Chapter: Mechanical : Strength of Materials : Transverse Loading On Beams And Stresses In Beam
Solved Answers: Transverse Loading on Beams and Stresses In Beam
Problem - 1
Draw the shear force & bending
moment diagram for given cantilever.
Problem - 2
Draw the shear force diagram for given
cantilever.
Solution :
To fine the support
reaction:
Taking moment at A = -RD 6+6 4+3 2
= 0
RD 6 = -30
RD = -5 KN
Upward force = Downward force
RA+5 = 6+3
RA = 4 KN
Shear Force
Calculation:
R.H.S
Shear force at A = 4 KN
Shear force at B = 4-3 = 1 KN
Shear force at C = 4-3-6
= -5 KN
Shear force at D = -5 KN
L.H.S
Shear force at D = -5 KN
Shear force at C = 6-5 =
1 KN
Shear force at B = 6+3-5
RA = B = 4 KN
To find the moment
Bending moment at A = 0
Bending moment at B = 4 2 = 8 KNm
Bending moment at C = (4 4) –(3 2)
= 10
KNm
Problem - 3
Draw the shear
force diagram for at bending moment for givern cantilever.
Solution
:
To fine the support
reaction:
Taking moment at A = (4 –(RE 4)
= 32 –4RE
RE = 8 KN
RA + 8 = 15
RA = 7 KNS
To fine Shear force:
Shear force at A = 7 KN
Shear force at B = 7-4 = 3 KN
Shear force at C = 7-4-5
= - 2KN
Shear force at D = 7-4-5-6 = -8 KN
Shear force at E = - 8 KN
To find bending moment
Bending Moment at A = 0
Bending Moment at B = 4 KNm
Bending Moment at C = (7 2) –(4
2)
= 10 KNm
Bending Moment D = (7 3) –(4
2)
= 8 KNm
Bending Moment E = 0
Problem - 4
A Beam of Total
length 8m is freely supported at a left end & at a point 6m from left end.
It carries 2 points floats of 15KN & 18KN. In which one is at the free end
and another is 3m from the left support. Draw the shear force and bending
moment diagram. Locate the point of contraflexture.
Solution
:
To fine the support
reactions:
Taking moment about A,
(Rc 6)
–(18 3) –(10 8) = 0
6 Rc = 54+120
Rc = 174/6
Rc = 29 KN
RA + Rc = 18+15
RA + 29 = 33
RA = 4 KN
To fine Shear force:
Shear force at D = 15 KN
Shear force at C = 29 KN +15 = -14 KN
Shear force at B = -14+18 =
4 KN
Shear force at A = 4 KN
To find bending moment
Bending Moment at D = 0
Bending Moment at C = -(15 2) = - 30 KNm
Bending Moment at B = -(15 5)+(29 3)
Bending Moment at A = -(15 8) +(29 6)-(18 3)
=
-120-54+174
=
0
Problem-5:
The
cross section of the beam is shown is beam is cantiliver type &carries a
UDL of 16KN/m. If the span of beam is 2.5m. Determine the maximum tension &
Compressible stress in the beam.
solution:
Section (1) :
Problem-6:
The
cast iron bracket subjected to bending has a cross section of I-shaped with
unequal flanges as shown. If the compressive force on the top of the flanges is
not to exceed 17mega pa. What is the bending moment of the section can take if
the section is subjected to a shear force of 90KN. Draw the shear stress
distribution over the depth of the section.
Solution:
Area of section (1) = lb
=
250
50
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