## LESSON 24.

24.1 MOMENT OF RESISTANCE

(A)

(B)

(C)

Fig.24.1

Bending equation gives  or ${M \over {I}} = {{{f_{all}}} \over {{y_{max}}}}$  or M = ${I \over {{y_{max}}}}$ fall

MR = Z fmax

This  quantity "ZF" is termed as "Moment of Resistence", which totally depends on allowable stress fall ans section modules Z.

• However, it would be necessary that moment of resistence is always greater than or equal to the external bending moment.

24.2 SECTION MODULUS

It is a property of sectional area.

We know that  ${M \over {I}} = {f \over y}$

Or f = ${M \over {I/y}}$  and fmax = ${M \over {I/{y_{max}}}}$

Where ${I \over {{y_{max}}}}$  is called section modulus ‘Z’.

Therefore  Z =  ${I \over {{y_{max}}}}$

fmax = ${M \over Z}$   which is similar to f =  ${P \over A}$    (in simple tension and compression)

• For same area of cross-section, but with different shapes, the section modulus will be different.

• In bending it would be desirable to choose sections which give the maximum value of section modulus.

• I sections have been found to give the maximum section modulus for a specific area of cross-section and hence should be preferred for bending loads.

7.8 SECTION MODULUS FOR VARIOUS SHAPES OF BEAM SECTIONS

(i) Rectangular Section

Fig.24.2 shows a rectangular section of width b and depth d. Let the horizontal centroidal axis be the neutral axis.

Fig.24.2

Section Modulus = Z = ${{Moment of inertia about the neutral axis} \over {Distance of most distant point from the neutral axis}}$

=  ${I \over {{y{max}}}}$

I =  ${{b{d^3}} \over {12}}$  and ymax =  ${d \over 2}$

Z ${{b{d^3}} \over {12}}$  . ${2 \over d}$  = ${{b{d^2}} \over {6}}$

Let f be the maximum stress offered by the beam section.

Therefore moment of resistance = M = f Z =${{b{d^2}} \over {6}}$

Or M =  ${1 \over 6}$  f bd2

(ii) Hollow Rectangular Section

Fig.24.3 shows a hollow rectangular section. Let the overall width and depth be B and D. Let the width and depth of the centrally situated rectangular hole be b and d.

Fig.24.3

Moment of inertia about the neutral axis = I =  ${{B{D^3}} \over {12}}$  -  ${{b{d^3}} \over {12}}$  =   ${1 \over 12}$  [BD3 - bd3]
ymax =   ${D \over 2}$

Therefore, Section Modulus = Z = ${I \over {{y_{max}}}}={1 \over {12}}(B{D^3}-b{d^3}$ )  ${2 \over D}$

Z = ${{B{D^3} - b{d^3}} \over {6D}}$

If f be the maximum bending stress the moment of resistance = M = f Z

M =  ${1 \over 6}f\left( {{{B{D^3} - b{d^3}} \over D}} \right)$

(iii) Circular Section

Fig. 24.4

Let the diameter of the section be d. Moment of inertia of the section about the neutral axis = I =  ${{\pi{d^4}} \over {64}}$

ymax ${d \over 2}$

Section Modulus = Z =  ${I \over {{y_{max}}}}$  = ${{\pi{d^3}} \over {32}}$

If  f be the stress offered by the section, the moment of resistance = M = f Z

= f  ${{\pi{d^3}} \over {32}}$

(iv) Hollow Circular Section

Fig.24.5

Fig.24.5 shows a hollow circular section of external diameter D and internal diameter d.

Moment of inertia about the neutral axis = I =  ${\pi\over {64}}\left[ {{D^4} - {d^4}} \right]$

ymax ${D \over 2}$

Therefore, section modulus = Z${I \over {{y_{max}}}}$   =  ${{\pi \left({{D^4} - {d^4}} \right)} \over {32D}}$

If f be the maximum stress offered by the section, the moment of resistance = M = f Z

M = f   ${{\pi \left({{D^4} - {d^4}} \right)} \over {32D}}$

Example 24.1: Fig.24.6 shows the section of a tube of aluminium alloy. Determine the maximum moment that can be applied to the tube if the permissible bending stress is 100 N/mm2. Find also the radius of curvature of the tube as it bends. Take E = 72800 N/mm2.

