kN at the center of span BC . Draw the shearing force and bending moment diagrams.

Fig(7-11)

7-3) Fig.(7-12) shows the curve of upward distributed water pressure reaction on a pontoon . This reaction is uniformly distributed over the central 9 m and decreases to zero at the two ends , the equation of this curve being y = kx2 , where k is a constant . The pontoon carries a uniformly distributed load of 50 kN/m run

over the central 7 m . Draw the S.F. & B.M diagrams due to this loading and find the max. values of S.F. and B.M.

Fig(7-12)


7-4) Fig(7-13) shows a fixed fixed beam with the load indicated in the Fig. .Plot the diagrams for the transverse force ( shearing force) and the bending moment .

Fig(7-13)

7-5) Fig(7-14) shows a simply supported beam with the load indicated . Draw the S.F. , B.M. & A.F. Diagrams

Fig(7-14) 7-6) Fig(7-15) shows the beam with the load indicated . Draw

the S.F. , B.M. & A.F. Diagrams .

Fig(7-15)

7-7) Fig.(7-16) shows a simply supported beam with the loaded indicated as shown . Find the value of x if the mid point of the beam is a point of inflexion ( or contra- flexure) and for this arrangement plot on square paper the bending moment diagram indicating the principal numerical value locate any other points of inflexion .


Fig(7-16) 7-

and symmetrically supported over a span of 8 m . It carries concentrated loads of 2 kN and 5 kN at the left hand and right hand ends respectively as well as a uniformly distributed load of 1.25 kN/m between the supports. Draw the shearing force and bending moment diagrams for this loading , state the the principal value of each and calculate the positions of the points of contra flexure .

7-9) Fig.(7-17) shows a beam under lateral loading, draw the

shear and bending moment diagrams .

Fig(7-17) 7-10) Prove relations between B.M. , S.F and intensity of loading on a

beam . A beam 8 m long is supported at the ends and carries a distributed load which varies uniformly in intensity from zero at one end to 30 kN/m at a section 2 m from the other end and over the remaining length is constant at 30 kN/m

Derive equations for S.F. and B.M. at any section of the beam , and draw the S.F. and B.M. diagrams , marking on the diagrams the max. values .

strength of materials- axial force shear force bending moment diagrams- hani aziz ameen

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