What is cantilever in architecture?
A cantilever beam is a flat, rigid structural support that is fixed at one end and hangs horizontally at the other, transferring all vertical load to the beam and the wall. A cantilever, like other structural parts, can be shaped into a beam, plate, truss, or slab. The top half of the thickness of such a beam is subjected to tensile stress, which tends to lengthen the fibers, and the lower half to compressive stress, which tends to crush them. Cantilevers are widely used in building construction and in machines. In construction, any beam that projects out of the wall and has free projection is considered a cantilever. Longer cantilevers in a building require more clear space. A cantilever building enables overhanging structures to be built without the need for extra support.
Use of Cantilever In Bridges and Buildings
Cantilevers can be widely seen in the construction of bridges, balconies, terraces, porches, etc. In cantilever bridges, cantilevers are commonly built in pairs, with each cantilever supporting one end of the center part. The Forth Bridge in Scotland, which is made for the railway, is a good example of a cantilever truss bridge.
In architecture, In Falling Water, FL Wright used cantilevers to project large balconies. Today, many contemporary buildings incorporate cantilevers in various ways to enhance the beauty of the building and provide open space.
Use of Cantilever In Structure (Some Solved Examples)
Example 1. A cantilever 120 mm wide and 200 mm deep is 2.5 m long. What is the uniformly distributed load which the beam can carry in order to produce a deflection of5 mm at the free end? Take E=200 GN/m².
Sol.
Given: Width, b= 120 mm.
Depth, d = 200 mm
Length, L= 2.5 m = 2.5 x 1000 2500 mm.
Deflection at free end, yb = 5mm
Value of E = 200GN/m2 = 200 x 109 N/m2 (Since G = Giga = 102)
= 200 x 109 N / (1000)2 mm2 {Since: 1m2 = (1000 mm)2 }
= 2 x 105 N/mm
Moment of Inertia,
I = bd3 /12
= 120 x 2003 /12
= 8 x 107 mm4
Let
w = uniformly distributed load per m length in N
W = Total Load
= w x L (Here L is in M)
= w x 2.5
= 2.5 x w N
Using eq. (13.6), we get (Refer strength of material by Dr. R.K. Bansal)
yb = WL3 /8EI
5 = 2.5w x 25003 / 8 x 2 x 105 x 8 x 107
w = 5 x 8 x 2 x 105 x 8 x 107 / 2.5 x 25003
= 16.384 kN/m. Ans (Source : Strength of material Page 560- 561 )
Example 2. A cantilever of length 2 m carries a uniformly distributed load of 2.5 kN/m run for a length of 1.25 m from the fixed end and a point load of 1 kN at the free end. Find the deflection at the free end of the section is rectangular 12 cm wide and 24 cm deep and E = 1 x 104 /mm².
Given:
Length, L= 2m = 200mm
U.d.l w = 2.5kN/m = 2.5 x 1000 N/m
= 2.5 x 1000/ 1000
= 2.5 N/mm
Point load at free end,
W = 1kN = 1000N
Distance AC, a = 1.25m = 1250mm
Width, b = 12mm
Depth, d = 24mm
Value of
I = bd3 /12
= 12 x 243 / 12
= 13824 cm4
= 1.3824 x 104 mm4
Value of
E = 1 x 108 N/mm2
Let y1 = Deflection at the free end due to point load 1 kN alone
y2 = Deflection at the free end due to u.d.l on length AC
(i)Now the downward deflection at the free end due to the point load of 1kN (or 100 N ) at the free end is given by eq. (13.2) as
(Refer strength of material by Dr. R.K. Bansal)
y1 = WL3 / 3EI
= 1000 x 20003 / 3 x 104 x 1.3824 x 108
= 1.929mm
(ii)The downward deflection at the free end due to uniformly distributed load of 2.5 N/mm on a length of 1.25m (or 1250mm) is given by eq. (13.8) as
(Refer strength of material by Dr. R.K. Bansal)
y2 = w a4 / 8EI + w .a3 /6 EI (L – a)
= 2.5 x 12504 / 8 x 104 x 1.3824 x 108 + 2.5 x 12503 / 6 x 104 x 1.3824 x 108 (2000 – 1250 )
= 0.5519 + 0.4415
= 0.9934
Therefore Total deflection at the free end due to point load and u.d.l
= y1 + y2
= 1.929 + 0.9934
= 2.9224 mm Ans
(Source: Strength of material Page 565- 566 )
More solved solutions will be added in coming days.
