9111-SRM MCET Mechanical Engineering

Unit- II

TRANSVERSE LOADING ON BEAMS AND STRESSES IN BEAM

Part – A (Q & A)

1. Define point of contraflexure (Nov/Dec’23, Apr/May’18, & Apr/May’19)

It is the point along a beam where the bending moment changes sign, meaning it
transitions from positive to negative or vice versa. At this point, the bending moment is
zero.

2. Define flitched beams. (Nov/Dec’23, (Nov/Dec’20, & Apr/May’21)

A flitched beam is a type of composite structural element that combines the properties of
wood and steel to enhance strength and stiffness. It typically consists of a vertical steel
plate (known as a flitch plate) sandwiched between two wooden beams. The layers are
secured together using bolts, allowing the beam to effectively carry loads while
minimizing deflection.

3. List any four assumptions made in the theory of simple bending. (Apr/May’23,
Apr/May’19)
1. Material Homogeneity and Isotropy; 2. Plane Sections Remain Plane; 3. Linear
Elastic Behaviour; 4. No shear deformation.

4. Classify beams (Apr/May’23)

Steel beams, Concrete Beams, Timber Beams, Composite Beams, Aluminium Beams.

5. Give the relationship between shear force and bending moment (Apr/May’24,
Nov/Dec’19)
Shear Force (V): The internal force that acts parallel to the cross section of a beam. It
results from external loads applied to the beam.
Bending Moment (M): The internal moment that causes the beam tp bend. It is the result
of external forces acting on the beam and is calculated about a specific point along the
beam.

6. What is the application of perpendicular axis theorem? (Apr/May’24)

Planer objects, Calculating Moment of Inertia in Engineering, Inertia Properties of
objects, Simplification in Comple Systems, Analysis of Thin Rings, Disk, and Plates.

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7. Describe the beam with its type (Nov/Dec’18, Nov/Dec’20)

A beam is a structural element designed to support loads and resist bending. It have
several types namely cantilever beam, Simply supported beam, overhanging beam, Fixed
beam, Continuous beam, Propped cantilever beam.

8. What is neutral axis of a beam section? How do you locate it when a beam is under
simple bending? (Nov/Dec’19)
The neutral axis of a beam section is a hypothetical line within the beam where there is
no longitudinal strain or no deformation due to bending. In simpler terms, it is the axis
about which the beam's fibers experience compressive and tensile strains, but at this
point, the strain is zero. For simple bending in beams, the neutral axis is located at the
centroid of the beam's cross-section.

9. Draw the shear force diagram and bending moment diagram for the cantilever
beam carries uniformly varying load of zero intensity at the free end and w kN/m at
the fixed end. (Apr/May’19)

Figure 2.1: Shear Force and Bending Diagram

10. What do you mean by shear stresses in beam? (Apr/May’18, Apr/May’19)

Shear stress in a beam refers to the internal stress that develops in response to a shear
force acting on the beam. Shear force is the force that acts parallel to the cross-sectional
surface of the beam, causing the material of the beam to shear or slide over itself.

11. Under what conditions two concentric shaft can act as composite shaft? State the
conditions (Nov/Dec’20)
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Material Compatibility, Load sharing, Bondin and connection, geometric alignment.
12. Define torsional rigidity (Nov/Dec’20)
Torsional Rigidity is defined as the measure of a structural member's resistance to
twisting under an applied torque. It quantifies how much torque is required to produce a
unit angle of twist per unit length of the member. Torsional rigidity is crucial in
engineering applications where components are subjected to torsional loads.

13. What do you mean by point of inflexion? (Apr/May’18)

A point of inflection (or inflection point) is defined as a point on a curve where the
curvature changes sign. This means that the curve transitions from being concave upward
(where it opens upwards) to concave downward (where it opens downwards), or vice
versa.

14. Mention and sketch any two supports and beams. (Nov/Dec’21)
A simply supported beam is supported at both ends, typically with one end on a pinned
support and the other on a roller support. This allows for rotation at the supports but does
not resist moments.

Figure 2.2: Simply supported beam

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15. A cantilever beam is fixed at one end and free at the other. The fixed end can resist both
vertical loads and moments, while the free end can rotate freely.

Figure 2.3: Cantilever beam

16. Write the equation of simple bending. (Apr/May’18)

σ = Bending stress
y = Distance from the neutral axis
M = Bending moment at the section of the beam
I = Moment of inertia of the beam’s cross section about the neutral axis.
E = Modulus of Elasticity
R = Radius of curvature of the bent beam.