Fig.24.6

Solution: Moment of Inertia of the section,

I = ${{60 \times {{100}^3}} \over {12}} - {{50 \times {{90}^3}} \over {12}} = 1962500$

= 1962500 mm4

Maximum moment on the section, M = ${f \over {{y_{max}}}}.I = {{100 \times 1962500} \over {50}}$ Nmm

= 3925000 Nmm

Radius of Curvature, R =  ${{EI} \over M} = {{72800 \times 1962500} \over {3925000}}$  mm

= 36400 mm

Example 24.2: A cast iron test beam 25mm × 25mm in section and 2 m long and supported at the ends fails when a central load of 700N is applied. What uniformly distributed load will break a cantilever of the same material 55mm wide, 110mm deep and 3m long?

Solution: Let us first consider the test beam.

Maximum bending moment = M = ${{WL}\over 4}={{700\times 2}\over 4}\times 1000Nmm$   = 35 × 104 Nmm

Moment of Resistance, R = ${1 \over 6}.fb{d^2}$  = ${1 \over 6}.f \times 25 \times {25^2}$ =  ${{15625} \over 6}f = 2604.17fNmm$

Fig.24.7

Equating the moment of resistance to the maximum bending moment

$2604.17f=$   35 × 104

f = ${{35{\rm{}} \times {\rm{}}{{10}^4}{\rm{}}} \over {2604.17}}$  = 134.39 N/mm2

Now let us consider the cantilever.

Let the distributed load on the cantilever be w N/m run so as to break it.

Therefore, maximum bending moment = M =  ${{w{l^2}} \over 2} = {{w \times {3^2}} \over 2} \times 1000Nmm$  = 4500w N mm

Moment of Resistance of the section =  ${1 \over 6}.fb{d^2} = {1 \over 6} \times 134.39 \times 55 \times {110^2}Nmm$

= 14.91 × 106 N mm

Fig.24.8

Equating the maximum bending moment to moment of resistance we have,

4500 w =14.91 × 106

w= 3313 N/m

Example 24.3: A machine component of semi circular section 300mm diameter acts as a beam of span 1.50m. It is placed with its base horizontal. If it carries a uniformly distributed load of 150kN/m run on the whole span, find the maximum stress induced.

Solution: Moment of inertia of the section about the neutral axis = I = 0.00686 d4

Extreme fibre distance from the neutral axis = r -  ${{4r} \over {3\Pi }} = {{3\Pi- 4} \over {4\Pi }} \times r = 0.5756$

r = 0.2878 d

Section modulus, Z =  ${{0.00686{d^4}} \over {0.2878d}} = 0.0238{d^3}$

Maximum bending moment, M =  ${{300 \times {{1.50}^2}} \over 8} = 84.375kNm$

Maximum bending stress  ${f_{max}}={M \over Z}={{84.375\times{{10}^6}}\over{0.0238\times{{300}^3}}}=131.30$  N/mm2

Example 24.4: Find the safe concentrated load that can be applied at the free end of a 2m long cantilever. The section of the cantilever is a hollow square of external side 50mm and internal side 40mm the safe bending stress for the material being 65 N/mm2.

Solution: Moment of inertia of the section of the cantilever,

I = ${{{{50}^4}} \over {12}}-{{{{40}^4}} \over {12}}=307500$  mm4

Section modulus, Z = ${I \over {{y_{max}}}}={{307500} \over {30}}$   = 10250 mm3

Let the safe concentrated load at the free end be W Newton

Maximum B.M, M = W × 2 × 1000 = 2000 W Nmm

M = f Z, 2000 W = 65 × 10250

W   = ${{65 \times 10250} \over {2000}}=$  333.125 N

Example 24.5: The moment of inertia of a beam section 450mm deep is 60.5 × 107 mm4. Find the span over which a beam of this section, when simply supported, could carry a uniformly distributed load of 45kN per meter run. The flange stress in the material is not to exceed 110 N/mm2.

Solution: Section modulus of section = Z =  ${I \over {{y_{max}}}} = {{60.5 \times {{10}^7}} \over {250}}$

Therefore, Z = 24.2 × 105 mm3

Let the maximum span be l meter.

Therefore, maximum bending moment = M = ${{w{l^2}} \over 8}={{45000\times{l^2}} \over 8} \times 1000Nmm$

= 56.25 × 105 l2 N mm

Moment of resistance of the section corresponding to the maximum bending stress of 110 N/mm2

= f Z = 110 × 24.2 × 105 N mm

Equating the maximum bending moment to the moment of resistance, we get

56.25 × 105 l2 = 110 × 24.2 × 105

l2 =  ${{110\times 24.2\times{{10}^5}} \over {56.25\times{{10}^5}}}=47.32$

Therefore, l = 6.88 m, say 7 m