Advertisements
Advertisements
प्रश्न
In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by `(Ypir^4)/(4R) . Y` is the Young’s modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
Advertisements
उत्तर
According to the problem, the elastic torque or the bending torque is given and we have to find the torque caused by the weight due to bending.
The diagram of the given situation is as shown.
The bending torque on the trunk of radius r of the tree = YπR4/4R where R is the radius of curvature of the bent surface.
`τ = Wd = (Ypir^4)/(4R)`
Here deforming torque is equal to elastic torque (restoring torque) caused by the bending of tree about its central axis when the tree is about to buckle.
Let h be the height of tree. If R >> h, then the centre of gravity of the tree is at a height of h/2 from the ground.
Refer to the figure in ΔABC
`R^2 = (R - d)^2 + (h/2)^2`
= `R^2 - 2Rd + d^2 + h^2/4`
Since, d << R, therefore the term d2 being very very small can be neglected.
∴ `R^2 = R^2 - 2Rd + h^2/d`
or `d = h^2/(8R)` ......(i)
If `omega_0` is the weight/volume, then
`(Ypir^4)/(4R) = omega_0 (pir^2h) h^2/(8R)` .....[∵ Torque is caused by the weight]
⇒ `h = ((2Y)/(omega_0))^(1/3) r^(2/3)`
Hence, critical height = `h = ((2Y)/(omega_0))^(1/3) r^(2/3)`.
APPEARS IN
संबंधित प्रश्न
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10–5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of Young’s modulus of steel to that of copper?
Read the following statements below carefully and state, with reasons, if it is true or false
The Young’s modulus of rubber is greater than that of steel;
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
A wire elongates by 1.0 mm when a load W is hung from it. If this wire goes over a a pulley and two weights W each are hung at the two ends, he elongation of he wire will be
A copper wire of cross-sectional area 0.01 cm2 is under a tension of 20N. Find the decrease in the cross-sectional area. Young modulus of copper = 1.1 × 1011 N m−2 and Poisson ratio = 0.32.
`["Hint" : (Delta"A")/"A"=2(Delta"r")/"r"]`
A uniform rectangular block of mass of 50 kg is hung horizontally with the help of three wires A, B and C each of length and area of 2m and 10mm2 respectively as shown in the figure. The central wire is passing through the centre of gravity and is made of material of Young's modulus 7.5 x 1010 Nm−2 and the other two wires A and C symmetrically placed on either side of the wire B are of Young's modulus 1011 Nm−2 The tension in the wires A and B will be in the ratio of:
Young's modulus of a perfectly rigid body is ______.
A rigid bar of mass M is supported symmetrically by three wires each of length l. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to ______.
The Young’s modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?
What is the Young’s modulus for a perfect rigid body ?
A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched? (Young’s modulus for steel is 2 × 1011 Nm–2.)
A steel wire of mass µ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m–3 (Young’s modules Y = 2 × 1011 Nm–2).
In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by `(Ypir^4)/(4R) . Y` is the Young’s modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
A metal wire of length L, area of cross section A and Young's modulus Y behaves as a spring of spring constant k given by:
Young's modulus is also known as ______.
What does the dimensional formula [L⁻¹M¹T⁻²] represent?
What is longitudinal strain?
In the formula Y = MgL/(πr²l), what does 'l' represent?