17. Sketch the bending stress distribution and shear stress distribution for the beam of
rectangular cross section (Nov/Dec’21)

Figure 2.4: Bending stress distribution diagram

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Figure 2.5: Shear stress distribution diagram

18. What is meant by transverse loading on beam? (Apr/May’21)

Transverse loading on a beam refers to loads that are applied perpendicular (or
transverse) to the longitudinal axis of the beam. This type of loading causes the beam to
bend and deform, resulting in internal stresses and strains.

19. Explain the term neutral axis (Nov/Dec’22)

It refers to an axis in the cross-section of a beam (or shaft) along which there are no
longitudinal stresses or strains when the beam is subjected to bending moments.

20. Calculate the moment of resistance of a beam subjected to a bending stress of 6

N/mm2 and section modulus is 2000 mm3 (Apr/May’23)
Moment of resistance of a beam can be calculated using the formula:
M=σxZ
Bending stress σ = 6 N mm2
M Section Modulus Z = 2000 mm3
M = 6 x 2000 = 12000 N mm = 12 N.m

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Part – B (Q & A)

1. Draw the S.F and B.M. diagrams for simply supported beam loaded as shown in
Figure. 2.7. (Nov/Dec’23)

Figure 2.7: Simply supported beam

Step 1: Determine the reactions at the supports
a) Summation of vertical forces = 0
RA + RB – 2 kN – 4 kN – 2 kN = 0
b) Summation of Moments about Point A = 0
(2 kN x 1 m) + (4 kN x 2 m) + (2 kN x 3 m) – (RB x 4 m) = 0
Solving these equations simultaneously, we get
RA = 2 kN
RB = 6 kN
Step 2: Draw the Shear Force (SF) Diagram
a) Start at Point A:
The SF at point A is equal to the reaction RA, which is 2 kN (upward)
b) Move towards Point C:
A downward force of 2 kN acts at point C. So, the SF drops by 2 kN to 0 kN.
c) Move towards Point D:
A downward force of 4 kN acts at point D. The SF further decreases by 4 kN,
reaching -4 kN.
d) Move towards Point E:
A downward force of 2 kN acts at point E. The SF decreases by another 2 kN,
reaching -6 kN.
e) Move towards Point B:
The SF at point B is equal to the reaction RB, which is 6 kN (upward). This brings the SF
back to 0 kN.
Step 3: Draw the Bending Moment (BM) Diagram
The BM diagram represents the variation of bending moment along the length of the
beam. To construct it, we'll follow these steps:
a) Start at Point A:
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The BM at point A is zero since it's a simple support.
b) Move towards Point C:
The SF between A and C is constant at 2 kN. This means the BM increases linearly
with a slope of 2 kN/m.
c) Move towards Point D:
The SF between C and D is constant at -2 kN. This means the BM decreases linearly
with a slope of -2 kN/m.
d)Move towards Point E:
The SF between D and E is constant at -6 kN. This means the BM decreases linearly
with a slope of -6 kN/m.
e) Move towards Point B:
The SF between E and B is constant at 6 kN. This means the BM increases linearly
with a slope of 6 kN/m, reaching zero at point B.
Final Diagrams
Here are the SF and BM diagrams for the given beam:

Figure 2.8: Shear Force and Bending Moment sample Diagram for simply
supported beam

2. A steel plate of width 120 mm and of thickness 20 mm is bent into a circular arc of
radius 10 m. Determine the maximum stress induced and the bending moment

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which will produce the maximum stress. Take E = 2 x 10 5 N/mm2. (Nov/Dec’23,
Apr/May’19)
Solution:
Step 1: Calculate the Radius of Curvature (R):
Given, R = 10 m = 10,000 mm
Step 2: Calculate the Bending Stress (σ):
The bending stress in a curved beam is given by:
σ=E*ε
where:
σ is the bending stress
E is the modulus of elasticity
ε is the strain
For a curved beam, the strain (ε) can be related to the radius of curvature (R) and the
thickness (t) as follows:
ε = t / 2R
Substituting this into the bending stress equation:
σ = E * (t / 2R)
Given:
E = 2 x 10^5 N/mm²
t = 20 mm
R = 10,000 mm
Calculating the bending stress:
σ = (2 x 10^5 N/mm²) * (20 mm / (2 * 10,000 mm)) = 200 N/mm²
Step 3: Calculate the Bending Moment (M):
The bending moment (M) can be related to the bending stress (σ) and the section
modulus (Z) as follows:
σ=M/Z
For a rectangular section, the section modulus (Z) is given by:
Z = (b * t^2) / 6
where:
b is the width of the section
t is the thickness of the section
Given:
b = 120 mm
t = 20 mm
Calculating the section modulus:
Z = (120 mm * (20 mm)^2) / 6 = 8000 mm³
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Now, calculating the bending moment:
M = σ * Z = 200 N/mm² * 8000 mm³ = 1.6 x 10^6 Nmm = 1600 Nm
Conclusion
Therefore, the maximum bending stress induced in the steel plate is 200 N/mm², and the
corresponding bending moment is 1600 Nm.

