3. Stress

Mechanics of Materials

Dr. Rami Zakaria

Reference:

1. Mechanics of Materials: R.C. Hibbeler, 9th ed, Pearson

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Term Definition Notes

Cohesive (material) Connected together / no fractions

homogeneous Made of the same material (with the same density)

Isotropic Has the same properties in all directions

prismatic (bar) Has the same cross-section along its axis

F.S. Factor of Safety

ASD Allowable Stress Design

LSD Limit State Design

Definition:

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Assumptions: 1. the material is continuous ( it consists of a uniform

distribution of matter).

2. the material is cohesive (all portions of it are connected

together, no breaks or cracks)

ΔF is

acting on

ΔA

Stress is the intensity of the internal

force acting on an area (plane) passing

through a specific point

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Normal Stress (σ): is the intensity of the force acting normal to the area.

If the normal force pulls on the area (tension) then we describe the stress as tensile stress.

If the normal force pushes on the area (compression) then we describe the stress as

compressive stress.

We can write:

Shear Stress (τ) : is the intensity of the force acting tangent to the area.

We can write the shear stress components:

The state of stress depends on the direction of the cross-section

The SI unit of stress (both normal and shear) is: N/m2

This unit is also called a Pascal (1Pa=1N/m2)

Note 1: A Pascal is a small unit, so we often use kPa, Mpa, … etc

Note 2: In the imperial unit system engineers express stress in psi (lb/in2), ksi (klb/in2)

Note 3: In some industrial applications (especially in Britain) you could see the unit Bar. 1 Bar = 100 kPa.

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Average Normal Stress

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Let’s determine the average stress distribution acting

on the cross-sectional area of an axially loaded bar.

Assumptions:

1. The bar is prismatic (all cross sections are the same)

2. the material of the bar is homogeneous (has the

same physical and mechanical properties throughout

its volume)

3. the material of the bar is isotropic (has these same

properties in all directions).

If these assumptions are true, then the bar will deform uniformly when a load P is applied.

Ref [1]

If we consider: dFFdAA ,

Note 1: As a graphical interpretation, the magnitude of the internal resultant force P is equivalent to

the volume under the stress diagram and passes through the centroid of this volume.

Note 2: Although this analysis is for prismatic bars, this assumption can be relaxed somewhat to

include bars that have small tapers or holes.

Consider equilibrium of the subject:

If P and A are constants, then so is the average stress. However if they are not constant then

it is important for an Engineer to find the Maximum Average Normal Stress by plotting the

normal force P versus its position x along the bar’s length.

Example:

9 Ref [1]

10 Ref [1]

11 Ref [1]

12 Ref [1]

13 Ref [1]

14 Ref [1]

Average Shear Stress

Large enough force F can deform the

material of this bar.

Shear forces will develop at the two

sides (double shear) V=F/2

The average shear stress distributed

over each sectioned area:

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From the equilibrium equations on a

volume element:

Check [ref.1, p.33]

How to analyse Shear Stress:

Step 1: Section the member at the

required point and draw the FBD.

Step 2: Calculate the overall shear

forces V that is necessary to hold

the part in equilibrium

(is it one shear or double shear? What is the

force direction? etc…)

Step 3: Calculate the sectioned area

A and determine the average shear

stress: τavg = V/A (this has the same direction as V)

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Problems:

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1. The supporting wheel on a scaffold is held in place on the leg

using a 4-mm-diameter pin as shown. If the wheel is subjected to

a normal force of 3 kN, determine the average shear stress

developed in the pin. (Neglect friction between the inner scaffold puller leg

and the tube used on the wheel).

3 kN

V V

3 kN+ 2V=0; V= 1.5 kN

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2. The lever is held to the fixed shaft using

a tapered pin AB, which has a mean

diameter of 6 mm. If a couple is applied to

the lever, determine the average shear

stress in the pin.

F

F

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3. (Classroom Exercise) The bar has a cross-

sectional area A and is subjected to the axial

load P. Determine the average normal and

average shear stresses acting over the

shaded section, which is oriented θ at from

the horizontal.

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4. (Classroom Exercise) If the turnbuckle is

subjected to an axial force of P= 900 lb,

determine the average normal stress

developed in section a–a and in each of the

bolt shanks at B and C. Each bolt shank

has a diameter of 0.5 in.

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5. (Classroom Exercise) If the block is

subjected to the centrally applied

force of 600 kN, determine the

average normal stress in the

material.

