Let us take two bars of equal lengths made by different materials. Bar (1) can support maximum axial load of 1000 N and Bar (2) can support maximum axial load of 10000 N. From the above data bar (2) supports 10 times load than that of Bar (1). Can we say ? the bar (2) is stronger than bar 1. The answer is ‘no’, because we have to consider the cross sectional areas of the two bars. The load carrying capacity per unit area is to be considered. Let the bar (1) has a cross sectional area of 20 mm2 and bar (2) has an area of 2000 mm^{2}.

Strength of bar (1) = σ_{1} = 1000 / 20 x 10^{-6} = 50 x 10^{6} N / m^{2}

and strength of bar (2) σ_{2 }= 1000 x 2000 x 10-^{6} = 5 x 10^{6} N / m^{2}

^{ }

Strength of bar (1) is 10 times that of bar (2).

Stress is the force per unit area and written as

σ (stress) = p/A

where P = force acting over area A

Equation (1.1) determines the average stress cr.

Let us take an elemental of area δA and the force δP is acting normally in the area oA.

Average stress = δP/ δA

In Fig. (1.5), the loads 1000 N and 10000 N must pass through the centroidal axis

of the bar (1) and (2) respectively. For proof let us take force dP is acting over area dA

as shown in Fig. 1.7.

**Σ** F_{Z} = 0 => P = ʃ dp

= ʃ σ dA

**Σ** M_{y }= 0 =>

Pb = x ʃ dp

= ʃ x (σ d A)

= σ ʃ x dA

b = σ ʃ x dA / p = σ ʃ x dA / σ A = ʃ x dA / A = _{xc}

x_{c} = x coordinate of the centroid of the section.

**1.3.1 Axial Loading (Normal Stress)**

Fig. 1.8 shows a bar is in tension under the action of the force P. The tensile stress

developed is given by σ_{1}= p/A , where A = cross sectional area of the bar.

The tensile force elongates the bar in axial direction (along y direction).

Fig. 1.9 shows a bar is in compression under the action of the force P. The compressive

stress developed is given by σ_{c} = P/A .

The compressive force shortens the bar in axial direction (y direction)

**UNIT OF STRESS** : The S.I. unit of stress is N/m2. Normal stresses are caused by

forces perpendicular to the areas on which they act.

**l.l.2 Shearing Stress**

Shear stress is caused by forces acting along or parallel to the area resisting the forces.

In that section of the body tends to slide past its adjacent section.

Fig. 1.10 (a) : The rivet resists shear across its cross-sectional area.

** Fig. 1.10 (b) : The bolt resists shear across two cross-sectional areas.**

A unifrom shearing stress will exist when the resultant shearing force V passe through the centroid of the cross section being sheared.

t = (shear stress ) = V / A [ t = Tau]

** Note.** The following conversion may be useful in problems involving mixed units.

1 psi = 6897 N / m^{2} = 6897 p_{a}

1 inch ^{2} = 6.4516 x 10^{-4} m^{2}

= 645.16 mm^{2}

1 ft = 0.3048 m = 304.8 mm

1 lb = 4.45 N

1 inch = 25.4 mm.

** **

** **

**1.3.3 Bearing Stress**

Bearing stress is a contact pressure between separate bodies. Refer to solved example 1.10, the bearing area is (152.4 / 1000 x c) m^{2} and the bearing force is the horizontal force P cos 0.

= 4.45 x 10^{4} cos 30° N.

**SOLVED EXAMPLES**

**(Problems in Normal Stress)**

**Example 1.1.** For the truss shown in Fig. P-1.1, determine the cross sectional areas of bars BD, BE, and CE so that the stresses wili not exceed 13794 x 10^{4} N/m2 in tension or 8276.4 x 10^{4 }N/m2 in compression. A reduced stress in compression is specified to avoid the danger of buckling.

**Solution.** METHOD OF SECTION : First let us find out the reaction RAy and RA x by

taking moment about point F

-R _{AY} x 7.32 + 7.32 + 160.2 x 4.88 = 0

R_{AY }= 160.2 x 4.88 / 7.32 = 106.8 KN

For the member forces i.e., F_{DB}, F_{EB} and F_{CE}, let us apply equilibrium conditions for a

rigid body, i.e.,

**Σ** F_{X} = 0, **Σ** F_{Y} = 0 and **Σ **M =0

Consider left half of the section for equilibrium.

Applying **Σ** F_{X} = 0,

-F_{DB} x 2.44 / 2.73 – fun x 1/2 + F_{CE}= 0

**Σ** F_{Y} = 0

-F_{DB} x 1.22 /2.73 + F_{EB} x 1/2 -160.2 + 106.8 =0

**Σ **M =0

106 x 2.44 – F_{CE} x 2.44 =0

F_{CE} = 106.8 =0

Equation (1) – F_{DB} x 2.44 / 2.73 – F_{EB} x ½ +106.8 =0

(2) x2 => – F_{DB} x 2.44 / 2.73 + 2 F_{EB} – 106.8 =0

From (3) and (4) => – F_{EB} (1/2 + 2 ) + 106.8 x 2 = 0

F_{EB} = 106.8 x 2 / 3 = 100.692 KN

F_{DB} x 2.44 /2.73 = 106.8 – 100.692 / 2 = 35.6

F_{DB} = 35.6 x 2.73 / 2.44 = 30.83 KN

Thus, Bar BD is in compression

Bar BE is in compression σ_{comp} = 8276.4 x 10_{4} N / m_{2}

Bar CE is in tension