3. Analyse the beam shown in Figure 2.9. Draw the shear force and bending moment
diagrams. (Apr/May’23)

Figure 2.9: Simply supported beam

Step 1: Determine the Reactions at the Supports
Since the beam is simply supported, we'll have vertical reactions at points A and B.
Let's denote these as RA and RB, respectively.
To find these reactions, we can use the equilibrium equations:
a. Summation of Vertical Forces = 0: RA + RB - 2 kN - 1 kN - 1 kN = 0
b. Summation of Moments about Point A = 0: (2 kN * 1 m) + (1 kN * 2.5 m) + (1
kN * 5 m) - (RB * 6 m) = 0
Solving these equations simultaneously, we get:
 RA = 1 kN
 RB = 3 kN
Step 2: Draw the Shear Force (SF) Diagram
The SF diagram represents the variation of shear force along the length of the beam. To
construct it, we'll follow these steps:
a) Start at Point A:
The SF at point A is equal to the reaction RA, which is 1 kN (upward).
b) Move towards Point C:
A downward force of 2 kN acts at point C. So, the SF drops by 2 kN to -1 kN.
c) Move towards Point D:

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A downward force of 1 kN acts at point D. The SF further decreases by 1 kN, reaching -
2 kN.
d) Move towards Point E:
A downward force of 1 kN acts at point E. The SF decreases by another 1 kN, reaching
-3 kN.
e) Move towards Point B:
The SF at point B is equal to the reaction RB, which is 3 kN (upward). This brings the
SF back to 0 kN.
Step 3: Draw the Bending Moment (BM) Diagram
The BM diagram represents the variation of bending moment along the length of the
beam. To construct it, we'll follow these steps:
a) Start at Point A:
The BM at point A is zero since it's a simple support.
b) Move towards Point C:
The SF between A and C is constant at 1 kN. This means the BM increases linearly with
a slope of 1 kN/m.
c) Move towards Point D:
The SF between C and D is constant at -1 kN. This means the BM decreases linearly
with a slope of -1 kN/m.
d) Move towards Point E:
The SF between D and E is constant at -2 kN. This means the BM decreases linearly
with a slope of -2 kN/m.
e) Move towards Point B:
The SF between E and B is constant at 3 kN. This means the BM increases linearly with
a slope of 3 kN/m, reaching zero at point B.
Final Diagrams

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Figure 2.10: Shear Force and Bending Moment sample Diagram for simply
supported beam

