lecture-23 (shear force diagram & bending moment diagram)

Lecture-23 (Shear force

diagram & bending moment

diagram)

Solid Mechanics

(CE 251)

Trigonometric functions:

(1) Line function: y = mx + c

y = dependent variable

x = independent variables

m = slope of line

c = interceptor on y-axis

y = mx

m = + ve

y = mx + c

m = +ve

c = +ve

y = mx + c

m = -ve

c = +ve

y = mx + c

m = +ve

c= -ve

y = mx + c

m = -ve

c= -ve

(1) Standard Equations of a Parabola with Vertex at (0, 0)

y = dependent variable

x = independent variables

1.1 y2 = 4ax

Vertex: (0, 0)

Focus: (a, 0)

Directrix: x = a

Symmetric with respect to the x axis

Axis the x axis

a < 0 (open left)

a > 0 (open right)

a < 0 (open down)

a > 0 (open up)

1.2 x2 = 4ay

Vertex: (0, 0)

Focus: (0, a)

Directrix: y = a

Symmetric with respect to the y axis

Axis the y axis

1.3 Convert quadratic equation to standard parabola

byYwhere

cbxaxy

:are Parabola ofVertex

Vertex:

Symmetric with respect to the y axis

Axis the y axis

a > 0 (open up)

1.4 y = ax3 + bx2 + cx +d

where a ≠ 0, and b, c and d (=0) are constants (can be

zero) and If a > 0 the graph of a cubic looks similar to

one of the following graphs:

yrxqxpx

))()((

IIICase

IICase

1.4 y = ax3 + bx2 + cx +d

where a ≠ 0, and b, c and d (=0) are constants (can be

zero) and If a < 0 the graph of a cubic looks similar to

one of the following graphs:

yrxqxpx

))()((

IIICase

IICase

Sign Conventions SF and BM:

Sign Conventions of SF & BM

Sagging

Hogging © CE 251 (Solid Mechanics) by Anil

Mandariya 10

Steps to Calculate SF and BM in Beams:

Step-1

Determine the reactions at the supports.

Step-2

Divide the beam span in different segments according

loading and support conditions.

Step-3

Consider left side end of beam as origin.

Step-4

Choose a section corresponding to each segment which

will be at ‘x’ distance from the left end of the beam

(origin). © CE 251 (Solid Mechanics) by Anil Mandariya

Step-5

Take a left segment of the beam from section Y-Y.

Step-6

Convert all UDL, UVL in point loads.

Step-7

Choose the sign conventions and write the static

equilibrium equations for the beam segment.

Step-8

Solve static equilibrium equations for the each beam

segment separately according to boundary condition of

segments.

Step-9

Consider a x and y axis below the beam such that length

of the x-axis will be equal to beam span.

Step-10

Mark the all points according to loading and supports

on the x-axis.

Step-11

Draw the curve separately according to SF and BM

equations for each segments with boundary values.

Example 1:

Step-1

Determine the reactions at

the supports.

ƩFx = 0 = HB= 0 ..........(1)

ƩFy = 0 = -RB - RD + 20 + 40= 0

RD + RB = 60 kN .............(2)

ƩM = 0 = ƩMB = 0

= - (3 x 40) + (5 x RD) + RB x 0

+ HA x 0 + (2.5 x 20)= 0

RD = 70/5 = 14 kN ..............(3)

From eq. (2) & (3)

RB = 46 kN © CE 251 (Solid Mechanics) by Anil Mandariya

Step-2

Segment-1: AB

Segment-2: BC

Segment-3: CD

Step-3

Step-4

Choose a section y-y’ corresponding to each segment

which will be at ‘x’ distance from the left end of the

beam (origin).

Step-5

Take a left segment of the beam from section Y-Y.

y’ Step-6

There is no UDL or UVL.

Step-7 & 8

equilibrium equations for the beam segment & solve them.

