USA: +1-585-535-1023

UK: +44-208-133-5697

AUS: +61-280-07-5697

Working Stress and Factor of Safety

The working stress or allowable stress is the maximum safe stress a material may carry.

The working stress should not exceed proportional limit. Since the proportional limit is difficult to determine accurately, we take yield point or the ultimate strength and divide this stress by a suitable number N, called the factor of safety. Thus,

σW ( Working stress) = σ p / Np

σW = σult / Nult

 

We use yield point stress for calculating σw for structural steel design. For other materials, the ultimate strength is considerea for calculating crw.

 

SOLVED EXAMPLES

 

Example 2.1. Dur ing a stress stra in test, the unit defor ma tion at a str ess of 34485 x 103 N/m2 was obser ved to b e 0.000167 m/m, and a t a stress of 13794 x 104 N/m2 it was 0.000667 m/m. If the pr opor tional limit is 20691 x 104 N/m2 ; (i) Wha t is the modulus of ela sticity ? (ii) Wha t is the stress corresponding to a strain of 0.0002 m/m ? (iii) Would these results be valid if the propor tional limit wer e 12414.6 x lcf N/m2 ?

 

Solution. Formula [ δ = PL /AE]

(i)                 Here given  δ / L = 0.000167

P /L = stress = σ = 34485 x 103 N /m2

E = ( P /L ) ( L /δ) = 34485 x 103 / 0.000167 =  206 .5 x 109 N /m2

Again for                       δ / L = 0.000667

and                              P / L = 13794 x 104

E = (P /A) (L / δ) = 13794 x 104 / 0.000667 = 206.5 x 109 N /m2

E = 2.065 x 1011 N/m2.Ans

(ii)                                     δ / L = 0.0002

(P / A) = ? , E = 2.065 x 1011 N/m2

p/A = (E) (δ / L) = 2.065 x 1011 x 0.0002

= 4.13 x 107 N/m2.Ans.

 

(ii)        (a) 12414.6 x 104 N/m2 < 13794 x 104 N/m2. Thus the stress of 13794 x 104 will not

follow the hook’s law. So for this case the result is invalid.

 

(b) 34485 x 103 N/m2 < 12414.6x 104 N/m2 In this case the result will be valid because

the stress level of 34485 x 103 is within the proportional limit.

 

Example 2.2. A steel rod having a cross sectional area of 3.2258 x l0-4 m2 and a length

of 182.88 m is suspended vertically. It supports a load of 22.25 x 103 N ut the lower end.if steel

weighs 7.7 x 104 N/m3 and E = 20691 x 107 N/m2. Find the total elongation in the rod.

 

Solution. Formula               δ = WL / 2 AE = wL2 / 2E

δ = wL2 / 2 E = 7.7 x 104 x (182.88)2 / 2 x 20691 x 107 = 0.06223 m.

 

Example 2.3. An aluminum bar having a cross-sectional area of 1.613 x 1 0-4 m2 carries the axial loads applied at the positions shown in Fig. P-2.3. If E = 6897 x 107 N/m2, compute the total deformation of the bar.

Solution. Formula [ δ = PL / AE]

 

δ1 = 35.6 x 103 x 1.22 / 1.613 x 10 -4 x 6897 x10 -3 m = 3.9 mm

 

δ2 = 22.25 x 103 x 1.52 / 1.613 x 10 -4 x 6897 x 107 m = 3 mm

 

δ3 = 8.9 x 103 x 0.92 / 1.613 x 10 -4 x 6897 x 107 m = 0.74 x 10-3 m = -   0.74 mm

 

total deformatio  =      δ = δ1 +  δ2 +  δ3

= (3.9 + 3- 0.74) mm

= 6.16 mm. Ans.

 

Example 2.4. A uniform concrete slab of total weight W is to be attached, as shown in Fig. P-2.4, to two rods whose lower ends are on the same level. Determine the ratio of the areas of the rods so tlwt the slab will remain level.

Taking moment about C,

 

-p1 x 1.83 +1.22 W =0

P2 = W –P1 = W – 0.67 W = 0.33 W =W

P1 /P2 = 0.67 / 0.33 = 2

δal  = p1 x 1.22 / Aal x 6897 x 107

δst  = p2 x 1.83 / Aal x 20691 x 107

 

given ,                                                 δal  = δst

 

p1 x 1.22 / Aal x 6897 x 107  = p2 x 1.83 / Aal x 20691 x 107

Aal / Ast  = p1 /p2 x 1.22 x 20691 / 1.83 x 6897 = 2 x 1.22 x 20691 / 1.83 x 6897

 

 

Example 2.5. The rigid bars shown in Fig. P-2.5 are separated by a roller at C and hinged at A and D. A steel rod at B helps to support the load of 53.4 x 103 N. Compute the vertical displacement of the roller at C.

We have drawn the free body diagrams of all parts.

For bar (1)                   Σ fx = 0

RAX = 0

Σ  Fy = 0

RAY + T – RC = 0

RC = T + RAY

Σ M A =0

- RC x 2.75 + 1.83 T  =0

RC = 1.83 / 2.75 T or T = 1.5 RC

For Bar (2),                   Σ MD = 0

-RC x 2.44 + 53.4 x 103 x 1.22 = 0

RC = 26.7 x  103 N

 

or  (2)                                           T = 1.5 RC = 1.5 x 26.7 x103 N = 40.05 x 103 N

δB  = 40.05 x 103 x 3.048 / 3.23 x10 -4 x 20691 x 107 = 1.826 x 10 -3 m.

δc =?

δc = δB = 1.83 + 0.2 / 1.83 = 2.75 /1.83

 

δc  = 2.75 / 1.83 x 1.826 x 10 -3 m = 2.744 x 10-3 m

 

 

Example 2.6. A specimen of any given material is subjected to a uniform triaxial stresses of magnitude σ each. Determine the theoretical maximum value of poisson’s ratio µ.

 

ε x + ε y + ε z = 1/ E [ σ x + σy z - 2µ (σ x + σy z) ]

ε + ε + ε  = 1 / E [ σ + σ + σ - 2µ x 3σ]

3ε = 1 / E [ 1 - 2µ] 3σ

ε = σ / E (1 – 2µ) All σ are tensile in nature

 

1 – 2µ = E ε / σ > 0 or 1 – 2µ > 0

µ = < ½

 

The theoretical maximum value of µ = 1/2 Ans.

 

Practically,                                      µ = 0.25 to 0.30 for steel,

µ = 0.33 for most other metals ; and

µ = 0.20 for concrete.

 

Example 2.7. A solid aluminum shaft of 76.2 mm dia fits concentrically in a hollow steel tube. Compute the minimum internal diameter of the steel tube so that no contact pressure exists when the aluminum shaft carries an axial compressive load 377360 N. Assume µ = 1/3  and Eal = 6897 x 1 07 N/m2.

 Solution.

εx = µ . σz / E = + 1 /3 x 82.75 x 106 /6897 x 107 = 4 x 10 -4 m/m.

εx = 4 x 10-4 m/m.

 

Therefore, the require diametrical clearance is

δ = ε L

δ = 4 x 10-4 mm = 0.0305 mm.

 

The required internal diameter of the steel tube is found by adding the clearance to

the original diameter of the aluminum shaft,

:. Minimum internal dia = 76.2 + 0.0305 mm

= 76.2305 mm. Ans.