4. A T-section beam with 100mm 15 mm flange and 150 mm x 15 mm web is subjected

to a shear force of 10 N at a section. Draw the variation of shear stress across the
depth of the beam and obtain the value of maximum shear stress at the section.
(Apr/May’23)
Solution:
Flange Width (b_f): 100 mm
Flange Thickness (t_f): 15 mm
Web Height (h_w): 150 mm
Web Thickness (t_w): 15 mm
Total Shear Force (V): 10 N
Step 1: Calculate the Cross-Sectional Properties
a. Total Height of the T-section:
h = tf + hw = 15 mm + 150 mm = 165 mm
b. Area of the Flange (A_f):
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Af = bf x tf = 100 mm x 15 mm = 1500 mm2
c. Area of the Web (A_w):
Aw = tw x hw = 15 mm x 150 mm = 2250 mm2
d. Total Cross-Sectional Area (A):
A = Af + Aw = 1500 + 2250 = 3750
Step 2: Calculate the centroid Area (A):
The position of the centroid from the bottom of the T-section can be calculated using:
Yc = Σ (AiYi) / Σ (Ai)
Where: Yf is the distance from the bottom to the centroid of the flange,
Yw is the distance from the bottom to the centroid of the web.
a) Centroid of Flange:
Distance from Bottom:
Yc = tf + hw – tf / 2 = 15 + 150
b) Centroid of Web:
Distance from Bottom:
Yw = hw / 2 = 150 / 2 = 75 mm
c) Calculating Centroid:
Total area contributions: Yc = [(AfYf) + AwYw)] / A
Step 3: Calculate Moment of Inertia (I)
I = If + Afd2f + Iw + Awd2w
a) Moment of Inertia for Flange
If = bf t3f / 12 = 100 (15)3 / 12 = 100 (3375) / 12
Distance from neutral axis to flange centroid:
df = | yc – yf | = | 108 – 157.5 | = 49.5 mm
Contribution to moment of inertia due to flange:
Af d2f = 1500 (49.5)2 = 1500 x 2450.25
b) Moment of Inertia for Web
Iw = tw h3w / 12 = [1500 (150)3] / 12 = [15 (3375000)] / 12
Distance from neutral axis to web centroid:
dw= | yc – yw | = | 108 – 75 | = 33 mm
Contribution to moment of inertia due to web:
Aw d2w = 2250 (33)2 = 2441 mm.
c) Total Moment of Inertia
Combining all contributions
I = If + Afd2f + Iw + Awd2w = 28125 + 367535 + 42187500 + 2441250 = 4621000 mm4
Step 4: Calculate Shear Stress Distribution
The shear stress at a distance y from the neutral axis is given by:
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Ty = VQ/ Ib
V=10 N=10 N /mm2
Q=Aabove ⋅ ybar
Calculate Q for different heights: For points above and below neutral axis:
For y = 108 mm (Neutral Axis):
Area above NA is area of flange only.
Aabov e= Af =1500 mm2
Centroid distance from NA is
Ybar = Yc – Yf = 108 – 157.5 = - 49.5 mm
Then,
Q = Aabov e .Ybar = 1500 (-49.5) = -74250 mm3
Now substituting into shear stress formula:
Ty = 10 x 74250 / 46281000 x 100 = 0.016 N/ mm2
Maximum Shear Stress Calculation
The maximum shear stress occurs at the neutral axis and can be calculated as follows:
At Neutral Axis (y = 108 mm): Substituting into shear stress formula:
For maximum shear stress at neutral axis:
Using maximum shear stress formula:
At Neutral Axis,
Tmax = VQ/ Ib
Substituting values,
Tmax = 10 x 74250 / ( 46281000 / 120) = 0.016 N /mm2
Step 5: Shear Stress Distribution Diagram

Figure 2.11: Shear Stress Distribution Diagram

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5. A 5 m simply supported beam is having a T-shaped cross section with the following
dimensions: flange width 125 mm, flange thickness 25 mm, depth of web 175 mm
and thickness of the web 25 mm. The beam caries a uniformly distributed load of
intensity 2.5 kN/m over the entire span. Sketch the shear stress distribution.
(Apr/May’24)
Given data:
Flange Width (b_f) : 125 mm
Flange Thickness (t_f): 25 mm
Depth of Web (d_w): 175 mm
Thickness of web (t_w): 25 mm
Area Calculation
A = Aflange + Aweb
Aflange = bf x tf = 125 mm x 25 mm = 3125 mm2
Aweb = tw x dw = 25 mm x (175 - 25) mm = 25 mm x 150 mm = 3750
mm2
Thus, A = 3125 + 3750 = 6875 mm2
Moment of Inertia Calculation
yNA = [Σ (Aiyi)] / A
Shear force and shear stress calculation
W = w x L = 2.5 kN / m 5 m = 12.5 kN = 12500 N
V (x) = W/ (2- wx)
Where x is the distance from one support.
The average shear stress Tavg can be calculated as
Tavg = V/ A
Shear Stress Distribution
For T-shaped sections, shear stress distribution is not uniform and typically
follows a parabolic shape across the depth. The maximum shear stress occurs at
the neutral axis, while it approaches zero at the top and bottom flanges.
a. Maximum shear stress occurs at the neutral axis.
b. Shear stress decreases linearly towards both flanges.

