FUNDAMENTALS OF THE FIRE HAZARDS OF MATERIALS

MECHANICS OF SOLIDS I CIVE 2200 - Lecture Notes Lectures 1 and 2: Stress and Strain Chapter 1 (1.3 1.7) and Chapter 2Carleton University Department of Civil and Environmental Engineering

(notes derived from Dr. Ehab Zalok / Dr. Heng Khoo) Wood and Wood Products1Chapter 1: StressBody in Static Equilibrium.Take a section across the body.Internal forces are shown on the cut surface.If the body as a whole is in static equilibrium, any part of the body is also in static equilibrium.When taking a section, internal forces of two parts always act in the opposite direction.When combined together, these internal forcescancel each other out.These internal forces vary from point to point. As the area of the surface reduces to a point,internal forces => stresses.Stress is the intensity of force per unit area ofsurface acting on infinitesimal area very small area as limit of Area 0.

2

Wood and Wood Products2

Take a section of the body normal to z-axis.Look at area A with internal force F.Resolve F into components in the x, y and z directions FX, FY, FZ.Divide components of F by A.In z-direction:

Take a limit for A goes to zero.As A reduces, Fz also gets smaller. the term approaches the force at a point.Normal stress in the z-direction

3

Wood and Wood Products3

4Normal Stress

The normal stress is the intensity of force component that is normal to the surface of section (or the surface). In this example Fz is normal to the surface of section. When it ispulling tensile stresspushing compressive stress

Wood and Wood Products45Shear Stress and

= shear stresses

They act parallel to the surface of the selection (plane).Similar to internal forces, these stresses vary from point to point.

Wood and Wood Products5General State of Stress:In addition to the section along X-Yplane, we take sections along Y-Z and X-Z planes. Limit as A 0, is the state of stress at a point.EquilibriumStresses on the back side of the surface always act in the opposite direction of those in front For 6

Wood and Wood Products6Average Normal StressAxially loaded prismatic bar (same cross-section)Normal stress

1) Load has to be applied at the centroid. No moment is generated about the centroid. Resultant load is along the longitudinal axis

Do not have axially loaded member 7

2) Assumptions: Material is homogeneous (i.e. same physical and mechanical properties throughout)Material such as wood or concrete lack homogeneity.The equation is then a measure of average stress, but it gives a good representation of the stress distribution.

3) Bearing stress:The average (nominal) stress over the surface of contact (footing, base plate).

8

Average Shear Stress Glued, Bolted, Welded Joints*Example for single shear (Shear is transferred through one surface)*Note: this is an approximation

1. Glued joint

Looking up

Looking Down 9

2. Bolted JointsExample of double shear.

Shear is transferred through two surfaces.Force is transferred by the bolt through shear over the cross-section area of the bolt.

10

double shear

d = diameter of bolt

Force is transferred between the bolt and plate through bearing. Bolt bear against the plate and vice versa

For top and bottom plates

11

Bearing stress on the bolt by the plate

Wood and Wood Products113)Welded Joints

shear along weld throat:

12

Minimum weld cross-section area occurs at the throat tt = w cos45

Allowable Stress (Section 1.6)When designing a structure, we want to ensure that it is safe against failure Failure = inability to serve its intended functionThere are many failure criteria used in designing a structure

13

So, must make sure that:stress in a member < max useful material strengthIn addition, must have safety margin to allow for:load that may be higher than anticipatedmaterial that is weaker than expected properties varies

Factor of Safety:

Max useful material strength depends on the failure criteria.

Required area 14

Example 1Two aluminum rods AB and AC have diameters of 10 mm and 8 mm respectively. Note: The allowable tensile stress for the aluminum is allow = 150MPa.Determine the largest vertical force P that can be supported.Assume that pins at A and C are adequate.Determine the minimum pin diameter at B for the force P from (a). Note: The allowable shear stress for the pin allow = 100 MPa. c) Calculate the average bearing stress on the plate D if the minimum pin diameter is used. The plate is 10 mm thick. All pinned connected.

15Assume AB fails

Assume AC fails16

FBD

From (1) and (2)

The smaller P = 7.54kN governs the design. AC will fail before AB.Therefore the largest vertical force P = 7.54 kN to ensure that both AB and BC will not fail.b) Pin at B

Minimum pin diameter, dp = 8.24mm17

18

c) Bearing on plate DExample 2Cross section area = 1000 mm2. Determine the normal and shear stress on the inclined surface.

Resolve resultant into components!Parallel and normal to surface.inclined surface area is differentfrom the cross-section area.On inclined:

19

20Example 3A round post is sitting on a 200 x 200 mm square base plate that is supported by a square footing.a) Calculate the bearing stress on the footing by the base plateb) The allowable soil bearing pressure is allow = 200 kPa, determine the size of the footing.

a) Bearing stress applied by the base plate.

b) Set b2 = allow = 200 kPa

Use footing 1600 x 1600 mm.

21Chapter 2: StrainStrain is the deformation per unit length.1) Normal strain,2) Shear Strain (Shear strain is a measure of change in angle due to deformation for two perpendicular lines (lines at right angle)

Shear strain

22ExamplePrismatic rod subjected to load P = 20 kN, and has a normal strain = 0.0002

What is the extension of the rod?

Similar to the general state of stress which has six components, the general state of strain also has six components: