Unit II – Shear and Bending in Beams 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 Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Unit II – Shear and Bending in Beams Objective: • To know the mechanism of load transfer in beams and the induced stress resultants.
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Unit II – Shear and Bending in Beams References: • 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.
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Unit II – Shear and Bending in Beams Types of Loads:
2 kN
1. Point loads: 2. Uniformly distributed load (UDL): The loads are uniformly applied over the entire length of the beam. 2 kN/m 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: 2 kN/m 2 kN/m 2 kN/m 1 kN/m
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Unit II – Shear and Bending in Beams Types of supports: 1. Roller support:
2. Hinged support:
3. Fixed support:
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Unit II – Shear and Bending in Beams Statically determinate beams: • 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. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Unit II – Shear and Bending in Beams Statically indeterminate 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.
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Unit II – Shear and Bending in Beams Shear force: • 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:
Section line
+ ve
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
Section line
- ve
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Unit II – Shear and Bending in Beams Bending Moment:
• 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: Sagging moment: Moments which Hogging Moment: Moments bend the beam upwards and cause which bend the beam downwards and cause compression in the compression in the top fibre and bottom and tension in the top tension in the bottom fibre are fibre are taken as negative. taken as positive. +ve
-ve
Sagging moment
Hogging moment
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 1. Cantilever beam subjected to a point load at the free end: • (i) Shear force (S.F.) Calculations: Sign convention: Section line
Section line
A
X 𝑥 W B
+ ve
S.F. @ B= + W S.F. @ XX= +W S.F. @ A= + W
X ‘L’ m
- ve W
W +ve
W
S.F.D. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 1. Cantilever beam subjected to a point load at the free end: • (ii) Bending Moment(B.M.) Calculations: Sign convention: +ve Sagging moment
-ve
X 𝑥 W
A
B X ‘L’ m
Hogging moment
B.M. @ B= 0 B.M. @ XX= -W𝑥 B.M. @ A= - WL
-ve Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
WL
0 W𝑥
B.M.D.
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Shear force and Bending moment diagrams for Statically determinate beams 1. Cantilever beam subjected to a point load at the free end: (iii) S.F.D. & B.M.D. Diagrams: A
W B
‘L’ m W +ve
W
S.F.D. -ve
WL
B.M.D
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 2. Cantilever beam subjected to point loads as shown in Fig. Draw S.F.D. & B.M.D.: • (i) Shear force (S.F.) Calculations: Sign convention: Section line Section line
A
3 kN
1m + ve
- ve
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
5 kN
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
D
2 kN C
2m
1m
B
5 kN 2 kN +ve S.F.D.
2 kN 0
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Shear force and Bending moment diagrams for Statically determinate beams 2. Cantilever beam subjected to point loads as shown in Fig. Draw S.F.D. & B.M.D.: • (ii) Bending Moment(B.M.) Calculations: Sign convention: -ve
+ve Sagging moment
A
2 kN
3 kN D
C
2m
1m
1m
B
Hogging moment
0
B.M. @ B= 0 B.M. @ C= 0
0
-ve
B.M. @ D = −2 × 2 = −4 kNm B.M. @ A= − 2 × 3 − 3 × 1 = −9 𝑘𝑁𝑚 Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
4 kNm 9 kNm B.M.D
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Shear force and Bending moment diagrams for Statically determinate beams 2. Cantilever beam subjected to point loads as shown in Fig. Draw S.F.D. & B.M.D.: 2 kN 3 kN A
• (i) S.F.D. & B.M.D.
D
1m 5 kN
C
2m
1m
B
5 kN 2 kN 2 kN +ve 0 S.F.D. -ve
9 kNm
4 kNm B.M.D.
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.: • (i) Shear force (S.F.) Calculations: Sign S.F. @ A= + Section line Section line : + ve
- ve
S.F. @ B= 0 S.F. @ XX= +𝑤𝑥
𝑤/𝑚 X 𝑥 A
B
‘L’ m X
𝑤𝐿 +ve
𝑤𝑥
S.F.D.
S.F. @ A= +𝑤𝐿 Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.: • (i) Bending Moment(B.M.) Calculations: Sign convention: -ve +ve Sagging moment
W /m X 𝑥 A ‘L’ m
X
Hogging moment
B.M. @ B= 0 𝑥 𝑤𝑥 2 B.M. @ XX= --𝑤𝑥 × = 2 2 𝐿 𝑤𝐿2 B.M. @ A= −𝑤𝐿 × = − 2 2 Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
-ve 𝑤𝐿2 2
𝑤𝑥 2 2 B.M.D. 17
B
Shear force and Bending moment diagrams for Statically determinate beams 3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.: • (iii) S.F.D & B.M.D:
W /m X 𝑥 A 𝑤𝐿
‘L’ m
+ve S.F.D. -ve 𝑤𝐿2 2
B
X 𝑤𝑥
𝑤𝑥 2 2 B.M.D.