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Figure 2.12: T Shaped cross section of simply supported beam

Figure 2.13: Shear stress distribution diagram

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6. Draw the shear force and B.M diagrams for a simply supported beam of length 8 m
and carrying a uniformly distributed load of 10 kN/m for a distance of 4m as shown
in figure 2.14. (Apr/May’19)

Figure 2.14: Simply supported beam

Step 1: Calculate Support Reactions

a. Total Load: The total load W due to the UDL can be calculated as:
W = w x L = 10 N m 4 m = 40 kN
b. Support Reactions
For a simply supported beam, the reactions at supports A and B can be found
using equilibrium equations. Let R A R A and R B R B be the reactions at
supports A and B respectively.
Taking moments about point A:
RB x 8 = W x (4+2) = 40 x 6
RB = 240 /8 = 330 kN
Using vertical force equilibrium:
R A + RB = W
RA + 30 = 40
RA = 10 kN

Step 2: Shear Force Calculation

The shear force V (x) V(x) at any section x x can be calculated as follows: From
A to C (0 to 4 m):
At x = 0: V(0) = RA = 10kN

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At x = 4:
Total load up to this point is 10 kN m x 4 m = 40 kN
Thus, V (4) = Ra – W = 10 -40 = - 30 kN
From C to B (4 to 8 m):
At x = 4: V (4) = -30 N
At x = 8:
No additional loads, so V (8) = V (4) + RB = - 30 + 30 = 0 kN
Step 3: Bending Moment calculation
The bending moment M(x) at any section can be calculated using:
From A to C:
At any point x < 4m:
M(x) = RA x – w (x2/2)
At x = 0m:
M(0) = 0
At x = 4 m:
M(4) = (10) (4) – (10) (42/2) = 40 – 80 = - 40 kNm
From C to B:
At any point x > 4m:
M (x) = M (4) + RB (x-4)
At x = 8 m:
M (8) = M (4) + (30) (8-4) + 120 = +80 kNm
Shear Force Bending Moment Diagrams
Shear Force Diagram (SFD)
Starts at +10 kN at A.
Drops linearly to -30 kN at C.
Returns to zero at B.
Bending Moment Diagram (BMD)
Starts at zero at A.
Reaches a maximum negative moment of -40 kNm at C.
Increases linearly to +80 kNm at B.

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Figure 2.15: Shear Force and Bending Moment Diagram

7. i) A simply supported beam AB of length 5 m carries point loads of 8 kN, 10 kN, and
15 kN at 1.50 m, 2.50 and 4.0 m respectively from left hand support. Draw shear
force and bending moment diagrams.
ii) A cantilever beam AB of length 2 m carries a uniformly distributed load of 12
N/m over entire length. Find the maximum shear stress and bending stress, if the
size is 230 x 300 mm. (Nov/Dec’19)
Step 1: Calculate support reactions
a) Total load: The total load on the beam is:
W = 8 kN + 10 kN + 15 kN = 33 kN
b) Moments about Point A:
RB x 5 = 8 x 1.5 +10 x 2.5 + 15 x 4
Calculating the moments:
RB x 5 = 12 + 25 + 60 = 97 kNm
Thus, RB = 97/5 = 19.4 kN
c) Vertical Force Equilibrium: RA + RB = W
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RA + 19.4 = 33
RA = 33 – 19.4 = 13.6 kN
Step 2: Shear force calculation
a) From A to load at 1.5 m:
At x = 0 m:
V (0) = RA = 13.6 N
At x = 1.5 m:
V (1.5) = RA - 8 = 5.6 kN
b) From load at 1.5 m to load at 2.5 m:
At x = 2.5 m:
V (2.5) = V (1.5) – 10 = 5.6 – 10 = - 4.4 kN
c) From load at 2.5 m to load at 4 m:
At x = 4 m:
V (4) = V (2.5) – 15 = - 4.4 – 15 = - 19.4 kN
d) At B (x=5 m):
V (5) = V (4) + RB = - 19.4 + 19.4 = 0
Step 3: Bending Moment calculation
The bending moment M(x) can be calculated as follows:
a) From A to Load at 1.5 m:
At x = 0 m:
M (0) = 0
At x = 1.5 m:
M (1.5) = RA . x – (8) (x – 1.5) = (13.6) (1.5) – (8) (0) = 20.4 kNm
b) From load at 1.5 m to Load at 2.5 m:
At x = 2.5 m:
M (2.5) = M (1.5) – (8) (2.5 – 1.5) = 20.4 – (8) (1) = 12.4 kNm
c) From load at 2.5 m to Load at 4 m:
At x = 4 m:
M (4) = M (2.5) – (10) (4– 2) = 12.4 – (10) (2) = - 7.6 kNm

d) At B (x=5 m):
At x = 5 m:
M (5) = M (4) + RB (x - 4) = - 7.6 + (19.4) (1) = 11 kNm
MB = 0