Ans: 5 MPa

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6. (Classroom Exercise) The chandelier is suspended

from the wall and ceiling using rods AB and BC,

which have diameters of 3 mm and 4 mm,

respectively. If the average normal stress in both

rods is not allowed to exceed 150 MPa, determine

the largest mass of the chandelier that can be

supported if θ = 45.

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7. (Classroom Exercise) The joint is subjected to an

axial force of P = 9 kN. Determine the average

shear stress developed in each of the 6-mm

diameter bolts, and the average shear stress

along each of the four shaded shear planes.

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29 Ans: 79.6 Mpa, 225 kPa

8. (Home Exercise): Determine the largest load P that can be

applied to the frame without causing either the average normal

stress or the average shear stress at section a–a to exceed σ =

150 MPa and τ=60 MPa, respectively. Member CB has a square

cross section of 25 mm on each side.

Allowable Stress Design

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When we design a structure, we should restrict the stress in the member material to a safe level.

The allowable stress restricts the maximum load on a structure.

The allowable stress should be less than the exact theoretic value using the expected loading.

There are many reasons for that:

1. The actual loadings placed on it could be different from the recommended loading

(accidently).

2. Sizes might not be exact, due to errors in fabrication or in the assembly of machine.

3. Unknown vibrations and changes in physical conditions.

4. Some materials, have high variability in mechanical properties.

We define the factor of safety (F.S.) :

; where Ffail is calculated in experimental testing.

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If the relationship between the loading and the stress is linear, as in the case σ=P/A and

τavg= V/A , then we can write the factor of safety:

or

The F.S. value depends on the application. But, of course, it should always be greater than 1

in order to avoid failure.

Allowable stress design (ASD) takes into consideration safety, environment, and economy.

Simple Connections Design

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If a member is subjected to normal force at a section, its

required area at the section is:

If a member is subjected to average shear force at a section,

its required area at the section is:

Limit State Design (LSD)*

It is a safety measure by separating between load uncertainties.

Load Factors (γ): for example let’s define possible loading types as: Dead load (D), Live

load (L), and Snow load (S).

Then we generate a load factor that reflects the separate probabilities of the different types

of load: R= 1.2D + 1.8L + 0.6S

Resistance Factors (ϕ): It relates to the materials quality, and determines the probability of

material failure. For example ϕsteel > ϕconcrete .

Design Criteria:

Φ Pn ≥ ∑ γi Ri

; Pn is the nominal strength (the load that cause failure of a member or deformation that

makes it unusable).

*Sometimes (in the USA), it is referred to as load and resistance factor design (LRFD) 34

Examples:

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1. Member B is subjected to a compressive force of

800 lb. If A and B are both made of wood and are

3/8 in. thick, determine to the nearest ¼ in. the

smallest dimension h of the horizontal segment so

that it does not fail in shear. The average shear stress

for the segment is τallow = 300 psi.

Problems:

V= 800 (5/13)= 307.7 lb

h

800 lb

V

N

3/8 in

5

12

13

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2. The lever is attached to the shaft A

using a key that has a width d and length

of 25 mm. If the shaft is fixed and a

vertical force of 200 N is applied

perpendicular to the handle, determine the

dimension d if the allowable shear stress

for the key is τallow = 35 MPa.

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3. (Classroom Exercise) The joint is fastened

together using two bolts. Determine the

required diameter of the bolts if the

failure shear stress for the bolts is τfail =

350 MPa. Use a factor of safety for

shear of F.S. = 2.5

4. (Home Exercise) Member B is subjected to a

compressive force of 600 lb. If A and B are both

made of wood and are 1.5 in. thick, determine

to the nearest 1/8 in. the smallest dimension a

of the support so that the average shear stress

along the blue line does not exceed τallow = 50

psi. (Neglect friction).

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5. (Classroom Exercise) The tension member is

fastened together using two bolts, one on

each side of the member as shown. Each

bolt has a diameter of 0.3 in. Determine the

maximum load P that can be applied to the

member if the allowable shear stress for the

bolts is τallow = 12 ksi and the allowable

average normal stress is σallow = 20 ksi.

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6. (Home Exercise): The aluminum bracket A is used to

support the centrally applied load of 8 kip. If it has a

constant thickness of 0.5 in., determine the smallest

height h in order to prevent a shear failure. The

failure shear stress is τfail = 23 ksi. Use a factor of

safety for shear of F.S. = 2.5 .

Answer: h=1.74 in