A simply supported beam of span 6 m carries two concentrated loads of 2N each at 1 m distance from each of the supports. If the flexural rigidity of the beam is constant. the slope at the support is
Slope at B = due to symmetry
=
=
=
A prismatic beam of length L is simply supported at its ends and subjected to a total UDL of W spread over its entire span. It is then propped at its Centre to neutralize the deflection. The net B.M. at its Centre will be
The propped reaction neutralize Downward deflection due to udl = upward Deflection due to support at C
∴ moment at C M_{c} =
R_{A} + R_{B} + R_{C} = W
∴ R_{c} =
R_{A} + R_{B} = W 
Due to symmetry R_{A} = R_{B} =
∴ M_{c} =
=
=
= (Clockwise)
A cantilever beam of uniform EI has a span equal to ‘L’. An upward force W acts at the midpoint of the beam and a downward force P acts at the free end. In order that the deflection at the free end is zero, the relation between P and W should be
Upward deflection at B due to W
=
=
=
Downward deflection due to P =
∴
W=
A point load ‘W’ is acting at the midspan of a cantilever of length ‘L’. If the free end is supported on a rigid prop, the reaction of the prop is
y_{b} =
y_{a} =
Equating upward and downward deflection
V_{B} =
A cantilever AB, 3 m long carries a uniformly distributed load of 2 kN/m throughout its length, rests over a similar cantilever of same crosssection and same material, 2 m long and shown below. What is reaction C?
Let the reactions at C is R. Deflection at C in cantilever CD,
δ_{C} =
=
= δ_{B} (in cantilever AB) ⋯ ①
Y_{B1} =
Y_{B2} =
Y_{B} = Y_{B1} + Y_{B2} =
From equation ① and ②
R = 1.736 kN
A beam ABC of length 3L has one support at the left end and the other support at a distance 2L from the left end. The beam carries a point load W at the right end. The deflection at the right end is
solving RA =
RB = 3/2W
Deflection at C,
δ_{c} =
A cantilever of Length ‘L’ carries a uniformly distributed load of w per unit run for a distance ‘a’ from the fixed end. The deflection at free end is
BM at B in conjugate beam = deflection at B in original beam
δn =
=
=
=
Method 2:
Deflection B = deflection at C + θc × (L − a)
=
=
A uniform beam of length ‘L’ is simply and symmetrically supported to a span ‘l’ find the ratio L/l so that the upward deflection at each end equals the downward deflection at mid span, due to a concentrated point load at mid span.
Deflection at midpoint = deflection at free end
A prismatic simply supported beam carries a point load at midspan section due to which a slope of 2° is produced at support sections. The slope at quarter span sections is
=
=
= 2  0.5 = 1.5°
A beam of span 6m and of uniform flexural rigidity EI = 40000 kNm^{2} is subjected to a clockwise couple of 300 kNm at a distance of 4m from the left end. The deflection at the point of application of the couple is _______ mm
(A) 9.12
(B) 9.72
Taking moments about A, V_{b} × 6 = 300
V_{b} = 50 kN ↑
∴ V_{a} = 50 kN ↓
The B.M at any section distant x from A is given by,
Integrating,
Integrating !gain,
EIy =
AT x = 0, t = 0
∴ C_{2} = 0
At x = 6, y = 0
∴ 0 =
∴ C_{1} = 200
Deflection at C. Putting x = 4m, in the deflection equation,
EIyc =
= 266.67
Max. deflection. This will occur in the larger segment. Equating the slope to zero, we get,
∴ x = 2√2 m
Putting x = 2√2 m in the deflection equation
EIy_{max} =
ymax =
= 9.43 mm
Question_Type: 5
A cantilever 3m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. The deflection at the free end is __________ mm (Take E = 200 GPa, I = 54000 cm4)
(A) 3.9
(B) 4.5
AB fixed at end B, Free at end A carrying a point load W at L_{1} from B end UDL of W throughout to length EI = flexural rigidity δ = deflection at free end
=
where, L_{1} = 1.2m, L_{2} = 1.8m, L = 3m,
W = 80 kN and W = 30 kN/m
EI = 200 × 106 kN/m2 × 54000 × 10^{−8} m^{4}
= 108000 kNm^{2}
δ =
= 0.42666 × 10^{−3} + 0.96 × 10^{−3} + 2.8125 × 10^{−3}
= 4.2 × 10^{−3}m = 4.2 mm
Question_Type: 5
A cantilever of length 6 m is loaded as shown in the figure below. The maximum deflection in the beam is
E = 200 MPa, I = 1 × 10−4 m−4, W = 1 kN.