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Figure 2.16: Shear Force and Bending Moment Diagram

(ii)
We have a cantilever beam with a uniformly distributed load (UDL) of 12 N/m over its
entire length of 2m. The beam's cross-section is rectangular, with dimensions 230mm x
300mm.
Step 1: Calculate Reactions and Maximum Bending Moment:
Reaction at the fixed end (A):
Total load = UDL * Length = 12 N/m * 2m = 24 N
Reaction at A = 24 N (upward)
Maximum Bending Moment:
For a cantilever beam with a UDL, the maximum bending moment occurs at the fixed end.
Maximum Bending Moment (Mmax) = (wL^2)/2 = (12 N/m * (2m)^2)/2 = 24 Nm
Step 2: Calculate Section Modulus (Z):
For a rectangular section, Z = (bd^2)/6
Z = (230mm * 300mm^2)/6 = 3450000 mm^3

Step 3: Calculate Maximum Bending Stress (σmax):

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σmax = Mmax / Z
σmax = (24 Nm * 1000 mm/m) / 3450000 mm^3 ≈ 0.00696 N/mm^2
Step 4: Calculate Maximum Shear Stress (τmax):
For a rectangular section, τmax = (3/2) * (Vmax * Q) / (Ib)
Vmax = 24 N (maximum shear force)
Q = (b * d^2)/8 = (230mm * 300mm^2)/8 = 2587500 mm^3 (first moment of area)
I = (b * d^3)/12 = (230mm * 300mm^3)/12 = 51750000 mm^4 (moment of inertia)
τmax = (3/2) * (24 N * 2587500 mm^3) / (51750000 mm^4 * 230 mm) ≈ 0.034 N/mm^2
solution
Therefore, the maximum bending stress is approximately 0.00696 N/mm² and the maximum
shear stress is approximately 0.034 N/mm².

8. Draw shear force diagram and bending moment diagram for the beam shown in
figure 2.17. (Nov/Dec’19)

Figure 2.17: Simply supported beam

We have a simply supported beam with a length of 4 meters. There is a 25 kN point load
acting at point E, which is 0.75 meters from the left support A.
Step 1: Calculate Reactions at Supports:
Let's denote the reaction forces at A and B as RA and RB, respectively.
Taking moments about point A, we get:
25 kN * 0.75 m = RB * 4 m
RB = 4.6875 kN
Summation of vertical forces:
RA + RB = 25 kN
RA = 25 kN - 4.6875 kN = 20.3125 kN
Step 2: Draw the Shear Force Diagram (SFD):
Section 0 to 0.75 m:

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Shear force at A = RA = 20.3125 kN (upward)
Shear force remains constant due to no load.
Section 0.75 m to 4 m:
Shear force at E = 20.3125 kN - 25 kN = -4.6875 kN (downward)
Shear force remains constant due to no load.
Step 3: Draw the Bending Moment Diagram (BMD):
Section 0 to 0.75 m:
Bending moment at A = 0
Bending moment varies linearly due to constant shear force.
Bending moment at E = 20.3125 kN * 0.75 m = 15.2344 Nm
Section 0.75 m to 4 m:
Bending moment varies linearly due to constant shear force.
Bending moment at B = 0

Figure 2.18: Shear Force and Bending Moment sample Diagram

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9. A cantilever beam is made up of a flitched section as shown in figure 2.19. if the

allowable bending stresses in steel and timer are respectively 165 N/mm2 and 8.5
N/mm2. Find the value of W. (Nov/Dec’19)

Figure 2.19: Cantilever beam

Given Data
We have a cantilever beam with a flitched cross-section, meaning it's composed of both
steel and timber. The goal is to determine the maximum allowable load (W) that the beam
can carry without exceeding the allowable stresses in either material.
Step 1: Calculate Section Modulus of Each Material:
Steel:
Area (As) = 150 mm * 10 mm = 1500 mm²
Distance from the neutral axis to the extreme fiber (ys) = 75 mm
Moment of inertia of the steel section about its centroidal axis (Is) = (150 mm *
(10 mm)^3) / 12 = 12500 mm^4
Section modulus of the steel section (Zs) = Is / ys = 12500 mm^4 / 75 mm =
166.67 mm³
Timber:
Area (At) = 40 mm * 150 mm = 6000 mm²
Distance from the neutral axis to the extreme fiber (yt) = 75 mm
Moment of inertia of the timber section about its centroidal axis (It) = (150 mm *
(40 mm)^3) / 12 = 300000 mm^4
Section modulus of the timber section (Zt) = It / yt = 300000 mm^4 / 75 mm =
4000 mm³
Step 2: Calculate the Modular Ratio (n):