Segment-1: AB (0 ≤ x ≤ 2.5)

ƩFx = 0 ..........(1)

ƩFy = 0

= 20 + Vx = 0

Vx = - 20 kN (constant).............(2)

ƩM = 0

= (x * 20) + Mx = 0

Mx = -20x (line equation with –ve

slop and zero intercept on y-axis)..(3)

From eq. (2) & (3)

At x = 0

VA = - 20 kN; MA = 0 kN-m

At x = 2.5 m

VB = - 20 kN; MB = - 50 kN-m © CE 251 (Solid Mechanics) by Anil Mandariya 18

y’ x

RB = 46 kN

(x - 2.5)

Segment-2: BC (2.5 ≤ x ≤ 5.5)

ƩFx = 0 ..........(1)

ƩFy = 0

= 20 + Vx – 46 = 0

Vx = 26 kN (constant value).............(2)

ƩM = 0

= (x * 20) - (x – 2.5)*46 + Mx = 0

Mx = 26x - 115 (line equation with +ve

slope and –ve intercept on y-axis)..........(3)

From eq. (2) & (3)

At x = 2.5 m

VB = 26.0 kN; MB = - 50.0 kN-m

At x = 5.5 m

VC = 26.0 kN; MC = 28.0 kN-m © CE 251 (Solid Mechanics) by Anil Mandariya 19

Segment-3: CD (5.5 ≤ x ≤ 7.5)

(x-2.5-3)

RB = 46 kN

ƩFx = 0 ..........(1)

ƩFy = 0

= 20 – 46 + 40 + Vx = 0

Vx = -14.0 kN (constant value).............(2)

ƩM = 0

= (x * 20) - (x – 2.5)*46 + (x – 2.5 – 3)*40

+ Mx = 0

Mx = -14x + 105 (line eq. with –ve slop

and +ve intercept)................................(3)

From eq. (2) & (3)

At x = 5.5 m

VC = -14.0 kN; MC = 28.0 kN-m

At x = 7.5 m

VD = -14.0 kN; MD = 0 kN-m © CE 251 (Solid Mechanics) by Anil Mandariya

Step-9

Consider a x and y axis below the beam such that length

of the x-axis will be equal to beam span.

Step-10

Mark the all points according to loading and supports

on the x-axis.

Step-11

Draw the curve separately according to SF and BM

equations for each segments with boundary values.

Segment-1: AB (0 ≤ x ≤ 2.5)

Vx = - 20 kN (constant)

Mx = -20x (line equation with

–ve slop and zero intercept on

y-axis)

At x = 0

VA = - 20 kN; MA = 0 kN-m

At x = 2.5 m

VB = - 20 kN; MB = - 50 kN-m

-20 kN SFD

Segment-1: AB (0 ≤ x ≤ 2.5)

Vx = - 20 kN (constant)

Mx = -20x (line equation with

–ve slop and zero intercept on

y-axis)

At x = 0

VA = - 20 kN; MA = 0 kN-m

At x = 2.5 m

VB = - 20 kN; MB = - 50 kN-m

-20 kN

-50 kN-m

0.0 kN-m

Segment-2: BC (2.5 ≤ x ≤ 5.5)

Vx = 26 kN (constant value)

Mx = 26x - 115 (line equation

with +ve slope and –ve

intercept on y-axis)

At x = 2.5 m

VB = 26.0 kN; MB = - 50.0 kN-m

At x = 5.5 m

VC = 26.0 kN; MC = 28.0 kN-m

Mx = 0

26x – 115 = 0

x = 4.42 m

-20 kN

-50 kN-m

0.0 kN-m

E x = 4.42 m 28 kN-m

Segment-2: CD (2.5 ≤ x ≤ 7.5)

Vx = -14.0 kN (constant value)

Mx = -14x + 105 (line eq. with –

ve slop and +ve intercept)

At x = 5.5 m

VC = -14.0 kN; MC = 28.0 kN-m

At x = 7.5 m

VD = -14.0 kN; MD = 0 kN-m

-20 kN

0.0 kN-m

-50 kN-m

E x = 4.42 m 28 kN-m

-14 kN

Example:2

Step-1

Determine the reactions at the supports.