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.: A 1.5 m
4 kN/m B 1.5 m
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.: A 1.5 m 6 𝑘𝑁
4 kN/m B 1.5 m 6 𝑘𝑁
+ve S.F.D. -ve 13.5 𝑘𝑁𝑚
4.5 kNm B.M.D.
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 4. A simply supported beam of span ‘L’ carries a central concentrated load ‘W’. Draw S.F.D. & B.M.D.: W ‘L’ m
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 4. A simply supported beam of span ‘L’ carries a central concentrated load ‘W’. Draw S.F.D. & B.M.D.: W ‘L’ m W/2
W/2 W/2
+ve
-ve S.F.D. 𝑊𝐿/4
W/2
+ve
B.M.D. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 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.: W 𝑎
𝑏 ‘L’ m
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams Solution: (i) Calculation of reactions: 𝑀𝐴 = 0
A 𝑎
+ve
−𝑅𝐵 × 𝐿 + 𝑊 × 𝑎 = 0 𝑾𝒂 ∴ 𝑹𝑩 = 𝑳 𝐹𝑉 = 0
W C
𝑅𝐴
B
𝑏 ‘L’ m 𝑅𝐵
+ve
𝑅𝐴 + 𝑅𝐵 − 𝑊 = 0 𝑊𝑎 𝑅𝐴 + −W=0 𝐿
𝑾𝒃 ∴ 𝑹𝑨 = 𝑳
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams Solution (contd…): (ii) Shear force (S.F.) Calculations: Sign Convention: Section line Section line + ve S.F. at S.F. at
S.F. at S.F. at
- ve
W C
A
𝑊𝑏 𝐿 𝑊𝑏 𝐿
𝑎
B
𝑏 ‘L’ m
𝑊𝑎 𝐿
+ve -ve
𝑊𝑏 A= 𝐿
𝑊𝑏 C (between A and C)= 𝐿 𝑊𝑎 C(between C and B)= 𝐿 𝑊𝑎 B= 𝐿 Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
𝑊𝑎 𝐿
S.F.D.
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Shear force and Bending moment diagrams for Statically determinate beams Solution: A (ii) Bending Moment(B.M.) Calculations: Sign Convention: +ve Sagging
-ve
𝑊𝑏 𝐿
W C
𝑎
Hogging
𝑀𝐴 =0 𝑊𝑏 𝑀𝐶 = 𝐿
×𝑎 =
𝑊𝑎𝑏 𝐿
𝑏 ‘L’ m 𝑊𝑎𝑏 𝐿 +ve
B 𝑊𝑎 𝐿
B.M.D.
𝑀𝐵 = 0 Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 5. A simply supported beam of span ‘L’ carries two point loads as shown in Fig. Draw S.F.D. & B.M.D.: 3 kN
4 kN
1.5 𝑚
3.5 𝑚
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
1m
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Shear force and Bending moment diagrams for Statically determinate beams 4 kN
Solution: (i) Calculation of reactions: 𝑀𝐴 = 0
+ve
3 kN
A
B 1.5 𝑚
𝑅𝐴
3.5 𝑚
1m
𝑅𝐵
−(𝑅𝐵 × 6) + (3 × 5) + (4 × 1.5) = 0 ∴ 𝑹𝑩 = 𝟑. 𝟓 𝒌𝑵 𝐹𝑉 = 0
+ve
𝑅𝐴 + 𝑅𝐵 − 4 − 3 = 0 𝑅𝐴 + 3.5 − 4 − 3 = 0
∴ 𝑹𝑨 = 𝟑. 𝟓 𝒌𝑵
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams (ii) Shear force (S.F.) Calculations: A
Sign Convention: Section line Section line
3.5 kN
+ ve
4 kN
C
1.5 𝑚
D 3.5 𝑚
B
1m 3.5 kN
3.5 kN
- ve
3 kN
+
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
-
0.5 kN S.F.D.
3.5 kN
S.F. at D (between D and B)=-3.5 kN
S.F. at B=−3.5 𝑘𝑁 Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams
(iii) B.M.Calculations: Sign Convention: +ve Sagging
A -ve Hogging
4 kN
C
1.5 𝑚 3.5 kN
D 3.5 𝑚
5.5 kNm
3 kN B 1m 3.5 kN
3.5 kNm 𝑀𝐴 =0
+
𝑀𝐶 =3.5 × 1.5 = 5.5 𝑘𝑁𝑚
B.M.D.
