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Chapter 9 - Simple StressesReading: Chapter 9 - Pages 323—337

9-1Introduction

The intensities of the internal resisting force are called stresses.

StaticsAll bodies are assumed to be rigid.

Strength of MaterialsBodies are considered deformable.Deformation per unit length is called the strain.

Two properties that affect how a body (member) will react to loading:1. The Material2. Shape

Definitions:Strength Load carrying capacity based on stresses inside a memberStiffness Ability to resist deformation. Deformation characteristics. (Serviceability)Stability The ability of a slender member to maintain its initial configuration without buckling while

being subjected to compressive loading.

9-2Normal and Shear Stresses

Stress ≡ the intensities of internal forces per unit area

There are two types of stresses:1. Normal Stresses (σ) - are caused by internal forces normal (perpendicular ) to the area in question.2. Shear Stresses (τ) - are caused by internal forces tangential (parallel ) to the area under question.

¾ Stresses are the most important concepts in the study of strength of materials. ¾ Whenever a body is subjected to external loads, stresses are induced within the body. ¾ Whether the material will fail and to what extent it will deform depends on the amount of stresses

induced within the body.

Units of Stress

U.S. Customary Units S.I. UnitsPounds per Square Inch (psi) Newton's per square meter (N/m2)Pounds per Square Foot (psf) N/m2 = Pascal (Pa)Kips per Square Inch (Ksi) 1 kPa = 103 Pa 1 MPa = 106 Pa 1 GPa = 109 Pa

P

A

B

C

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9-3Direct Normal Stresses

Normal Stress (σ) - intensity of internal force perpendicular to the area under question.

(a) If the bar is in equilibrium, any segment must be in equilibrium when cut by a transverse plane.

(b) Equilibrium conditions requires that the internal force in the section be equal to the external force P.

(c) The internal force is normal (perpendicular ) to the section, the stress induced is the normal stress.

Normal stresses due to axial loads through the centroids of the cross-sections are usually distributed uniformly over a cross section.

Direct Normal Stress Formula

Where σ = the normal stress in the cross-section P = the internal axial force at the section A = the cross-sectional area of the rod

Tensile Stress - induced by tensile forces (T)Compressive Stress - induced by compressive forces (C)

Tension (T) Compression (C)

σ = PA

P

P

AP

P

A

P

(a) (b) (c)

External Force

External Force

InternalForce

Normal Stress

σ = PA

Member is elongating (stretching) Member is shortening

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Rope Construction � Rope is sometimes generally referred to as cordage � Rope construction involves twisting fibers together to form yarn.

For twisting rope, the yarn is then twisted into strands, and the strands twisted into rope.

� Three-stranded twisted rope is the most common construction. � For braided rope, the yarn is braided rather than being twisted

into strands. � Double-braided rope has a braided core with a braided cover. � Plaited rope is made by braiding twisted strands. � Other rope construction includes combinations of these three

techniques such as a three-strand twisted core with a braided cover.

� The concept of forming fibers or filaments into yarn and yarn into strands or braids is fundamental to the rope-making process.

A 1-½ in diameter fitness rope is composed of three strands. Each strand is made of 10 yarns and each yarn is made of 100 fibers. The tension in the rope is 240 lb. Determine the normal stress in the rope, strand, yarn, and fiber.

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Fibe

rYa

rnSt

rand

Rope

169

Allowable Axial Load

Consider the simply supported truss shown.

B

C

P

P

Member BC is in axial tension

Cut the member along its length

B

C

P

P

P

P

A

B

C

D

P

Rotate the member 90°

Internal Normal Stress ≡ σ = PA [Intensity of Force / unit area]

Also, P = σ A≡ Resultant of stress [product of the stress x Area = Force]

Upper Limit of Allowable StressIf we know the strength of a material, σFailure, can divide by a Factor of Safety (F.S. > 1) to give safe or allowable stress,

σAllow = σFailure / F.S.

the allowable force, PAllow = σAllow A

Required AreaRequired minimum cross-sectional area A of a member designed to carry a maximum axial load P without exceeding the allowable stress σallow.

These two equations apply only if the compressive member length is relatively short compared to the lateral dimensions of the member. For now, the discussion is limited to short compression members that do not buckle.

A =P

σAllow

P PA

P

Internal Forceσ

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Internal Axial Force DiagramVariation of internal axial force along the length of a member can be depicted by an internal axial force diagram whose ordinate at any section of a member is equal to the value of the internal axial force at that section.

