unit ii shear and bending in beams of... · syllabus: • beams and bending- types of loads,...

Syllabus: • Beams and Bending- Types of loads, supports - Shear Force and

Bending Moment Diagrams for statically determinate beam with concentrated load, UDL, uniformly varying load.

• Theory of Simple Bending – Analysis of Beams for Stresses – Stress Distribution at a cross Section due to bending moment and shear force for Cantilever, simply supported and overhanging beams with different loading conditions

• Flitched Beams

Unit II – Shear and Bending in Beams

Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE

1

Objective:

• To know the mechanism of load transfer in beams and the induced stress resultants.



2

• 1. Rajput.R.K. “Strength of Materials”, S.Chand and Co, New Delhi, 2007.

• Bhavikatti. S., "Solid Mechanics", Vikas publishing house Pvt. Ltd, New Delhi, 2010.

• Junnarkar.S.B. andShah.H.J, “Mechanics of Structures”, Vol I, Charotar Publishing House, New Delhi,1997.



3

References:

Types of Loads:

1. Point loads:

2. Uniformly distributed load (UDL):

The loads are uniformly applied over the entire length of the beam.

It can be shown as follows:

3. Uniformly varying load (UVL):

Triangular or trapezoidal loads fall under this category. The variation in intensities of such loads is constant.

It can be shown as follows:



4

2 kN

2 kN/m

2 kN/m 2 kN/m 2 kN/m 1 kN/m

Types of supports:

1. Roller support:

2. Hinged support:

3. Fixed support:



5

• 1. cantilever beams:

A cantilever beam which is fixed at one end and free at the other end.

• 2. Simply supported beams:

A simply supported beam rests freely on hinged support at one end and roller support at the other end.

• 3. Overhanging beams:

If a beam extends beyond its supports it is called an overhanging beam.

Over hanging portion could be either any one of the sides or both the sides.



6

Statically determinate beams:

• 1. Propped cantilever beam:

A cantilever beam with propped support at the free end.

• 2. Fixed beam:

A beam with both supports fixed.

• 3. Continuous beam:

A beam with more than two supports.



7

Statically indeterminate beams:

• Shear force at a section of a loaded beam may be defined as the algebraic sum of all vertical forces acting on any one side of the section.

• Sign Convention:



8

Shear force:

+ ve

Section line

- ve

Section line

• Bending moment at a section of a loaded beam may be defined as the algebraic sum of all moments of forces acting on any one side of the section.

• Sign Convention:



9

Bending Moment:

Sagging moment

+ve

Sagging moment: Moments which bend the beam upwards and cause compression in the top fibre and tension in the bottom fibre are taken as positive.

Hogging Moment: Moments which bend the beam downwards and cause compression in the bottom and tension in the top fibre are taken as negative.

Hogging moment

-ve

1. Cantilever beam subjected to a point load at the free end:

• (i) Shear force (S.F.) Calculations:

Sign convention:

Shear force and Bending moment diagrams for Statically determinate beams


10

W

‘L’ m

A B

+ ve

Section line

- ve

Section line

S.F. @ B= + W S.F. @ XX= +W S.F. @ A= + W

𝑥 X

X

W W W

S.F.D.

+ve


• (ii) Bending Moment(B.M.) Calculations:

Sign convention:



11

W

‘L’ m

A B

B.M. @ B= 0 B.M. @ XX= -W𝑥 B.M. @ A= - WL

𝑥 X

X

B.M.D.

Sagging moment

+ve

Hogging moment

-ve

0

W𝑥

WL

-ve




12

W

‘L’ m

A B

W W

S.F.D.

+ve

WL

-ve

B.M.D

(iii) S.F.D. & B.M.D. Diagrams:

2. Cantilever beam subjected to point loads as shown in Fig. Draw S.F.D. & B.M.D.:


Sign convention:



13

+ ve

Section line

- ve

Section line

S.F. @ B= 0 S.F. @ C= +2 kN S.F. @ D (without Pt. Load at D)=+ 2 kN S.F. @ D (with Pt. Load at D)=+ 5 kN S.F. @ A= + 5 kN

0

2 kN

5 kN

S.F.D.

