MECHANICS OF SOLIDS _ UNIT_01_16 MARK QUESTIONS:

16 MARK QUESTIONS:

1. A circular rod of diameter 20 mm and 500mm long is subjected to a tensile force of 45 KN. the modulus of elasticity  may be taken as 200 KN/mm². Find stress, strain and elongation of the bar due to the applied load.  [APR/MAY 2011]

Given data:

                                                         

 Load p =45 KN = 45 x 10³ N                             

Young’s modulus E = 200 KN /mm²

Length of the rod, l = 500 mm

Diameter of the rod = 20 mm

                                                   

                                                                                             

Data asked:

Stress, strain, elongation                                                    CIRCULAR ROD

Solution:

Cross sectional area; A = πd²/4 =π x 20²/4 = 314.159 mm²

Stress σ =load / area =45x10³/314.5 = 143. 24 N/mm²

Strain, e = stress / young’s modulus =143.24/200x10³ = 0.0007162

Elongation δl=Pl/AE = (45x10³ x 500) / (314.159 x 200x10³⁾ = 0.358 mm

Result:

Stress σ = 143.24 N/mm²

Strain = 0.0007162

Elongation δl = 0.358 mm

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2. A hollow steel rod of tube is to be used to carry an axial compressive load of 140 KN. The yield stress for steel is 250 N/mm². A factor of safety of 1.75 is to be used in the design. The following three classes of the tubes of external diameter 101.6 mm are available. Which section do you recommended from 3.65 mm, 4.05mm and 4.85 mm?

Given data:

Yield stress = 250 N/mm²

Factor of safety = 1.75

External diameter of the tube = 101.6 mm

Load P = 140 KN = 140x10 N       

                                                                                                       

                                                                    

                                                                                      

                                                                                                        HOLLOW STEEL ROD

Data asked:

Which section do you recommended from the following given thickness 3.65 mm, 4.05mm, 4.85mm

Solution:

Working stress or permissible stress σ = yield stress/FOS = 250/1.75 = 142.85 N/mm²

Cross sectional area A = load / working stress = 140x10 /142.85 = 980 mm²

Also area of the hollow tube = π/4(D²-d²) = 980

π/4 (101.6 – d²) = 980  or  d² = 9074.78

d = 95.26 mm, thickness   t = (D – d)/2 = (101.6 – 95.2)/2 =3.169 mm

Hence light section, t = 3.65 mm is enough.

Result:

Thickness t = 3.65 mm is enough

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3. The safe stress for a hollow steel column which carries an axial load of 2.1x10 KN is 125 MN/m². If the external diameter of the column is 30 cm. determine the internal diameter. [NOV/DEC-2007]

Given data:

Axial load P = 2.1 x 10 KN

Stress = 125 MN/m² = 125 N/mm²

External diameter of the column=30 cm = 300mm

                                                                                              

                                                                            

Data asked:

Internal diameter of the column d

Solution:

Area   A = load/stress =2.1x10⁶/125 =16800 mm2                 HOLLOW STEEL COLUMN

But area of the hollow A = (π/4) (D² – d²) = 16800

                                          = (π/4) (300 ²- d²) = 16800

                                          d² = 68598.73;

                                           d=262mm

Result:

Internal diameter of the column d = 262 mm

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4. Define the elastic constants and write the relationship between them Modulus of elasticity, modulus of rigidity and bulk modulus are the three elastic constants. [NOV/DEC-2007]

Modulus of elasticity:   it is defined as the ratio of stress to strain within the elastic limit and is usually denoted by letter ‘E ‘

Modulus of elasticity E = stress/strain = σ/e, N/mm²

Modulus of rigidity:     it is defined as the ratio of shearing stress to shearing strain within the elastic limit and is denoted by the letter ‘G’ or ‘N’

Modulus of rigidity G = shear stress/ shear strain = τ / Ф

Bulk modulus:                                                                                                                             

It is defined as the ratio of identical pressure ‘P’ acting in three mutually perpendicular directions to corresponding volumetric strain is denoted by letter ‘K’

Bulk modulus K = pressure ‘p’ acting in three mutually perpendicular direction/volumetric strain=p/e_v

Relationship between the three elastic constants:                                                      

Wkt the relationship between elastic constant and rigidity modulus E = 2G (1+1/m) ------- (1)

The relationship between elastic constant and bulk modulus K = 3K (1- 2/m) --------------- (2)

From the equation (1) we get    1/m = (E/2G)-1   substituting it in equation (2)

We get, E = 3K (1- 2((E/2G)-1))

                  = 3K (1-(E/G) +2)

                  = 3K (3-(E/G)) = 9K-(3KE/G)

E (1+3K/G) = 9K   ,   E ((E+3K)/G) = 9K

E = 9GK/ (G+3K)

