A beam of rectangular section (12cm wide × 20cm deep) is simply supported over a span of 12m. It is acted upon by a concentrated load of 80 kN at the mid span. The maximum bending stress induced is
Moment of inertia of crosssection
= 8 x 10^{3}cm^{4}
Maximum moment under given loading,
Maximum bending stress
= 300 MPa
A beam with a rectangular section 120mm × 60mm, designed to be placed vertically is placed horizontally by mistake. The maximum stress is to be limited, the reduction in load carrying capacity would be
So load carrying capacity would be half if placed horizontally by mistake in place of vertically.
The ratio of the section moduli of a square beam (Z) when square section is placed
(i) With two sides horizontal (Z1 ) and
(ii) With a diagonal horizontal (Z2 ) as shown is
Moment of inertia
Two beams, one having a square crosssection and another circular crosssection, are subjected to the same amount of bending moment. If the crosssectional area as well as the material of both the beams are the same, then
1. Both the beams will experience the same amount of deformation
2. The circular beam experience more extreme flexural stress than the square one
Which of the above is/are correct?
Area = a^{2} = A, material is same this means that E is the same.
Assuming same span length and loading beam having larger moment of inertia will have smaller deflection.
⇒ δ_{rec} < />_{cir }
⇒ Statement 1 is wrong (Flexural strength)max = Mymax / 1 For the same bending moment, a beam having larger ymax / 1 will experience large flexural stress.
⇒ Circular section experience more flexural stress.
A cantilever of length 1.2m carries a concentrated load of 12kN at the free end. The beam is of rectangular crosssection with breadth equal to half the depth. The maximum stress due to bending is not to exceed 100 N/mm^{2}. The minimum depth of the beam should be
At point A, M_{max }= WL
= (12 × 10^{3})Nm × (1.2)
= 14.4 × 10^{3}Nm = 14.4 × 106Nmm
bh^{2} = 6 × 14.4 × 10^{4}
(h / 2) × h2 = 86.4 × 10^{4}
h3 = 172.8 × 104
h = 120 mm
A bar of rectangular crosssection (b × 2b) and another bar of circular crosssection (diameter = d) with the same length, are made of same material, and are subjected to the same bending moment and have the same maximum bending stress developed. The ratio of weights of rectangular bar and circular bar will be
An equal flange Isection of 300mm depth and 200mm wide flanges has a uniform thickness of 10mm. it is subjected to a bending moment of 100kNm. The bending stress at top most fiber is (in MPa)
Due to symmetry ̅X = 100 mm ̅
Y = 150 mm
Bending stress
M = fZ
100 × 10^{6} = f × 682844.44
f = 146.44 N/mm^{2}
A cantilever has length of 2.5cm. It is of T section with IX = 2127cm4 and fibre distance (tensile) from neutral axis is 7cm. if the maximum allowable tensile stress is 300kgf/sq cm. Then the maximum UDL that can be applied in kg/cm is
Bending equation,
M = 91.157.143kgcm
= 911.571kgm
MA = 3.125w
Substituting M value, 3.125 w = 911.571
w = 291.7kg/m
A beam AB simply supported at its ends A and B, 3m long, carries a uniformly distributed load of 1kN/m over its entire length and a concentrated load of 3kN, at 1m from A.
If ISMB 150 with IXX = 300 cm4 is used for the beam, the maximum value of bending stress is
Taking moment about point A
3R_{B} = 3 + 4.5
R_{B} = 2.5 kN R_{A} + RB = 3 + (1 × 3)
R_{A} + R_{B} = 6
R_{A} = 3.5 kN
M_{max} = (B M)c
= 75 MPa
The maximum shearing stress induced in the beam section at any layer at any position along the beam length (shown in figure) is equal to
Crosssection of beam
= 30 kgf/cm2
Simply supported beam of 3m span is subjected to loads as shown in figure below the beam is of Isection and all of the dimensions are in mm. The maximum compressive stress at point D in the web is (The section is located at a distance of 1m for right hand support)
Reactions, 2 × 3 × 1.5 + 4 × 2 = 3R_{B}
17 = 3R_{B}
Reactions, R_{B} = 5.667 kN
Reactions, R_{A }= 2 × 3 + 4 − 5.667
= 10 kN − 5.667 = 4.333 kN
M_{C} = Bending moment at C
= 4.333 × 2 − 2 × 2 × 1
= 8.666 − 4 = 4.666 kNm
= 4.666 × 10^{6} Nmm
(Sagging) Shear force at C
F_{C} = 4.333 − 2 × 2 = 0 = 0.33 3 (just left to C)
F_{C }′ = 0.333 − 4 = −3.667 kN (just right to C)
Jump shear force at C = 4 kN ↓
= 50 × 10^{6} − 29.16 × 10^{6}
= 20.84 × 106 mm4
= −11.20 N/mm^{2} (Sagging moment) compressive
Shear stress at D
p_{1} = 11.385 N/mm^{2}
p_{2} = +0.19 N/mm^{2}
Two beams of equal crosssection area are subjected to equal bending moment. If one beam has square crosssection and the other has circular section, then
Given, A_{square} = A_{circular}
Stress generated in circular section,
⇒ Stress in square crosssection < stress="" in="" circular="" />
Hence, the square beam is stronger.
