Quick reviewT i & T DiTorsion & Torque Diagrams

upward"point " that P, ofleft the to torquesAllintTInternal Torque:

Shear Stress:

TrGr

downward"point " that P, ofleft the to torquesAll

JTr

LGr

Angle of twist (Engineers’ Theory of Torsion):

TLLL is the length of the shaft

GJTL

GrL

G is the shear modulusJ is the polar moment of arear is the radius at which we want to calculate the shear stress

TTorque diagram:

B andA between x vsJGT ofgraph aunder Area/ AB

Quick reviewBM & SF di b E ilib iBM & SF diagrams by Equilibrium

1 Cutting the beam1. Cutting the beam2. Draw the free body diagram3 Use equilibrium3. Use equilibrium

0M0F0 F

4 Find internal bending moment and shear force

0M ,0F ,0 Ay xF

4. Find internal bending moment and shear force

Note: Identify every point where the loading on the beam changes.

Lecture 4

Chapter 9: BM & SF diagrams by inspection (Section 6.2)Two Rules

Using Rules to Construct SF & BM diagrams

End Conditions

Principle of Superposition

Chapter 10: BM & SF diagrams by Singularity functions

(Section 12.3)Macaulay Brackets & Singularity functions

Applications

Summary y

Chapter 9 BM & SF Diagrams by Inspection

Shear force Bending moment 2m

3kN/m

Shear force

vx=2.7

vx

g

Mx=2.7xMx Mx=-1.5x2+8.7x-6

5m

2m

x

A B

C

x

8 7 3x

vx=8.7-3x

The important features that we want to determine are: 1. The shape of the diagrams in each region—horizontal, linear,

parabolic or cubic.2. If it is parabolic or cubic, is it concave up or concave down.3 The values of bending moment and shear force at each change of3. The values of bending moment and shear force at each change of

loading.4. The maximum values of bending moment and shear force.

Relating BM & SF diagrams to distributed forces, point forces and point couples

2m

3kN/mωxωx=3

2m

x

A B

C

Region ωx (kN/m) vx (kN) Mx (kNm)0<x<2m 0 2.7 2.7x x

5m2m<x<5m 3 -3x+8.7 -1.5x2+8.7x-6

Shear forceBy differentiating, the relationship between ωx, vx and Mx is as follows.

Shear force

vx=2.7

vx

By integratingdx

dMv x

x

dxvM xx

x

vx=8.7-3x

dxdvx

x dxv xx Bending moment

M 2 7

Mx Mx=-1.5x2+8.7x-6

Mx=2.7x

dxv xx

dxvM

Rule 1

Rule 2x dxvM xx Rule 2

Rule 1: The change in vx between any two points A and B is given by

fddi t ib td thAleft the toforcespoint downward All-

forceddistribute under theArea xAxB vv

F2

F3 F4 F5 areaarea

vx

A BF1 F6 43 FFareavv xAxB

vxA

v BvxB

xA xB

x

Rule 2: The change in Mx between any two points A and B is given by

diagramforceshearunder theAreaMMleft the tocouplespoint clockwise All

diagramforceshear under theArea xAxB MM

F2

F3 F4 F5

A BF1 F6

C

vx

A BF1 F6

area CareaMM xAxB

xA

xB x

MMxB

Mx

x MxA

Note that an area below the x axis is negative.

Example 3: Draw the shear force and bending moment diagrams for the beam shown.

A

B

12kN A

B 12kN

4m 2m 4m 2mRA RB

Fi d th t tivx(kN)

Shear diagram

Find the support reactions

veAM

0

06

8

veyF 0 012 BA RR

kNRR BB 402126 0-4

2x (m)

kNRA 8

6

Mx(kNm)

16 Sl 8

Slope=-4

06

2x (m)

Slope=8

Some further points to help with constructing shear force and bending moment diagrams are:

1. The change of vx at any point is equal to F1 F6

minus the value of ωx at that point.

dv

vx

1 6

Slope=-ωx

xx

dxdv

x

2. The change of Mx at any point is equal to the value of v at that point vxthe value of vx at that point.

x vdM

xvx

xx v

dx

Mx

Slope=vx

x

x

p x

Some further points to help with constructing shear force and bending moment diagrams are: 3. vx will have a local maximum or minimum value where its derivative is

equal to 0. This happens when ωx=0.The value of v at this point is given by Rule 1The value of vx at this point is given by Rule 1