+ve

2 kN 3 kN A

B 1 m 2 m 1 m

C D

2 kN

5 kN


• (ii) Bending Moment(B.M.) Calculations:

Sign convention:



14

B.M. @ B= 0 B.M. @ C= 0

0

2 kN 3 kN

A B

1 m 2 m 1 m

C D

Sagging moment

+ve

Hogging moment

-ve

0

4 kNm B.M. @ D = −2 × 2 = −4 kNm

B.M. @ A= − 2 × 3 − 3 × 1 = −9 𝑘𝑁𝑚 9 kNm

-ve

B.M.D


• (i) S.F.D. & B.M.D.



15

2 kN 3 kN A

B 1 m 2 m 1 m

C D

0

2 kN

5 kN

S.F.D.

+ve 2 kN

5 kN

4 kNm

9 kNm

-ve

B.M.D.

3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.:


Sign S.F. @ A= +

:



16

+ ve

Section line

- ve

Section line

S.F. @ B= 0 S.F. @ XX= +𝑤𝑥

𝑤/𝑚

‘L’ m A B

𝑥 X

X

𝑤𝑥

S.F. @ A= +𝑤𝐿

𝑤𝐿

+ve

S.F.D.


• (i) Bending Moment(B.M.) Calculations:

Sign convention:



17

B.M. @ B= 0

W /m

‘L’ m

Sagging moment

+ve

Hogging moment

-ve

A B 𝑥 X

X

B.M. @ A= −𝑤𝐿 ×𝐿

2= −𝑤𝐿2

2

𝑤𝑥2

2 B.M. @ XX= --𝑤𝑥 ×

𝑥

2=𝑤𝑥2

2

𝑤𝐿2

2 B.M.D.

-ve


• (iii) S.F.D & B.M.D:



18

𝑥 X

X

A B W /m

‘L’ m

𝑤𝑥2

2

𝑤𝐿2

2 B.M.D.

-ve

𝑤𝑥

𝑤𝐿

+ve

S.F.D.




19

A B 4 kN/m

1.5 m 1.5 m




20

A B 4 kN/m

1.5 m

13.5 𝑘𝑁𝑚 B.M.D.

-ve

6 𝑘𝑁

+ve

S.F.D.

1.5 m

4.5 kNm

6 𝑘𝑁

4. A simply supported beam of span ‘L’ carries a central concentrated load ‘W’. Draw S.F.D. & B.M.D.:



21

W

‘L’ m

4. A simply supported beam of span ‘L’ carries a central concentrated load ‘W’. Draw S.F.D. & B.M.D.:



22

W

‘L’ m

W/2 W/2

+ve

W/2

W/2

-ve S.F.D.

𝑊𝐿/4

B.M.D.

+ve

4. A simply supported beam of span ‘L’ carries a eccentric concentrated load ‘W’ as shown in Fig. Draw S.F.D. & B.M.D.:



23

W

‘L’ m 𝑎 𝑏

Solution:

(i) Calculation of reactions:



24

W

‘L’ m 𝑎 𝑏

A B C

−𝑅𝐵 × 𝐿 +𝑊 × 𝑎 = 0

𝑀𝐴 = 0 +ve

∴ 𝑹𝑩 =𝑾𝒂

𝑳

𝐹𝑉 = 0 +ve

𝑅𝐴 𝑅𝐵

𝑅𝐴 + 𝑅𝐵 −𝑊 = 0

𝑅𝐴 +𝑊𝑎

𝐿−W = 0 ∴ 𝑹𝑨 =

𝑾𝒃

𝑳

Solution (contd…):

(ii) Shear force (S.F.) Calculations:



25

W

‘L’ m 𝑎 𝑏

A B C

𝑊𝑏

𝐿

𝑊𝑎

𝐿

+ve

-ve

𝑊𝑏

𝐿

𝑊𝑎

𝐿

S.F.D.