Couple: arm = 2 m and

force = 8 kN

Mo = 2 x 8 = 16 kN-m

UDL: 2 Nos.

q1 = q2 = q = 4 kN/m on 2 m

Resultant point load = 4 x 2 = 8 kN

Will act at centroid of the UDL

= 1 m from A and B © CE 251 (Solid Mechanics) by Anil Mandariya

1.0 m 3.0 m 1.0 m

8.0 kN 8.0 kN

ƩFx = 0 = HA= 0 ..........(1)

ƩFy = 0 = -RA - RB + 8.0 + 8.0 = 0

RA + RB = 16.0 kN .............(2)

ƩM = 0 = ƩMB = 0

= RB x 0 + (1 x 8) + - (Mo = 16) + (7 x 8) - (8 x RA) + HA x 0 = 0

RA = 48/8 = 6.0 kN ..............(3)

From eq. (2) & (3)

RB = 10.0 kN

So Reactions are HA= 0; RA = 6.0 kN; RB = 10.0 kN

Step-2

Segment-1: AC

Segment-2: CD

Segment-3: DE

Segment-4: EB © CE 251 (Solid Mechanics) by Anil Mandariya 30

Step-3

Step-4

Choose a section y-y’ corresponding to each segment

which will be at ‘x’ distance from the left end of the

beam (origin). 1st choose for segment AC and so on.

Step-5

Take a left segment of the beam from section y-y'.

6.0 kN

Step-6

UDL: 1 Nos.; q = 4 kN/m & length = x m

Resultant load of UDL = 4x kN

Will act at centroid of UDL: x/2 m

Step-7 & 8

equilibrium equations for the beam segment & solve them.

Segment-1: AC (0 ≤ x ≤ 2.0 m)

ƩFx = 0 ............................................(1)

ƩFy = 0 = - 6.0 + 4x + Vx = 0

Vx = (6- 4x) kN ............................................(2)

ƩM = 0

= - 6*x + (x/2 * 4x) + Mx = 0

Mx = - 2x2 + 6x ...........................................(3)

From eq. (2) & (3)

At x = 0 m

VA = 6.0 kN; MA = 0.0 kN-m

At x = 2.0 m

VC = -2.0 kN; MC = 4.0 kN-m © CE 251 (Solid Mechanics) by Anil

Mandariya 34

Segment-2: CD (2.0 ≤ x ≤ 4.0 m)

6.0 kN x

6.0 kN

8.0 kN

ƩFx = 0 ..........................(1)

ƩFy = 0 = - 6.0 + 8 + Vx = 0

Vx = -2.0 kN .............................. (2)

ƩM = 0

= - 6*x + (x – 1) * 8 + Mx = 0

Mx = - 2x + 8 .............................(3)

From eq. (2) & (3)

At x = 2.0 m

VC = -2.0 kN; MC = 4.0 kN-m

At x = 4.0 m

VD = -2.0 kN; MD = 0.0 kN-m

Segment-3: DE (4.0 ≤ x ≤ 6.0 m)

y’ 6.0 kN

8.0 kN

ƩFx = 0 .....................................(1)

ƩFy = 0 = - 6.0 + 8 + Vx = 0

Vx = -2.0 kN ....................................... (2)

ƩM = 0

= - 6*x + (x – 1) * 8 – Mo (= 16) + Mx = 0

Mx = - 2x + 24 ....................................(3)

From eq. (2) & (3)

At x = 4.0 m

VD = -2.0 kN; MD = 16.0 kN-m

At x = 6.0 m

VE = -2.0 kN; ME = 12.0 kN-m

Segment-4: EB (6.0 ≤ x ≤ 8.0 m)

y’ 6.0 kN

8.0 kN

4.0 kN/m

ƩFx = 0 .....................................(1)

ƩFy = 0 = - 6.0 + 8 + 4*(x – 6) + Vx = 0

Vx = 22 – 4x kN ................................ (2)

ƩM = 0

= - 6*x + (x – 1) * 8 – Mo (= 16) + 4*(x – 6)*(x – 6)/2 + Mx = 0

Mx = - 2x2 + 22x - 48 ....................................(3)

y’ 6.0 kN

8.0 kN

4.0 kN/m

4*(x – 6)

(x – 6)/2 m

From eq. (2) & (3)

At x = 6.0 m

VE = -2.0 kN;

ME = 12.0 kN-m

At x = 8.0 m

VB = -10.0 kN;

ME = 0.0 kN-m

A B D E C

C D E B

Consider a x and y axis below the beam such that length of the x-

axis will be equal to beam span and mark the all points according

to loading and supports on the x-axis.