)leftthetoforcespointdownward(Allleft) the to ofgraph aunder Area( x/ MINMAXv

4. Similarly, Mx will have a local maximum or minimum value where its

)left the toforcespoint downward(All-

derivative is equal to 0. This happens when vx=0. From Rule 2, the corresponding maximum (or minimum) value for Mx is given by

l ft)thtfhdA(M)left the tocouplespoint clockwise All(

left) thetovofgraph aunder Area( x/

MINMAXM

Some further points to help with constructing shear force and bending moment diagrams are:

5. Since , where ωx is positive (acting down), the graph of Mx will

be concave down and whenever wherever ωx is negative (acting up), x

x

dxMd

2

2

x g ( g p)

the graph of Mx will be concave up.

ω acting downω acting downω acting up

MMx

Concave up

x

Concave down

Some further points to help with constructing shear force and bending moment diagrams are: 6. It is useful to remember the general shape of vx and Mx for the common

types of loading.

Loading Shear force diagram Bending moment diagram

P i t C l ( l k i ) N ff t U d tPoint Couple (clockwise) No effect Upward step C Mx

vx

x x

Point Force (Upward) Upward step Kink

M vx

x

Mx

x

When integrating,

)1(yCubic)1y(Parabolakx)y(lineStraightk)y(Constant 32 kxkx

This allows us to anticipate the shape of the bending moment and shear force diagrams under different distributed loads.

)6

(yCubic)2

y(Parabolakx)y(lineStraight k)y(Constant kxkx

Loading Shear force diagram Bending moment diagram

U if l di t ib t d St i ht li P b l ( dUniformly distributed force

Straight line Parabola (concave down if ωx points down)

Mxvx

x x

Linearly changing distributed force

Parabola Cubic (concave down if ωx points down)

vx

x

Mx

x

Some of the common loading cases are shown below.

Example 4: Construct shear force and bending moment diagrams for a beam with the loading shown.

12N/m12N 12N/m

A B

12N 20Nm

4m 4m1m1m

Feq=12×4=48N 12N 20Nm

44R R11

20Nm

4m4mRA RB1m1m

024882010120

Ave

B RMStep 1: Find the support reactions,

NRA 5.29

048125.290 Bvey RF NRB 5.30

1 2N /n 12 N 2 0N m

v (N)

4m 4m29 .5N 30 .5N 1m 1m

vx(N)-12+29.5=17.5

Slope 12

4m 4m1m 1m

xSlope=-12

Shear Force Diagram

-1217.5-12×4=-30.5

26 17 5 1 46/2 38 76 Mx(Nm) -44+17.5×4=2626+17.5×1.46/2=38.76

10 62 1

x(m)-12

-32

Bending moment Diagram-12×1=-12 7.46

-32

-44-32-12×1=-44

Example 5: Draw the shear force and bending moment diagrams for the beam shown.

A

B

6kN 12kNA

B 6kN 12kN

2m 2m 2m 2m 2mRA RB2m

Fi d th t tivx(kN)

Shear diagram

Find the support reactions

veAM

0

0 6

8

kNRR

B

B

100412266

0

-102

x (m)

Slope=2

veyF 0 0612 BA RR

Mx(kNm)

16 Sl 8

Slope=-10Slope=2

8×2=16

16+2×2=20

kNRA 80 6

2x (m)

Slope=8 8×2=16

End conditions12N/m12N

20Nm

J t t id th fi t d4m 4m

A B

1m1m

Just outside the first and last force or couple on the beam, Mx=0 and

vx(N)

17.5

vx=0.

102

7.46

6( )

17.5

-12x(m)

-30.526

38.76Mx(Nm)26

10621

( )

x(m) -12

-32

7.46

-44

The Principal of SuperpositionThe Principal of SuperpositionThe shear force and bending moment on a beam under the action of two or more loads is equal to the sum of the shear force and bending moment due to each of loads appliedforce and bending moment due to each of loads applied individually.

