Chapter 1 - University of Toledo

THE UNIVERSITY OF TOLEDO CIVE 1160:001 Engineering Mechanics: Strength of Materials Instructor: Dr. Andrew G. Heydinger

Text: Mechanics of Materials, 9th edition by R.C. Hibbeler (Required) Prerequisite: Must have passed Engineering Mechanics: Statics Lectures: Attendance and participation in all lectures is strongly recommended. Assignments: Follow the instructions carefully. Course Grade: Homework problems 15% Quiz and exam problems 85% [Homework problems are graded on the basis of 10 pts each; Quiz and exam problems are graded on the basis of 20 pts each.] 1

Definition of Mechanics of Materials

Study of the interactions between bodies and the forces acting on them. 2

Description of Mechanics of Materials

• Derive relationships between applied loads and internal load intensities and deformations • Study failure mechanisms, material behavior and material properties 3

Name: Please Print THE UNIVERSITY OF TOLEDO CIVE 1160 Engineering Mechanics: Strength of Materials Instructor: Dr. Andrew G. Heydinger

The steel bar AB of the frame is pin-connected at its ends. If P = 18 kN, determine the factor of safety with respect to buckling about the y-y axis due to the applied loading. E = 200 GPa, σy = 360 MPa. 4

Depiction of Forces Acting on a Body

Fig. 1-1

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External Forces Acting on a Body Surface Forces Concentrated forces (F) Linear distributed forces (F/L) Distributed forces (F/L2) Body Forces Force due to gravity (F) (Acts through the centroid of the body.) 6

Types of Supports Table 1-1

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Example of a Mechanical System

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External Support Reactions (Q)

Question: When you studied Statics, how did you compute the support reactions?

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External Support Reactions (A) Answer: Equilibrium equations are used compute forces acting on a body. SF = 0 SMo = 0

to

Support reactions are treated as external loads acting on a body. 10

Determination of Reactions (Q&A)

Question: In Statics, what useful means did you use to show all the forces or moments? Answer: Free-body diagrams (FBD) are used to show all the forces and moments acting at a joint or on a member. 11

Application of Statics to Mechanics of Materials • In Statics, you determined external loadings acting on a body. Knowing the external loads does not tell us if the member can safely carry the loads. • For Mechanics of Materials, we will determine internal resultant loadings which are the loadings acting within a body. 12

Determining Internal Resultant Loadings (IRL) • Cut a convenient section though the body where the IRL are to be determined.

• Sketch free-body diagram showing all the external loads and the IRL acting both normal and parallel to the section. Directions of IRL are assumed. • Compute the magnitudes and the directions of the IRL. 13

Magnitudes and Directions of Internal Resultant Loadings (Q&A) Question: How can you determine the magnitudes and directions of the internal resultant loadings? Answer: Use equilibrium equations to compute the resultant internal loadings. If the resultant is negative, the assumed direction is incorrect. 14

Cutting a Section Through a Body

Fig. 1-2

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Internal Resultant Loadings

Fig. 1-2

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Types of Internal Resultant Loadings Reactions acting perpendicular to section: 1. Normal Force, N SFN = 0 (F) 2. Bending Moment, M SMN = 0 (F•L) Reactions acting parallel to section: 1. Shear Force, V SFV = 0 (F) 2. Torsional Moment or Torque, T SMT = 0 (F•L) 17

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Magnitude of IRL (Q&A) Question: Is it sufficient to determine the magnitudes and directions of the internal resultant loadings acting on a section? Answer: It is not sufficient. You need to know the intensity of the resulting forces associated with IRL. 21

Stress Definition: Stress is defined as the force intensity acting on a section. Assumption: Materials are continuous (without voids) and cohesive (all points are connected). Therefore we assume that the force intensities are distributed across the section. 22

Stress Components For every IRL, there is an either an associated normal stress (stress acting perpendicular to a section) or shear stress (stresses acting parallel to a section).

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Stress at a Point Normal Stress:   lim A0

FN dFN  A dA

Shear Stress:   lim A0

(F/L2)

Fs dFs  A dA FN

(F/L2) A

FS 24

Generalized State of Stress at a Point

Fig. 1-11

Fig. 1-12 25

Stress Notations Normal Stress, x, y or z The subscript refers to the direction that is normal to the section or plane. Shear Stress, xy, yz or zx The first subscript refers to the direction that is normal to the section or plane. The second subscript refers to the direction of the stress. 26

Sign Convention for Normal Stresses Tensile stresses are positive. Compressive stresses are negative.

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Average Normal Stress in an Axially Loaded Member The member is prismatic that is the cross section does not change so the member will deform uniformly.

Assumptions: 1. There are no localized distortions at the point of application of the load. 2. The material is homogenous (same properties throughout) and isotropic (properties same in all directions). 30

Average Normal Stress, , in an Axially Loaded Member dF   dA

 dF



 dA A

P  A P   A

P is the total force that acts through the centroid of the cross section with area A. 31

Uniform Stress Distribution What is necessary to have uniform,  = constant, stress? Moment about the x-axis or the y-axis (M R ) x   M x

(M R ) y   M y

0   ydF   ydA    ydA 0  A

A

A

 xdF   xdA    xdA A

A

A

Since both  ydA  0 and  xdA  0 , the force must act A through the centroid of the area so the stress will be uniform, equal to the force divided by the area. A

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Magnitude of the Internal Resultant Force For a member in tension or compression, the force P is equal to the volume of the stress diagram and it acts through the centroid of the area.

Fig. 1-15

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Members With Multiple Loadings For members with multiple loadings, take sections perpendicular to the longitudinal axis at various locations to determine the maximum loading. The maximum stress will be equal to the maximum force divided by the area.

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Average Shear Stress The average shear stress is computed by dividing the shear force V acting over a section by the area A.  avg

V  A

Fig. 1-20

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Single Shear Single shear occurs when a force F acts only through one section.

Fig. 1-21 38

Double Shear Double shear occurs when a force F has to act through two sections.

Fig. 1-22 39

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Force Equilibrium of a Volume Element Pure shear Volume Element Fig. 1-23 The forces acting in the z-direction are:

 yz xz  and  (xz ) ' yz

Summing forces in the z-direction:

F

z

0

 yz (xz )   yz' (xz )  0  yz'   yz  

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Moment Equilibrium

x

Fig. 1-23 Volume Element Taking moments about the x-axis:

M

x

0

  yz (xz )y   zy (yx)z  0

 zy   yz  

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Representation of Shear Stress To satisfy equilibrium:

• All four shear stresses must be equal in magnitude. • All shear stress must either point towards each other or away from each other at opposite corners of the element. 43

Complete Representation of State of Stress at a Point xy = yx yz = zy zx = xz

Fig. 1-12

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Allowable Stress And Factor of Safety Allowable stress the stress that can be used safely for design. Factor of Safety give the margin of safety expressed as a ratio. F .S . 

F fail Fallow

 fail F .S .   allow  fail F .S .   allow

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Design of Connections 1. Required cross-sectional area for members in tension. 2. Required cross-sectional area for members subjected to shear. 3. Required area for compression members supported on bearing surfaces, uses (b)allow. 4. Required area of members with axial loads to resist shear along surfaces. 46

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