𝑀𝐷 = 3.5 × 1 = 3.5 𝑘𝑁𝑚 𝑀𝐵 = 0 Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 5. A simply supported beam of span ‘L’ carries u.d.l throughout the . Draw S.F.D. & B.M.D.: 𝑤 𝑘𝑁/𝑚 ′𝐿′ 𝑚
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
31
Shear force and Bending moment diagrams for Statically determinate beams 5. A simply supported beam of span ‘L’ carries u.d.l throughout the . Draw S.F.D. & B.M.D.: 𝑤 𝑘𝑁/𝑚 X B A 𝑥 X ′𝐿′ 𝑚 S.F. @ XX=0 wL/2 wL/2 𝑤𝐿 − 𝑤𝑥 = 0 wL/2 2 + ∴ 𝑥 = 𝐿/2 M @ L/2
𝑤𝐿2 = 8
S.F.D. 2 𝑤𝐿 8
-
wL/2
+ B.M.D. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
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Shear force and Bending moment diagrams for Statically determinate beams 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. 15 kN/m 25 kN A
C 3m
1m
20 kN/m B
D 4m
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
E 2m
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Shear force and Bending moment diagrams for Statically determinate beams (i) Support Reactions: 𝑀𝐴 = 0
+ve
15 kN/m 25 kN A
C 3m
B
D
1m
4m
20 kN/m E 2m
𝑅𝐴
𝑅𝐵 1 2 − 𝑅𝐵 × 8 + 20 × 2 × 9 + 25 × 4 + × 3 × 15 × × 3 = 0 2 3 ∴ 𝑹𝑩 = 𝟔𝟑. 𝟏𝟐𝟓 𝒌𝑵 +ve 𝐹𝑉 = 0 1 𝑅𝐴 + 63.125 − × 3 × 15 − 25 − 20 = 0 2 ∴ 𝑹𝑨 = 𝟐𝟒. 𝟑𝟕𝟓 𝐤𝐍 Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
34
Shear force and Bending moment diagrams for Statically determinate beams 15 kN/m 25 kN
(ii) S.F. calculations:
X
𝑆. 𝐹. @ 𝐴 = +24.37 𝑘𝑁
A
C
B
D
3m X 1m 𝑆. 𝐹. @ 𝑋𝑋 = +24.37 24.37 kN 1 − × 𝑥 × 5𝑥 2 24.37 Parabola = 24.37 − 2.5 𝒙𝟐 1.88 S.F. @ C=
4m
20 kN/m E 2m
63.13 kN
1 2
24.37− × 3 × 15 =1.88 kN
S.F. between C&D=1.88 kN Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
35
Shear force and Bending moment diagrams for Statically determinate beams 15 kN/m 25 kN
(ii) S.F. calculations: S.F. between D&B= 1 2
A
C
24.37− × 3 × 15 −25 3m =−23.13 𝑘𝑁 𝑅𝐴 =24.37
S.F. @B (including reaction at B)= -23.13+63.13=+40 kN
24.37
+
B
D
1m
20 kN/m E
4m
2m 𝑅𝐵 = 63.13 40
Parabola
+
1.88
-
S.F. @ E=40 − 20 × 2 = 0 23.13 S.F.D. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
23.13 36
Shear force and Bending moment diagrams for Statically determinate beams 15 kN/m 25 kN
(iii) B.M. calculations: 𝑀𝐴 = 0
A
𝑀𝐶 = + 24.37 × 3 − 1 1 × 3 × 15 × × 3 2 3 =50.68 kNm
𝑀𝐷 =24.37× 4 − 1 2
C 3m
B
D
1m
4m
𝑅𝐴 =24.37
20 kN/m E 2m
𝑅𝐵 = 63.13 50.68 52.5 kNm
1
× 3 × 15 × 1 + 3 3 =52.5 kNm
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
37
Shear force and Bending moment diagrams for Statically determinate beams 15 kN/m 25 kN
(iii) B.M. calculations: 𝑀𝐵 = −(20 × 2 × 1) = −40 kNm 𝑀𝐸 =0
A
D
C 3m
X
1m
B
20 kN/m
4m
𝑅𝐴 =24.37
E 2m
𝑅𝐵 = 63.13
𝑥 50.68 52.5 kNm 𝑀𝑋𝑋 = 0 Cubic X − 20 × 2 × 𝑥 − 1 + parabola Point of 63.13 × (𝑥 − 2) =0 contraflexure + ∴ 𝒙 = 𝟑. 𝟕𝟑 𝒎
B.M.D.
Parabola
40 kNm Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
38
Shear force and Bending moment diagrams for Statically determinate beams 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.
A
2 kN/m
3 kNm C D
1.5 m 0.5 m
2 kN B
E 1m
1m
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
39
Shear force and Bending moment diagrams for Statically determinate beams S.F.Diagram: A S.F. @ XX=0
2 kN/m
2.18 kN
2.18 − 2𝑥 = 0 ∴ 𝑥 = 1.09 𝑚
X
3 kNm C D
1.5 m 0.5 m 𝑥
2.18
2 kN E 1m
1m
B 2.82 kN
X
+ 0.82
0.82 -
S.F.D 2.82 . Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
2.82 40
Shear force and Bending moment diagrams for Statically determinate beams B.M.Diagram: A
2 kN/m
2.18 kN M @ 1.09 m from A =1.18 kNm
X
2 kN
3 kNm C D
1.5 m 0.5 m
E 1m
1m
B 2.82 kN
1.09 𝑚
X 1.18 1.05
3.62 kNm 2.82 kNm
0.62
+
B.M.D. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
41
Shear force and Bending moment diagrams for Statically determinate beams 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.
5 kN A
20 kN 5 kN/m B C
1m
2m
D
2 kNm
3m
Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE
E
2m
42