The Principal of SuperpositionThe Principal of Superposition

A

3kN/m 8kN 3kN/m

8kN

2m

A

B A

3kN 3kN 4kN 4kN

vx vx vx 7

3 4

x

4

x-3

x -4 x

-4 x

-7 Mx

Mx Mx

1.5

x

4

5.5

Mx

x xx

Expressions for the shear force and bending moment

12kN 3kN/m

AB

12kN

A B

x4m 2m

4m

x

3xwxv 36

208 xvxdMxvx 36

xxM x 65.1 2

624 xx

62244

208xx

xxM x

dxdM

v xx

62244 xx

Polynomial functions--

1d

naxB )( Could we use one single

1)())(( nn axnBaxBdxd

B 1

equation for shear force and bending moment?

Caxn

BdxaxB nn

1)(1

)(

Macaulay BracketsWe define a new notation consisting of angular brackets with

zeroornegativeisa)-(xif0

positive is a)-(x if )( axax

zeroor negativeisa)-(xif 0

Singularity functions-- naxB

1)( nn axnBaxBdxd

Caxn

BdxaxB nn

1

1

positive is 2)-(x if12

zeroor negative is 2)-(x if0212 0x

positive is 2)-(x if)2(12zeroor negative is 2)-(x if0

212 1

xx

positive is 2)-(x if)2(12zeroor negative is 2)-(x if0

212 22

xx

Example 1: 12kN

A Mx

K12kN

A B

‘Cut’ 2mx8kN

vx 4kN

M 04m2m

8kN 4kN

veKM

0

0)6(4)2(128 xxxM x

62244

208xx

xxM x 6421208 xxxM x

)6(4)2(128 xxxM x

208 x

v

62244 xx

000 6421208 xxxvx 624 x

vx x

Example 2: Determine expressions for shear force and bending moment being carried in the beam shown.

MA

5N/m

being carried in the beam shown. A

BC

5N/m Feq1=6×5/2=15N 2m

4m

MA B

8Nm

C

4N/mRA

7m 4m

0 5N

8Nm

10m 4N/m

7m 0.5N 10m 4N/m 0.5N

Feq2=3×4=12N

Step 1: Support Reactions

Feq2 3 4 12N

05.012150

Avey RF

NRA 5.2

veAM

0

0105.05.8128158 AM A

NmM A 5

Feq1=5(x-4)/6×(x-4)/2Step 2: Cut the beam at a point just short of the right hand end.

x

-5 KC

ωxmax=5(x-4)/6

Mx(x-4)/3 ωxmax

ωmaxLx

x maxmax

7m4m 8Nm

4N/m2.5

vx

xmax

0.5Nx

Feq2=(x-7)×4x

L

veKM

0

7)4()4()4(5 xxxx 0)10(5.02

7)7(43

)4()2

)4(6

)4(5(8)5(5.2

xMxxxxxxx

)10(50)7(2)4(58552 23

xxxxM )10(5.0)7(236

855.2 xxxM x

5Use Macaulay brackets,

105.0724365480505.2 2300 xxxxxxM x

Step 3: Differentiate the expression for bending moment to give an expression for shear forcep

Bending moment,

105072454805052 2300M

Shear force,

105.0724365480505.2 2300 xxxxxxM x

020 105.074412505.2 xxxx

dxdM

v xx

Example 3: Find expressions for shear force and bending moment being carried in the beam shown.carried in the beam shown.

A

10N 6Nm

7N/mA

10N 6Nm

7N/m

Feq=3×7/2=10.5N

3m

B6N

C

3m

AB

6NmC

RA RB 3m/37m

10m 7m

10m

Step 1: Support Reactions

ve

AM

0 015.10671010 AR NRA 45.7

05.10100

BAvey RRF NRB 05.13

Step 2: Cut the beam at a point just short of the right hand end.

10N 7N/

)7(37

max x

7 A

B

10N 6Nm

C

7N/m

wxmax

A10N

6Nm

2/)7()7(37

xxFeq

7m 3m

10m 3m

A 6NmC

7.45 vx

Mx

7m x 13.05

(x-7)/3

ve

BM

0

0)10(0513)7()7()7(76)3(10457

Mxxxxxx 0)10(05.13323

6)3(1045.7 xMxxx

)10(05.13)7(1876)3(1045.7 3 xxxxM x

1005.13718736310045.7 30 xxxxxM x

Use Macaulay brackets,

18

Step 3: Differentiate the expression for bending moment to give an expression for shear forcep

10051377363100457 30 xxxxxM

Bending moment,

1005.13718

36310045.7 xxxxxM x

Shear force,

0200 1005.13767310045.7 xxxx

dxdM

v xx

Note: