(3.7)=> µ = E / 2G -1

µ = (1 – E /3K) ½

E / 2G -1 = 1/2 – E /6K

E /2G + E /6K = 1+1/2 = 3/2

E (3K +G) / 6KG = 3/2

**SOLVED EXAMPLES**

**Example 3.1.** The short concrete post shown in Fig. P-3.1, is reinforced axially with six symmetrically placed steel bars, each 6.4516 x 10^{-4} m^{2} in section. If the applied load P is

1014.6 x 10 N, compute the stress developed in each material. Assume Est = 20691 x 107 N/m^{2}

and E,0 = 13794 x 10^{6} N/m^{2}

** **

**Solution. ****Σ**** **f_{x} = 0

** Σ **f_{y} =0 => p_{st} + p_{cv} = 1014.6 x 10^{3 }

Here, number of unknowns = 2 (P_{st }and P_{cv}).

Number of useful equation = 1

Degree of indeterminacy = 2 – 1 = 1.

Thus, the problem is statically indeterminate to the first degree. The supplementary equation is to be set up from the consideration of the deformation of the structure.

The deformation of the concrete of length L is δ_{co}=the deformation of steel for the same length L, δ_{st }

δ_{co } + δ_{st }

P_{ST} L_{ST} / A_{ST} E_{ST }= P_{CO} L_{CO} / A_{CO}E_{CO}

= 0.651 P_{CO}

P_{ST} = 0.651 P_{CO}

P_{ST} + P_{CO} = 1014.6 x 10^{3}

0.651 P_{CO }+ P_{CO } = 1014.6 x 10^{3}

P_{CO }= 1014.6 x 10^{3} / 1.651 = 614.5 x 10^{3}

σ_{co} = 614.5 x 10^{3} / [ ( 0.305)^{2} – 6 x 6.4516 x 10 ^{-4} ] N / m^{2}

= 6893 x 10^{3} N/m^{2 }.Ans

σ_{st} = p_{st} / A_{st} = 0.651 / 614.5 x 10^{3} / 6 x 6.4516 x 10 ^{-4 }

= 10.33 x 10^{7} N/m^{2}

^{ }

** **

**Example 3.2.** A copper rod is inserted into a hollow aluminum cylinder. The copper rod projects 0.127 mm, as shown in Fig. P-3.2 what maximum load P may be applied to the bearing plate ? Use the data given below.

**Solution.** We have, δ_{cu} = δ_{al} + 0.127 /1000

P_{CU} L_{CU} / A_{CU} E_{CU} = P_{al} L_{al} / A_{a}l E_{al} + 0.127 / 1000

σ_{cu }L_{CU }/ E_{CU} = σ_{al} L_{al} / E_{a}l + 0.127 / 1000

σ_{cu } = 1.7 σ_{al} + 5.85 x 10^{7}

Equation (1) is the stress relationship between copper and aluminum.

Allowable stress for aluminum.= 6.897 x 10^{7} N/m^{2}, putting this value in (1), we get

σ_{cu} = 1.7 x6.897x 10^{7} + 5.85 x 10^{7}

= 17.57 x 107 N/m2 > 13.8 x 10^{7}

Thus copper governs the equation (1).

13.8 x 10^{7 }= 1.7 σ_{al} + 5.85 x10^{7}

σ_{al }= 4.68 x 10

Thus the total safe load P = P_{cu} + P_{al}

σ_{cu} x A_{CU} + σ_{al }x A_{al}

= 13.8 x 10_{7} x 12.9 x 10 _{-4} + 4.68 x 10_{7} x 19.35 x 10 _{-4}

= 268.6 x 10_{3} N,Ans.

**Example 3.5.** A steel rod 10 m long is at a temperature of 10° C. Find the free expansion of the lengih when the temp is raised to 60° C. Find also the thermal stress produced when (i) the expansion of the rod is prevented (ii) The rod is permitted to expand by 5 mm. Take

a = 12 x 10^{-6} per °C and E = 200 GN / m^{2}

**Solution.** Free expansion δ_{st} = la (Δt)

=10 x 12 x 10 ^{-6} x (60 ^{-10})

= 6 x10 ^{-3 }m =6 mm.Ans.

(l) When the expansion is fully prevented

Thermal stress = σ_{th}

δ_{ts} = σ_{th} l / E

σ_{th} = δ_{th} E / L = 6 x 10^{-3} x 1200 x10^{9} / 10

= 12 x10^{7} N/m^{2}. Ans.

(ii) When the rod is permitted to expand by 5 mm

σt_{h} = ΔE / l = (6 -5) x 10 ^{-3} x 200 x 10^{9} / 10

= 1 x10^{7} N/m^{2}.Ans.

**Example 3.6.** A copper tube 30 cm long having a cross sectional area 18.75 cm^{2} is placed between two very rigid caps made of invar. Four 20 mm dia bolts are symmetrically arranged parallel to the axis of the tube and lightly tightened. Find the stress in the tube if the temp the assembly is raised from 15° C to 70° C. Take E,cu = 1.2 x 10^{6} kg/cm2 Es = 2.1 x 10^{6} kg/cm^{2} , acu = 16 x 10 / C and as= 12 x 10/ c. Refer Fig. P-3.13(a) & (b) of Chapter 3.

**Solution. **p_{c} = 4p_{s}

L_{C} a_{c} Δt_{c} = p_{c} L_{c }/ A_{a} . E_{c} + P_{s} L_{s} / A_{s} E_{s} + L_{s} as Δt

16 x10 ^{-6} x 55 = P_{c} / 18.75 x 1.2 x 10^{6} +

880 = P_{c} / 22.5 + P_{s} / 6.6 + 660

P_{c} 22.5 + Pc / 26 .4 = 220

P_{c} = 2672 / 18.75 = 142.5 Kg / cm^{2}.Ans.

**Example 3.7.** An invar tube with a cross sectional area of 3000 mm^{2 }and a length of 500 mm and a copper tube of identical dimensions are placed end to end at a temp of 0° C. They are then heated to 45° C and compressed by Jorces preuenting any change in their total length. What is the = 200 x 10^{3} N / mm^{2} and for copper a, = 1.65 x 10 ^{-5} / C and E, = 100 x l0^{3} N/mm^{2}

** **

**Solution**.

δ_{tCB} – δ_{CB} = δ_{CA}

or δ_{tCB} – δ_{CB} = δ_{CA}

la ( Δt) = pl _{CB} / A_{CB} E_{CB} + Pl_{CA} / A_{CA} E_{CA}

500 x 1.65 x 10 -5 x 45 = p x500 / 3000 x200 x10^{3}

P = 148 x 10^{3} N = 148 KN each forceAns.

Length of C_{A} at 45^{0} C =500 – Δca = 500 – 0.123 = 499.877 mm.Ans

Length of C_{B} at 45^{0} C = ( δtCB – Δcb) +500

= ( 500 x 1.65 x 10 ^{-5} x 45 – 0.247 ) + 500 mm

= 500.123 mm.Ans.