A simply supported beam of span L and constant width b carries a point load W at midspan. The depth of the beam required at the mid span for maximum extreme fiber stress P.
A link is under a pull which lies on one of the faces as shown in the figure below. The magnitude of maximum compressive stress in the link would be
Let P_{1 }= P_{2} = P at neutral axis So P and P_{2} gives bending moment and P_{1} give axial tensile force
Bending stress,
Maximum compressive stress at lower fiber
= σ_{b} − σ_{a}
= (16 − 5.33)
= 10.67 N⁄mm^{2} = 10.7 MPa
Beam A is simply supported at its end and carries UDL of intensity ‘w’ over its entire length. It is made of steel having Young’s Modulus E. Beam B is cantilever and carries a UDL of intensity w/4 over its length. It is made of brass having young’s modulus E/2. The two beams are of the same length and have the same crosssectional area. If σA and σB denote the maximum bending stresses developed in beams A and B, respectively, then which one of the following is correct?
σ = My / I
σ = M / Z
Where Z is section modulus For beams with the same cross section, Z is the same. It is given that the cross sectional area is the same. We cannot infer that Z is same
Since, Z_{1} and Z_{2} depends on shape of cross section, hence σ_{A} / σ_{B} also depends on shape of
cross section.
At a section of a beam, shear force is F with zero BM. The crosssection is square with side ‘a’. Point ! lies on the neutral axis and point B is midway between the neutral axis and top edge, i.e. at distance a/4 above the neutral axis. If τA and τB denote shear stresses at points A and B, then what is the value of τA⁄τB?
For square beam, shear stress,
If the area of crosssection of a circular section beam is made four times, keeping the loads, length, support conditions and material of the beam unchanged, then the quantities (ListI) will change through different factors (ListII). Match the ListI with the ListII and select the correct answer using the code given below the lists.
ListI ListII
a. Maximum BM 1. 8
b. Deflection 2. 1
c. Bending Stress 3. 1/8
d. Section Modulus 4. 1/16
Codes: a b c d
Assuming, d_{2} = 2d_{1} A case of cantilever beam loaded with a point load P is used.
A shaft is subjected to torsion as shown in the figure below.
Which of the following figures represents the shear stress on the element LMNOPQRS?
Shear stress
When a shaft is subjected to torsion, shear stresses act over the crosssection and on longitudinal planes. This shaft is under pure shear whose direction depends upon the directions of the applied torque T.
So, in the given figure
A cast iron water pipe 300 mm outer diameter with 15 mm wall thickness is supported over a span of 8 m. The maximum stress in cast iron when the pipe is running full of water is ___________MPa.
Take Density of CI = 71.6 kN/m^{3}, Density of water = 9.8 kN/m^{3}
(A) 12.9
(B) 13.7
Outer diameter, D = 0.3 m
t = 0.015 m
Inside diameter, d = 0.27 m
Area of crosssection (CI)
= π /4(0.32 − 0.272) = 0.01343 m2
Area of crosssection of water
= = 0.05725 m^{2}
Weight of CI per meter length
= 0.01343 x 71.6 = 0.9616 kN
Weight of water per meter length
=0.05725 x 9.8 = 0.5610 kN
Total weight of beam per meter length,
w = 0.9616 + 0.5610 = 1.5226 kN/m
L = 8 m
Maximum bending moment
M_{max} =
= 12.1808 kNm
= 12.1808 × 10^{6} Nmm
Section modulus of CI pipe,
= 91.158 × 10^{4} mm^{4}
Maximum stress in CI pipe,
σ =
= ± 13.36 N/mm^{2}
Question_Type: 5
The shear force diagram for rectangular cross section beam is shown in the figure below. Width of the beam in 100 mm and depth is 250 mm, the maximum bending stress in the beam is ___________ MPa.
(A) 25.8
(B) 27.0
Maximum bending moment occurs at fixed end A
M_{A} = 5(2.5) + 5 × 2 + 5 × 1
= 12.5 + 10 + 5 = 27.5 kNm
= 27.5 × 106 Nmm = σb × Z
= 26.4 N/mm2 (Max. bending stress in beam)
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