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The modular ratio (n) is the ratio of the modulus of elasticity of steel (Es) to the modulus
of elasticity of 1 timber (Et). It's a measure of the relative stiffness of the two materials.
Let's assume the following values for Es and Et:
Es = 200 GPa (typical for steel)
Et = 10 GPa (typical for timber)
Therefore, n = Es / Et = 200 GPa / 10 GPa = 20
Step 3: Calculate the Equivalent Section Modulus (Zeq):
The equivalent section modulus (Zeq) takes into account the relative stiffness of the
two materials:
Zeq = Zs + n * Zt = 166.67 mm³ + 20 * 4000 mm³ = 80166.67 mm³
Step 4: Calculate the Maximum Bending Moment (Mmax):
The maximum bending moment occurs at the fixed end of the cantilever beam and is
equal to:
Mmax = W * L = W * 2 m
Step 5: Determine the Allowable Load (W):
The maximum bending stress in the composite section is given by:
σmax = Mmax / Zeq
To ensure that neither the steel nor the timber exceeds their allowable stresses, we need
to consider the limiting condition for both materials.
For Steel:
σmax = Mmax / Zeq ≤ σs (allowable stress in steel)
For Timber:
σmax / n = Mmax / (n * Zeq) ≤ σt (allowable stress in timber)
Since the steel has a higher allowable stress and a smaller section modulus, it will likely
be the limiting factor. Let's use the steel's allowable stress to calculate W:
W * 2 m / 80166.67 mm³ ≤ 165 N/mm²
W ≤ (165 N/mm² * 80166.67 mm³) / 2 m ≈ 6613.83 N
Solution
Therefore, the maximum allowable load (W) is approximately 6613.83 N.

10. A simply supported beam of span 5 m carries a uniformly distributed load of 10

kN/m over the entire span and also a point of 10 kN at 2 m from the left support. if
the permissible bending stress is 8 Mpa, design a suitable rectangular section taking
the depth twice the width. (Apr/May’22)
Given Data
Span of the beam (L): 5 m
Uniformly distributed load (w): 10 kN/m
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9111-SRM MCET Mechanical Engineering
Point load (P): 10 kN at 2 m from the left support
Permissible bending stress in concrete ( σ c σ c ): 8 MPa
Relationship between depth and width: Depth h = 2 b h=2b
Step 1: Calculate Reactions at Supports:
Total UDL = 10 kN/m * 5 m = 50 kN
Reaction at each support (RA = RB) = (50 kN + 10 kN) / 2 = 30 kN
Step 2: Determine the Maximum Bending Moment:
We can use the principle of superposition to find the maximum bending moment.
Bending Moment due to UDL:
M_udl = (wL^2) / 8 = (10 kN/m * 5^2 m^2) / 8 = 31.25 kNm
Bending Moment due to Point Load:
M_point = P * x * (L - x) / L = 10 kN * 2 m * (5 m - 2 m) / 5 m = 12 kNm
Maximum Bending Moment (M_max):
The maximum bending moment occurs at the point of application of the point load.
M_max = M_udl + M_point = 31.25 kNm + 12 kNm = 43.25 kNm
Step 3: Design the Rectangular Section:
Permissible bending stress (σ) = 8 MPa = 8 N/mm²
Depth (d) = 2 * Width (b)
Section Modulus (Z): Z = (bd^2) / 6 = (b * (2b)^2) / 6 = (4b^3) / 6 = (2b^3) / 3
Bending Stress Formula:
σ=M/Z
Substituting values:
8 N/mm² = (43.25 * 10^6 Nmm) / ((2b^3) / 3)
Solving for b:
b^3 = (43.25 * 10^6 Nmm * 3) / (8 N/mm² * 2) ≈ 8109375 mm^3
b ≈ 200 mm
Therefore, the width (b) is approximately 200 mm.
Depth (d):
d = 2 * b = 2 * 200 mm = 400 mm
solution
Hence, the required rectangular section is 200 mm x 400 mm.

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