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Relationship Between E, G, and K

(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 m2 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/m2

and E,0 = 13794 x 106 N/m2

 

Solution.             Σ fx = 0

     Σ fy =0  => pst + pcv = 1014.6 x 103

 

Here, number of unknowns = 2 (Pst and Pcv).

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 δco=the deformation of steel for the same length L, δst

δco  + δst

PST LST / AST EST = PCO LCO / ACOECO

= 0.651 PCO

PST = 0.651 PCO

PST + PCO = 1014.6  x 103

0.651 PCO + PCO  = 1014.6  x 103

PCO = 1014.6 x 103 / 1.651 = 614.5 x 103

σco = 614.5 x 103 / [ ( 0.305)2 – 6 x 6.4516 x 10 -4 ] N / m2

= 6893 x 103 N/m2 .Ans

σst = pst / Ast = 0.651 / 614.5 x 103 / 6 x 6.4516 x 10 -4

= 10.33 x 107 N/m2

 

 

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,                                      δcu = δal + 0.127 /1000

PCU LCU / ACU ECU = Pal Lal / Aal Eal + 0.127 / 1000

σcu  LCU / ECU = σal Lal / Eal + 0.127 / 1000

σcu  = 1.7 σal + 5.85 x 107

 

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

Allowable stress for aluminum.= 6.897 x 107 N/m2, putting this value in (1), we get

σcu = 1.7 x6.897x 107 + 5.85 x 107

= 17.57 x 107 N/m2 > 13.8 x 107

 

Thus copper governs the equation (1).

13.8 x 107 = 1.7 σal + 5.85 x107

σal = 4.68 x 10

 

Thus the total safe load P = Pcu + Pal

σcu  x ACU + σal x Aal

= 13.8 x 107 x 12.9 x 10 -4 + 4.68 x 107 x 19.35 x 10 -4

= 268.6 x 103 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 / m2

 

Solution. Free expansion  δst = 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                             = σth

δts = σth l / E

σth = δth E / L = 6  x 10-3 x 1200 x109 / 10

= 12 x107 N/m2.  Ans.

 

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

σth = ΔE / l  = (6 -5) x 10 -3 x 200 x 109 / 10

= 1 x107 N/m2.Ans.

 

Example 3.6. A copper tube 30 cm long having a cross sectional area 18.75 cm2 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 106 kg/cm2 Es = 2.1 x 106 kg/cm2 , acu = 16 x 10 / C and as= 12 x 10/ c. Refer Fig. P-3.13(a) & (b) of Chapter 3.

 

Solution.                        pc = 4ps

LC ac Δtc = pc Lc / Aa . Ec + Ps Ls / As Es + Ls as Δt

16 x10 -6 x 55 = Pc / 18.75 x 1.2 x 106  +

880 = Pc / 22.5 + Ps / 6.6 + 660

Pc 22.5 + Pc / 26 .4 = 220

Pc = 2672 / 18.75 = 142.5 Kg / cm2.Ans.

 

Example 3.7. An invar tube with a cross sectional area of 3000 mm2 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 103 N / mm2 and for copper a, = 1.65 x 10 -5 / C and E, = 100 x l03 N/mm2

 

Solution.

δtCB  – δCB = δCA

or                                              δtCB  – δCB = δCA

la ( Δt) = pl CB / ACB ECB + PlCA / ACA ECA

500 x 1.65 x 10 -5 x 45 = p x500 / 3000 x200 x103

P = 148 x 103 N = 148 KN each forceAns.

 

Length of CA at 450 C                         =500 – Δca = 500 – 0.123 = 499.877 mm.Ans

Length of CB at  450 C                  = ( δtCB – Δcb)  +500

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

= 500.123 mm.Ans.