S.F. at A=𝑊𝑏

𝐿

S.F. at C (between A and C)=𝑊𝑏

𝐿

S.F. at C(between C and B)=𝑊𝑎

𝐿

S.F. at B=𝑊𝑎

𝐿

Sign Convention:

+ ve

Section line

- ve

Section line

Solution:

(ii) Bending Moment(B.M.) Calculations:



26

W

‘L’ m 𝑎 𝑏

A B C

𝑊𝑏

𝐿

𝑊𝑎

𝐿

B.M.D.

𝑀𝐴=0

𝑀𝐶=𝑊𝑏

𝐿× 𝑎 =

𝑊𝑎𝑏

𝐿

𝑀𝐵 = 0

𝑊𝑎𝑏

𝐿

+ve

Sagging

+ve

Sign Convention:

Hogging

-ve

5. A simply supported beam of span ‘L’ carries two point loads as shown in Fig. Draw S.F.D. & B.M.D.:



27

3 kN 4 kN

1.5 𝑚 3.5 𝑚 1 m

Solution:



28

3 kN 4 kN

1.5 𝑚 3.5 𝑚 1 m 𝑅𝐴 𝑅𝐵

(i) Calculation of reactions:

𝑀𝐴 = 0 +ve

−(𝑅𝐵 × 6) + (3 × 5) + (4 × 1.5) = 0

A B

∴ 𝑹𝑩 = 𝟑. 𝟓 𝒌𝑵

𝐹𝑉 = 0 +ve

𝑅𝐴 + 𝑅𝐵 − 4 − 3 = 0

𝑅𝐴 + 3.5 − 4 − 3 = 0 ∴ 𝑹𝑨 = 𝟑. 𝟓 𝒌𝑵



29

3 kN 4 kN

1.5 𝑚 3.5 𝑚 1 m 3.5 kN

3.5 kN

+ -

S.F.D.

(ii) Shear force (S.F.) Calculations:

S.F. at A=3.5 𝑘𝑁

S.F. at C (between A and C)=3.5 𝑘𝑁 S.F. at C(between C and D)=−0.5 𝑘𝑁 S.F. at D (between D and B)=-3.5 kN

A C D B

3.5 kN

0.5 kN

3.5 kN

Sign Convention:

+ ve

Section line

- ve

Section line

S.F. at B=−3.5 𝑘𝑁



30

3 kN 4 kN

1.5 𝑚 3.5 𝑚 1 m

3.5 kN 3.5 kN

B.M.D.

(iii) B.M.Calculations:

A C D B

Sagging

+ve

Sign Convention:

Hogging

-ve

𝑀𝐴=0 𝑀𝐶=3.5 × 1.5 = 5.5 𝑘𝑁𝑚 𝑀𝐷 = 3.5 × 1 = 3.5 𝑘𝑁𝑚 𝑀𝐵 = 0

5.5 kNm

3.5 kNm

+

5. A simply supported beam of span ‘L’ carries u.d.l throughout the . Draw S.F.D. & B.M.D.:



31

′𝐿′ 𝑚

𝑤 𝑘𝑁/𝑚

5. A simply supported beam of span ‘L’ carries u.d.l throughout the . Draw S.F.D. & B.M.D.:



32

′𝐿′ 𝑚

𝑤 𝑘𝑁/𝑚

wL/2 wL/2

wL/2

wL/2

+

- S.F.D.

A B

S.F. @ XX=0

X

X 𝑥

𝑤𝐿

2− 𝑤𝑥 = 0

∴ 𝑥 = 𝐿/2

M @ L/2 =𝑤𝐿2

8

𝑤𝐿2

8

+

B.M.D.

Over hanging beams:

Problem:

An over hanging beam of length 10 m is loaded as shown in Fig. Draw the S.F.D. and B.M.D. Mark the values at salient points.