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5. A bar of 20 mm diameter is tested in tension. It is observed that when a load of 37.7 KN is applied, the extension measured over  a gauge length of 200 mm is 0.12 mm and contraction in diameter is 0.0036 mm. find the poison’s ratio and elastic constants E, G and K. (MAY 2002)

Given data:

Load    P = 37.7 KN = 37.7x10³ N

Length   l = 200mm

Extension    δl =0.12mm                                  

And contraction in diameter δd = 0.0036 mm

Data asked:

Poison’s ratio and the value of three elastic constants                                                                                
   BAR

Solution:

Area A = (π/4) d² = (π/4) x20²= 314.5 mm²

Linear strain, e = δl/l =     0.12/200 = 0.0006

Lateral strain e_l = δd/d = 0.0036/20 =0.0018

Wkt,   δl = Pl/AE    or we can change it as E = Pl/A δl = (37.77x10³x200) / (314.5x0.12)

                                                                                              =200004.71 N/mm²

E = 2G [1+ (1/m)]    or G = E/2[1+ (1/m)] = 200004.71/2(1+0.3)

Rigidity modulus G = 76924.89 N/mm²

Also we know that    E = 3K (1 – (2/m))   or K = E /(1 – (2/m))

K = 200004.71/3(1- 2x0.3) = 166670.59 N/mm²

Result:

Young’s modulus E = 200004.71 N/mm²

Rigidity modulus G = 76924.89 N/mm²

Bulk modulus K = 166674.59 N/mm²

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6.A circular rod of 100 mm diameter and 500mm long is subjected to a tensile force of 1000 KN .Determine the modulus of rigidity , bulk modulus and change in volume if poison’s ratio = 0.3 and young’s modulus E = 2x10⁵ N/mm² (APR/MAY 2005)

Given data:

Diameter of rod d = 100 mm

Length of the rod l = 500 mm

Tensile force P = 1000 KN = 1000 x 10³ N       

Poisson’s ratio 1/m = 0.3                                                   

Young’s modulus E = 2x10⁵ N/mm²

Data asked:

Modulus of rigidity,

bulk modulus and

change in volume                                                              

                          

                                                      

Solution:                                                                            CIRCULAR ROD

Wkt   E = 2G [1+ (1/m)] = 3K (1 – (2/m))

Rigidity modulus G = E/2[1+ (1/m)] = 2x10 ⁵/2(1+0.3) = 0.7692x10 ⁵ N/mm²

 Bulk Modulus K = 2x10⁵/3(1- 2x0.3) = 1.6667x10⁵ N/mm²

Longitudinal stress = P/A =1000x10³/ (π/4) x 100²  =127.32 N/mm²

Linear strain e_x = stress/young’s modulus =127.32/2x10 ⁵ = 63.66x10 ^-5

Lateral strain    e_y = - (1/m) e_x and ez = - (1/m) e_x

wkt    volumetric strain  e_v  =  e_x + e_y + e_z = e_x(1 – (2/m))

Volumetric strain e_v = 63.66x10^-5(1 – 2x0.3) = 25.46x10^-5

Also wkt    volumetric strain e_v   = δV/V

Therefore change in volume, δV = e_v x V =25.66x10^-5 x (π/4) x d²x l

 = 25.66x10^-5 x (π/4) x100² x500 = 1000 mm³

Result:

Modulus of rigidity G = 0.7692x10⁵ N/mm²

Bulk modulus K = 1.6667x10⁵⁵ N/mm²   and

Change in volume δV = 1000 mm³  

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7. An axial pull of 35000N is applied on a bar consists of three lengths as shown in fig . the young’s modulus E = 2.1x`10⁵ N/mm².determine  (1) stresses in each section (2) total extension of the bar [NOV/DEC-2007 & 2012]

Given data:

Axial pull P = 3500 N

Young’s modulus E = 2.1x10⁵ N/mm²

Data asked:

Stresses in each section

and total extensions of the bar

                                                                                                      

                                                    

Solution:

Tensile stresses section 1

= load/area at section 1                                              

= P/A = 35000/ (π/4) x20²                                                              STEPPED BAR

= 111.46 N/mm²

Tensile stresses in section 2

= load/area at section 2

= P/A = 35000/ (π/4) x30²

 = 49.53 N/mm²

Tensile stress in section 3

= load/area at section 3

= 35000/ (π/4) x50²

= 17.83 N/mm²

Total extension of the bar δl

= (P/E) [l₁/A₁+l₂/A₂+l₃/A₃]

= (35000/2.1x10⁵) [(200/314) + (250/706.5) + (220/1962.5)]

δl = 0.1838 mm

Result:

Tensile stress in section1 = 111.46 N/mm²

Tensile stress in section2 = 49.53 N/mm²

Tensile stress in section3 = 17.43 N/mm²

Total extension δl = 0.1838 mm

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8. The load ‘p’is applied on the bars as shown in fig. find the safe load ‘p’ of the stresses in the brass and steel are not to exceed 60 N/mm² and 120 N/mm² respectively.  E –for the steel = 200N/mm² and  E- for brass = 100 KN/mm².the copper rods are 40mm x 40 mm in section and the steel rod is 50mm x 50mm in section length of the steel is 250mm and copper rod is 150mm [NOV/DEC-2009]

  

COMPOSITE BAR

Given data:

Stress in brass σ_c = 60 N/mm²                                                

Stress in steel σ_s = 120 N/mm²

E –for steel E_s = 200KN/mm² =200x10³ N/mm²

E –for brass E_c = 100KN/mm² = 100x10³ N/mm²

Cross section of the copper rod = 40mmx40mm

Cross section of the steel rod = 50mmx50mm

Length of the copper rod l_c = 150mm

Length of the steel rod   l_s = 250mm                                                   

Asked:

Find the safe load ‘P’

Solution:

Area of the copper component A_c = 2(40x40) = 3200mm²

Area of the steel component A_s = 50x50 = 2500 mm²

Decrease in length of copper = decrease in length of steel

δls =δl_c    ,  e_cl_c = e_sl_s    or    (e_s /e_c) = (l_c/l_s)= 150/250=0.6

 (σ_s/σ_c)  = (e_s/e_c)(E_s/E_c) = 0.6x(200x10³/100x10³) = 1.2

 σ_s=1.2σ_c

Where P_c-reaches 60 N/mm² P_s- will reach 1.2x60 N/mm² which is less than the permissible value.

Where         P = σ_c A_c + σ_s A_s = (60x3200) + (72x2500) = 372000 N = 372 KN

Result:

The load is applied on the bar ‘P’ = 372 KN

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9. A steel rod of 25mm diameter is placed inside a copper tube of 30mm internal diameter and 5mm thickness and the ends are rigidly fixed. The assembly is subjected to a compressive load of 250KN. Determine the stresses induced in the steel rod and copper tube. Take the modulus of elasticity of the steel and the copper as 200 GPa and 80 GPa respectively.   [NOV/DEC-2006]

Given data:

Diameter of the steel rod = 20mm

 Inside diameter of the copper tube   d = 30mm

Outside diameter of the copper tube D = 40mm 

(ie calculated from the given thickness  t = 5 mm)

                                                                                                            

                                                   

Asked:

Stresses induced in the steel rod and copper tube                                   STEEL ROD

Solution:                                                                              

Area of the steel rod A_s = (π/4) d² = (π/4) x25² = 490.6mm²

Area of the copper tube A_c = (π/4) (D² – d²) = (π/4) (40² – 30²) = 549.5 mm²

From the static equilibrium condition

P_s + P_c = P ----------- (1)

Where Ps –load on steel and P_c—load on copper

Wkt,     δl_s = δl_c      or (P_s l/A_sE_s) = (P_c l/A_cE_c)

Both sides length are same since we can eliminate both side ‘l ’

(P_s/A_sE_s) = (P_c//A_cE_c)   or Ps  = Pc [(A_sE_s)/ (A_cE_c)]

P_s  = Pc [(490.6x200x10³)/ (549.5x80x10³)] = 2.23P_c

P_s   = 2.23Pc   substitute this on equation (1) we get

2.23P_c + P_c = 250x10³

3.23P_c = 250x10³   or Pc = 77.39KN 

The remaining value from ‘P’ is   P_s       i.e., Ps = P – P_c = 250x10³ – 77.29 = 172.61KN

Stress in steel rod σs = load on steel /area of steel rod = P_s /A_s = 172.61x10³/490.6= 351 N/mm²

Stress in copper tube σc = load on copper tube/area of copper tube= P_s/A_s

=77.39x10³/549.5=140.8 N/mm²

Result:

The stress on the steel rod   σ_c= 351 N/mm²

The stress on the copper rod   σ_c = 140.8 N/mm²

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10. Find the total strain energy stored in a steel bar of diameter 50 mm and length 300mm when it is subjected to an axial load of 150 KN. Take the modulus of elasticity of steel as 200x10 3MPa [NOV/DEC-2006]

Given data:

Diameter of the steel rod d = 50mm

Length of the steel rod   l = 300mm

Axial load on the steel rod P = 150 KN= 150x10³ N                                 

                                                                  STEEL BAR

Modulus of elasticity of steel E = 200x10³MPa =200x10³ N/mm²

Asked:

Total strain energy stored

Solution:

Total strain energy stored = (σ²/2E) xAl

Stress in the bar σ =load /Area= 150x10³/ (π/4) d²= 150x10³/(π/4)x50² =76.43 N/mm²

                                               (76.43)²                                        

Total strain energy stored =                  x (π/4) x50²x300 =8598 N mm

                                             2x200x10³

Result:

Total strain energy stored = 8598 N mm

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