Convention Used in TextbookTensile force is plotted as postiveCompressive force is plotted as negative

Example 1

Solution.

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Example 2

Solution.

FBD - Entire RodEquilibrium Equations

FBD - Left Portion of Section 1-1

If the rod is in equilibrium, each section of the rod is in equilibrium.

172

FBD - Left Portion of Section 2-2

FBD - Left Portion of Section 3-3

FBD - Right Portion of Section 3-3

Internal Axial Force Diagram

173

Normal Stresses in Each Segment

174

9-4Direct Shear Stresses

Shear Stress (τ) — intensity of internal force tangential (parallel ) to the area under question.

Normal Stress Shear Stress

P

P

P

Normal Stress = σ = PA

Perpendicular to direction of applied force

Shear Stress = τ = PA

Parallel to direction of applied force

Direct ShearConsider a block with a protruded part as shown in the figure below.

¾ A horizontal force P is applied to the block's protruded part as shown above in (a). ¾ The force P tends to shear the part off the block along the shear plane abcd. ¾ The body resists the force P by developing resisting shear stresses in the shear plane. ¾ The resultant of the shear stresses must be equal to the applied force P, as shown in (b). ¾ The shear stress may not be uniformly distributed over the shear area As.

The average shear stress, however, can be calculated from:

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9-5Bearing Stresses

If two bodies are pressed against each other, compressive forces are developed on the area of contact. The pressure caused by these surface loads is called bearing stress.

Examples of bearing stress are the soil pressure beneath a pier and the contact pressure between a rivet or bolt and the side of the hole it is in.

If the bearing stress is large enough, it can locally crush the material, which can lead to more serious problems. To reduce bearing stresses, engineers sometimes employ bearing plates, the purpose of which is to distribute the contact forces over a larger area.

Soil Bearing CapacityThe Bearing Capacity of Soils is perhaps the most important all all the topics in soil engineering. Bearing Capacity is the capacity of the soil to support the loads applied to the ground. The bearing capacity of soil is the maximum average contact pressure between the foundation and the soil which should not produce shear failure in the soil.

σb = Load (lb)

Area of Footing (ft2)

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Clevis Applications

177

Shearing Stress in Pin

Example - Normal, Shear, Tearout, and Bearing StressesClevis and Pin

CLEVIS

TONGUE

PIN

P

P

TONGUE

P

P

P

"Sheared' off Pin

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Normal Stress in Tongue

TONGUE

P

P

P/2

P

P/2

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Tearout Stress

P

TONGUE

P

P

180

Normal Stress in Clevis

CLEVIS

P/2

PIN

P

CLEVIS

TONGUE

P

181

TONGUE

P

Average Bearing StressP

P

P

TONGUE TONGUEP

P

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Lap Joint - Bearing StressConsider the lap joint formed by the two plates that are bolted together as shown.

The bearing stress caused by the bolt is not constant; it actually varies from zero at the sides of the hole to a maximum behind the bolt.

This complicated stress distribution is avoided by assuming that the bearing stress σb is uniformly distributed over a reduced area.

The reduced area Ab is take to be the projected area of the bolt:

Ab = td

where t is the thickness of the plate and d represents the diameter of the bolt.

PP

P P

FBD - Top PlateThe Bearing Stress P is Assumed to be Uniform on Projected Area td

P

Pt

d

Projected Area of the Bolt

σb = = PAb

Ptd

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τavg = = P/2As

P2As

where As is the cross-sectional area of the rivet.

Since the shear stress occurs in only one section of the rivet, it is said to be in single shear.If there are several rivets in the joint, the load is assumed to be shared equally by the rivets.

The Butt JointConnects nonoverlapping tension plates using connecting plates.Rivet A in the joint is subjected to shear stresses at sections m-m and n-n.Assuming the shear force is shared equally by the two sections, the average shear stress is

where As is the cross-sectional area of the rivet.

Since the shear stresses occurs in two sections of the rivet, the rivet is said to be in double shear.If there are several rivets on each side of the joint, the load is assumed to be shared equally by the rivets.

Lap Joint - Shear StressConnects two overlapping tension plates with a rivet (or a bolt).The average shear stress is

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ExampleConsider the block below. The rod is being acted on by the force P on both sides. The rod is pushing against the block.

Given that the block is 1 in thick and the rod diameter is ¾ in determine the bearing stress between the block and the rod. The force P = 50 lb

P

P

Solution.