P1 = 3W P2 = 2W P3 = 1W
Slope at free end
I =
=
=
= 0.002 radian
Deflection at free end
y =
=
=
=
= 0.00893 m = 8.93 mm
A horizontal steel rod ABC of diameter 50 mm and length 400 mm carrying a uniformly distributed load of 1.8 kN/m is simply supported at its ends A and C and is also supported at its midpoint B by its connection to a hanging vertical steel wire of 5 mm diameter and length 360 mm as shown in figure below. The deflection at point B is _________ mm (Take E = 2 × 105 N/mm^{2}).
(A) 0.0076
(B) 0.0082
l_{1} = 400 mm, l_{2} = 360 mm,
W = 1.8 kN/m = 1.8 N/mm
D = 50 mm, I = = 306796.875 mm
d = 5 mm, A = = 19.635 mm^{2}
T = = 86.228 N
∴ Deflection of B
= = 0.0079 mm
Question_Type: 5
A horizontal beam (I = 8.616 × 107mm4) carries a uniformly distributed load of 50 kN over its length of 3m. The beam is supported by three vertical steel tie rods, each 1.80m long. One at each end and one at in the middle, the end rods having a diameter of 25mm and the centre rod a diameter of 30mm. The deflection of the centre of the beam is _________mm (Take E = 2 × 105 N/mm^{2})
(Take E = 2 × 105 N/mm^{2})
(A) 0.131
(B) 0.139
Area of each rod
= A1 = = 490.87 mm^{2}
Area of the middle rod
= A2 = = 706.86 mm^{2}
Let, P_{1} = Tension in each end rod
P_{2} = Tension in the middle rod
f_{1} = Stress in the rods
f_{2} = Stress in the middle rod
δ_{1} = Extension of each end rod
δ_{2} = Extension of the middle rod
Deflection of the midpoint of the beam
δ2 − δ1 =
EI(δ2 − δ1) =
Deflection of beams
8.616 × 10^{7} × 1800(f_{2} − f_{1})
=
2205.696(f_{2} − f_{1}) = 250 − 5654.88f_{2}
2205.696f_{2} − 2205.696f_{1} = 250 − 5654.88f_{2 }
7860.576f_{2} − 2205.696f_{1} = 250 ⋯ ①
2f_{1}A_{1} + f_{2}A_{2} = 50
2f_{1}(490.87) + f_{2} (706.86) = 50
981.74f, + 706.86f, = 50 ⋯ ②
From equation ① and ②,
7860.576f_{2} − 2205.696f_{1} = 5(981.74f,1 + 706.86f_{2})
7860.576f_{2} − 2205.696f_{1} = 4908.70f_{1} + 3534.3f_{2 }
4326.76f_{2} = 7114.396f_{1}
∴ f_{2} = 1.644f_{1}
Substituting in equation ②
981.74f_{1} + 706.86 × 1.644f_{1} = 50
2143.8178f_{1} = 50
f_{1} = 0.02332 kN/mm^{2} = 23.32 N/mm^{2}
f_{2} = 1.644 × 23.32 = 38.33 N/mm^{2}
Deflection of the midpoint of the beam
δ_{2} − δ_{1} =
=
=0.135 mm
Question_Type: 5
determine the slope and deflection at a point in a beam is suitable for beams subjected to concentrated loads and can be extended to uniformly distributed loads Reason(R): Macaulay’s method is based upon the modification of moment of area method. This is applicable to a simple beam carrying a single concentrated load but by superposition, this method can be extended to cover any kind of loading.
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