Step-9

Draw the curve separately according to SF and BM equations for

each segments with boundary values.

Step-11

Segment-1: AC (0 ≤ x ≤ 2.0 m)

Vx = (6- 4x) kN (line eq. with –ve slop and +ve intercept on y-axis and SF =0 at

x = 1.5 m).

Mx = - 2x2 + 6x (quadratic eq. of parabola with vertex (1.5, 4.5))

At x = 0 m

VA = 6.0 kN; MA = 0.0 kN-m

At x = 2.0 m

VC = -2.0 kN; MC = 4.0 kN-m

Select this part

of parabola for

BMD according

to boundary

A B D E C

C D E B

-2 kN 1.5 m

4 kN-m

Segment-2: CD (2.0 ≤ x ≤ 4.0 m)

Vx = -2.0 kN (constant value, horizontal axis parallel to x-

Mx = - 2x + 8 (line eq. with –ve slop and +ve intercept on y-

At x = 2.0 m

VC = -2.0 kN; MC = 4.0 kN-m

At x = 4.0 m

VD = -2.0 kN; MD = 0.0 kN-m

-2 kN 1.5 m

2 m C E D

A B D E C

C D E B

4 kN-m

2 m 2 m

Segment-3: DE (4.0 ≤ x ≤ 6.0 m)

Vx = -2.0 kN (constant value, horizontal axis parallel to x-

Mx = - 2x + 24 (line eq. with –ve slop and +ve intercept on

y-axis)

At x = 4.0 m

VC = -2.0 kN; MC = 16.0 kN-m

At x = 6.0 m

VD = -2.0 kN; MD = 12.0 kN-m

-2 kN 1.5 m

2 m C E D

A B D E C

C D E B

4 kN-m

16 kN-m

12 kN-m

2 m 2 m 2 m

Segment-4: EB (6.0 ≤ x ≤ 8.0 m)

Vx = 22 – 4x kN (line eq. with –ve slop and +ve intercept

on y-axis)

Mx = - 2x2 + 22x - 48 (quadratic eq. of parabola with vertex

(5.5, 36.5))

At x = 6.0 m

VE = -2.0 kN;

ME = 12.0 kN-m

At x = 8.0 m

VB = -10.0 kN;

ME = 0.0 kN-m

Select this part

of parabola for

BMD according

to boundary

-2 kN 1.5 m

2 m C E D

A B D E C

C D E B

4 kN-m

16 kN-m

12 kN-m

2 m 2 m 2 m

-10 kN

SFD & BMD for different types of beam and

loading condition:

(1) Simply supported beam:

(2) Cantilever beam:

BM will be maximum at:

1. At sections where a concentrated load acts and the

shear force changes its sign.

2. At sections where shear force = 0

3. At supports providing vertical reactions.

4. At sections where a couple are applied.

Point of contra flexure:

A point where BM changes its sign.

Anil Mandariya 52

Important points for BM:

The slop of the SFD at any section of the beam is equal

to the intensity of the distributed load over

corresponding section.

Presence of the concentrated loads are indicated by

abrupt changes in SFD. The abrupt change (change in

ordinate) of SF at a section is equal to the

concentrated load at the section.

The slop of the BMD at any section is equal to the SF

at the corresponding section.

At concentrated load point, the slop of BMD is abruptly

chnges.

Anil Mandariya 53

Properties of SFD & BMD:

Total load on a beam between two sections =

difference between the SF at these two sections

The area of SFD between two section = difference

between BM at these two sections.

Anil Mandariya 54

Thanks

lecture-23 (shear force diagram & bending moment diagram)

Documents

shear force and bending moment

bending torsion tension shear tutorial

shear & non-uniform bending

tripura public service commission -...

analysis of beams _ shear force & bending moment diagram ~...

shear force and bending moment diagram

unsymmetrical bending & shear center hst21

shear blades panel bending tools bending

bending moment and shear force

unsymmetric bending and shear centre

unsymmetrical bending shear stress distribution

shear force & bending moment diagram lecture

combined shear and bending

4. bending moment and shear force...

chapter #3 shear force & bending moment introduction types...

shear forces and bending moments

shear and bending moment diagrams.ppt

shear and bending moments beams

shear-force & bending-moment diagrams distributed...

shear force and bending