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ExampleThe lap joint shown is fastened by four rivets of ¾-in. diameter. Find the maximum load P that can be applied if the working stresses are 14 ksi for shear in the rivet and 18 ksi for bearing in the plate. Assume that the applied load is distributed evenly among the four rivets, and neglect friction between the plates.

Solution.Calculate P using each of the two design criteria. The largest safe load will be the smaller of the two values.

Design for Shear Stress in Rivets

Design for Bearing Stress in Plate

P

P

P

4 in.

⅞ in.

⅞ in.FBD - Lower Plate

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Required Shear AreaThe allowable shear stress τallow is the upper limit of shear stress that must not be exceeded.

The minimum shear area As required to carry a design shear load P without exceeding the allowable shear stress τallow is

The Shaft keyConnects a gear to a shaft.The moment M on the gear is transmitted to the shaft through the key.The key is subjected to forces (labeled P).These forces are assumed to be concentrated on the rim of the shaft.

The moment Pr of P about the center of the shaft must be equal to the transmitted moment M.

The average shear stress at section m-m of the key is

where b is the width of the key, L is the length of the key, and r is the radius of the shaft. For a square key, the width b is approximately equal to one-quarter of the shaft diameter.

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Bearing Stress in Shaft KeyBearing stress occurs between the key and the gear and between the key and the shaft.The compressive force P is assumed to be uniformly distributed over an area (h/2)L

The bearing stress is, therefore,

where M = the moment transmitted by the key h = the height of the key L = the length of the key r = the radius of the shaft

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Bearing Stress Between Rivet and PlateIn the lap joint and butt joint, bearing stresses occur between rivets or bolts and the plates.

The maximum bearing stress is approximately equal to the value obtained by dividing the compressive force by the projected area of the rivet onto the plate.

The bearing stress between the rivet or the bolt and the plate is computed by

Required Bearing AreaThe allowable bearing stress (σb)allow is the upper limit of compressive stress that must not be exceeded.

The minimum bearing area Ab required to carry a design bearing load without exceeding the allowable bearing stress (σb)allow is

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Solution.

Example 9-5 (textbook)A circular blanking punch is operated by causing shear failure in the plate. The thickness of the steel plate is 0.5-in. and the ultimate shear strength of the steel (the greatest shear stress a material can withstand before failure) is τu = 40,000 psi. Determine the minimum force P required to punch a hole 2-in. in diameter.

Solution.

0.5 in.

2 in.

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Solution.

Shear Stress in the Pin

9-18 The clevis shown is connected by a pin of ¾ in. diameter. Determine the shear stress in the pin and the bearing stress between the pin and the plates if P = 10 kips and t = ¼ in.

Bearing Stress Between the Pin and the Plates

Bearing Stress Between the Pin and the Tongue

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Solution.

(a) Shear Stress in the Key

(b) Bearing Stress between the Key and the Shaft

9-18 A force F = 600 lb is applied to a crank and is transmitted to a shaft through a steel key, as shown. The key is ½ in, square and 2½ in. long. Determine (a) the shear stress in the key and (b) the bearing stress between the key and the shaft.

192

Example - Single Shear StressConsider a 1 in diameter rod passing through the two connections as shown. Force P is applied to each connection equal and opposite in direction. If P = 20 lb determine the shear stress in the rod.

P

PA

B

Free-Body Diagram —Top View of the Rod

A

20 lb

A B

20 lb

20 lb

1

1

1

1

Free-Body diagram of the left portion of section 1-1

Cut through the rod at section 1-1 and sketch the free-body diagram of the left portion of section 1-1

20 lbA 20 lb internal shear force must be pushing down.Since the object is three-dimensional and the rod is circular the 20 lb shear force gets distributed over the entire surface area of the shear plane, we call this the average shear force, τavg.

Sketch the free-body diagram of the top view of the rod.

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P/2

PA

B

P/2

Example - Double Shear StressConsider the same part as previously discussed, but add a third plate as shown. If P is applied to the middle connection by equilibrium the force in the other connections must be in the opposite direction and each force is equal to P/2. If P = 20 lb determine the shear stress in the rod.

Sketch the free-body diagram of the top view of the rod.

A B

P

P/2

1

1

P/2

Free-Body Diagram —Top View of the Rod

Cut through the rod at section 1-1 and sketch the free-body diagram of the left portion of section 1-1

A 10 lb internal shear force must be pushing down.

A

P/2 = 10 lb

1

1

10 lb

Free-Body diagram of the left portion of section 1-1