33

A

3 m

15 kN/m 25 kN

1 m 4 m 2 m

C D B E

20 kN/m

(i) Support Reactions:



34

𝑀𝐴 = 0 +ve

𝑅𝐴 𝑅𝐵

− 𝑅𝐵 × 8 + 20 × 2 × 9 + 25 × 4 +1

2× 3 × 15 ×

2

3× 3 = 0

A

3 m

15 kN/m 25 kN

1 m 4 m 2 m

C D B E 20 kN/m

∴ 𝑹𝑩 = 𝟔𝟑. 𝟏𝟐𝟓 𝒌𝑵 𝐹𝑉 = 0 +ve

𝑅𝐴 + 63.125 −1

2× 3 × 15 − 25 − 20 = 0

∴ 𝑹𝑨 = 𝟐𝟒. 𝟑𝟕𝟓 𝐤𝐍

(ii) S.F. calculations:



35

A

3 m

15 kN/m 25 kN

1 m 4 m 2 m

C D B E 20 kN/m

𝑆. 𝐹.@ 𝐴 = +24.37 𝑘𝑁

𝑆. 𝐹.@ 𝑋𝑋= +24.37

−1

2× 𝑥 × 5𝑥

= 24.37 − 2.5 𝒙𝟐

S.F. @ C=

24.37−1

2× 3 × 15

=1.88 kN

24.37

1.88

S.F. between C&D=1.88 kN

Parabola

24.37 kN 63.13 kN

X

X

(ii) S.F. calculations:



36

𝑅𝐴=24.37 𝑅𝐵 = 63.13

A

3 m

15 kN/m 25 kN

1 m 4 m 2 m

C D B E 20 kN/m

S.F. between D&B=

24.37−1

2× 3 × 15 −25

=−23.13 𝑘𝑁

24.37

1.88

Parabola S.F. @B (including reaction at B)= -23.13+63.13=+40 kN

S.F. @ E=40 − 20 × 2 = 0

+ +

-

S.F.D.

40

23.13 23.13

(iii) B.M. calculations:



37

𝑅𝐴=24.37 𝑅𝐵 = 63.13

A

3 m

15 kN/m 25 kN

1 m 4 m 2 m

C D B E 20 kN/m

𝑀𝐴 = 0

𝑀𝐶 =+ 24.37 × 3 −1

2× 3 × 15 ×

1

3× 3

=50.68 kNm

𝑀𝐷=24.37× 4 −1

2× 3 × 15 × 1 +

1

33

=52.5 kNm

50.68 52.5 kNm

(iii) B.M. calculations:



38

𝑅𝐴=24.37 𝑅𝐵 = 63.13

A

3 m

15 kN/m 25 kN

1 m 4 m 2 m

C D B E 20 kN/m

𝑀𝐵 = −(20 × 2 × 1) = −40 kNm

𝑀𝐸=0

+

- B.M.D.

X

X 𝑀𝑋𝑋 = 0

− 20 × 2 × 𝑥 − 1 + 63.13 × (𝑥 − 2) =0

𝑥

∴ 𝒙 = 𝟑. 𝟕𝟑 𝒎

Point of contraflexure

40 kNm

52.5 kNm 50.68 Cubic parabola

Parabola

Simply supported beam:

Problem:

A simply supported beam is loaded as shown in Fig. Draw the S.F.D. and B.M.D. Mark the values at salient points.



39

A

1.5 m

2 kN/m 2 kN

0.5 m 1m 1 m

D B E 3 kNm

C



40

A

1.5 m

2 kN/m 2 kN

0.5 m 1m 1 m

D B E 3 kNm

C

2.18 kN 2.82 kN

2.18

0.82

2.82 2.82

0.82

+

-

S.F.D.

S.F.Diagram:

S.F. @ XX=0

X

X

𝑥 2.18 − 2𝑥 = 0

∴ 𝑥 = 1.09 𝑚



41

A

1.5 m

2 kN/m 2 kN

0.5 m 1m 1 m

D B E 3 kNm

C

2.18 kN 2.82 kN

1.18

3.62 kNm 2.82 kNm

0.62

B.M.D.

B.M.Diagram:

1.05

M @ 1.09 m from A =1.18 kNm

X

X

1.09 𝑚

+

Over hanging beam:

Problem:

Draw the shear force and bending moment diagram for the overhanging beam shown in Fig. Indicate the salient values on them.



42

B

2 m

5 kN/m

20 kN

3m 2 m

D E 2 kNm

C

5 kN

A

1 m

unit ii shear and bending in beams of... · syllabus: • beams and bending- types of loads,...

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