Level 200 Bridging

Question One

Solution

∑ = 0

RB x 10 – 3 x 4x 12-3x 8 x4 -24+ 6 x 4 = 0

RB = 24kN

RA = 3 x 4+ 3x 8 + 6 – 24

RA = 18kN

Portion AB: Measuring x from A and considering left hand side; F= -3x

At x =0, F= 0

At x = 4m, F= -3x 4, = -12kN

M – 3x = -3x2/2

At x = 0, M = 0

At x = 4m, M = -3x 42/2 = -24kN-m

Portion BC: Measuring x from A and considering forces on left hand side of the section,

F = -12 + 24 - (x -6) = 30 – 3x

2m 4m

3kN/m6kN

24kNm

4m 4m

3kN/m

4m

A B C D EF

126

12

12

-ve

+ve

-ve

+ve

+ve-ve-ve 0

24

SFD

BMD

24

24

SPACE DGM

At x = 6m, F = 30 – 3 x 6 = 12 kN

At x = 10m, F = 30 – 3 x 10 = 0

M = - 12 (x- 2) + 24 (x - 4) – 3(x – 6) (x-6)/2

= -1.5x2 + 30x -126

At x = 6m, M = - 1.5 x 62 + 30 x 6 -126 = 0

At x = 10m, M = - 1.5 x 102 + 30 x 10 -126 = 24kN-m

This maximum, since at this point SF=0

Portion DE: Measuring x from E and considering forces on right hand side of the section

F= 6 – 18 + 3(x-4) = 3x-24

At x = 4m, F = 3 x 4-24 = - 12kN

M = -6x + 18( x – 4) -3(x-4) (x-4)/2, = - 1.5x2 + 24 x -96

At x = 4m, M = - 1.5x 42 + 24 x 4 -96 = -24kN-m

At x = 8m, M = M = - 1.5x 82 + 24 x 8 -96 = 0

Portion EF: SF = 6 kN, M = -6x.

Question Two

Most metals are ductile and fail due to yielding. Hence, the yield strength

characterises their failure. Ceramics and some polymers are brittle and rupture or

fracture when the stress exceeds certain maximum value. Their stress–strain behaviour

is linear up to the point of failure and they fail abruptly.

The stress required to break the atomic bond and separate the atoms is called the

theoretical strength of the material. It can be shown that the theoretical strength is

approximately equal to E/3 where, E is Young’s modulus. However, most materials

fail at a stress about one–hundredth or even one–thousandth of the theoretical strength.

This enormous discrepancy could be explained as follows. Ceramics and some

polymers are brittle and rupture or fracture when the stress exceeds certain maximum

value. Their stress–strain behaviour is linear up to the point of failure and they fail

abruptly.

In ductile material yielding occurs not due to separation of atoms but due to sliding of

atoms (movement of dislocations). Thus, the stress or energy required for yielding is

much less than that required for separating the atomic planes. Hence, in a ductile

material the maximum shear stress causes yielding of the material.

In brittle materials, the failure or rupture still occurs due to separation of atomic planes.

However, the high value of stress required is provided locally by stress concentration caused

by small pre-existing cracks or flaws in the material. When this process becomes unstable,

the material separates over a large area causing brittle failure of the material.

b) Suppose that P is the failure load of a column of a given material and of any length.

Assuming also that Ps is the failure load in compression of a short column of the same

material and that Pcr is the buckling load of a long slender column, again of the same

material. The Rankine theory proposes that = +

Let the σs be the yield stress in compression of the material of the column and A is the crosssectional area, then,

Ps = σs A

But, Pcr = 2EI/Le2

Substituting for Ps and Pcr

= + 1 /∏2EI/Le2

= ∏2EI/Le

2 + σs A

σs A∏2EI/Le

2

so that

P = σs A∏2EI/Le

2

∏2EI/Le

2 + σs A

Dividing top and bottom of the right hand side of this equation by ∏2EI/Le

2 , we have ,

P = σs A

1 +σs AL2e

∏2EI

But I = Ar2, so that,

P = σs A

1 +k(Le/r)2

K is a constant that depends on the material of the column

The failure stress in compression of any column is thus given by

σc = = σs A

1 +k(Le/r)2

c)

Given, concrete strength = 30N/mm2, steel strength = 460N/mm2

Area of steel = x d2 = x 400 = 100 x 8 = 800 = 2513.274mm2

Area of concrete = 4002- 2513.274 = 157,486.726mm2

Axial load = 20000 Litres, = 20000 kg, Assuming g = 10 m/s2, Axial load = 200,000N = 200kN.

Hence 200, 000 N = Axial Load to be carried by the Four columns.

Column capacity:

Fs + Fc = Axial Load

But, σsAs + σcAc = Axial Load

(460 x 2513.274) + (30 x157, 486.726) = 200,000/4

400

mm

1156106.04 + 4724601.78 = 50,000

4, 840, 9217 > 50,000

Thus the column can adequately support the polytank’s load.

Question Three

a) Since maximum shearing stress is proportional to applied torque we have; = =

constant for a given shaft. Where Τ = applied torque,τ = shearing stress, J = Polar secondmoment of area and R = radius

Now the condition specified is that = = constant

of solid = of hollow = constant

Substituting the values,˄ ∏

= [( 1.5d)4 –d4] x .= 2252 = ( )4 –d4

65d2/24

65d3/24

d3 =ˆ

= 0.420 x 106

The required inside diameter d = 162mm

The required outside diameter D = 162.5mm = 243mm

The mass of material contained in each shaft is proportional to its cross sectional area

Asolid = x 2252 = 39760.782mm2

Ahollow = x 2432 – 1622 = 25764.987mm2

. .. x 100% = 35%

b) Solution

..........................1

............................2

...............................3

Boundary Conditions

The equation for slope at x is

........................4

The equation for deflection at x is

..........................5

The deflection at mid span can be determined from..........5

Question Four