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MECHANICS OF DEFORMABLE BODIES

1/37 Engr. Divina R. Gonzales

MECHANICS OF DEFORMABLE BODIES- Study of the relationship between externally applied loads and their internal effects on deformable bodies. RIGID BODY Bodies which neither change in shape and size after the application of forces FREE BODY DIAGRAM Sketch of the isolated body showing all the forces acting on it. THREE MAJOR DIVISIONS OF MECHANICS

1. Mechanics of Rigid Bodies Engineering Mechanics 2. Mechanics of Deformable Bodies Strength of Materials 3. Mechanics of Fluids Hydraulics

SIMPLE STRESS TENSILE STRESS AND COMPRESSIVE STRESS

STRESS unit strength of the body

Which bar is stronger A or B? Assume that the given loads are the maximum loads each can carry.

AP

S = Where: S Uniform internal stress P Axial force A Uniform cross-sectional area

Perpendicular Passing through the centroid

1000N 500N

A=50mm2 A=20mm2

BAR A BAR B

P

P

Compressive stress

Tensile stress



PROBLEMS ON SIMPLE STRESS

1. Determine the weight of the heaviest traffic lighting system that can be carried by the two wires shown if the allowable stress on wire AB is 90MPa and on wire AC is 110MPa given that the cross sectional areas of wire AB is 50mm2 and that of AC is 80 mm2.

2. Determine the required cross sectional areas of members BE, CD and CE of the given truss shown, if the allowable stress in tension is 120MPa while in compression is 105MPa. A reduced allowable stress in compression is given to reduce the danger of buckling.

3. An bronze rod is rigidly attached between a aluminum rod and a steel rod as shown. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in steel of 140MPa, in aluminum of 90MPa or in bronze of 100MPa.

35 70 A

B C

3m

4m

3m 3m

3m 3m 3m A B

C

H

G

F E

D

50KN 75KN 50KN

2P P 3P

Bronze A= 200mm2

Aluminum A= 500mm2 Steel

A= 150mm2

L alum= 3.5m L br = 2m L st = 1.2m



A

C B

3m

9m

8m

4. Determine the weight of the heaviest cylinder that can be supported by the structure shown if the cross sectional area of the cable is 120mm2 and its allowable stress is 80MPa.

5. The figure shows the landing gear of a light

airplane. Determine the compressive stress in strut AB caused by the landing reaction R=30KN. Neglect the weights of the members. The strut is a hollow tube, with 40mm outside diameter and of uniform thickness of 7mm.

6. In the recently opened World Trade Center, a

showcase of designers work is being featured. There is a piece of marble table which is just supported by three legs. If the weight of the table is 500N, find the stress in each leg if its cross-sectional section is 50mmx50mm.

7. The tripod shown supports the total station which weighs 10N. Find the required diameter of the leg if the maximum allowable stress in each leg is 25MPa.

R

55

200mm

A

600mm

B

5m

A

C

B 3m

9m

A

6m

3m 3m

4m

C

D

B

8m



SHEARING STRESS This arises whenever the applied loads cause one section of the body to slide past its adjacent section. The force acts parallel to the area.

TYPES OF SHEARING STRESS

1. SINGLE SHEAR

2. DOUBLE SHEAR

AP

Ss =

Where: Ss Shearing stress P Force acting parallel to the area A Surface area

P P

Shearing area parallel to the load

Rivet under Single shear LAP JOINT

t

Width of plate 130mm

P P

P P

BUTT JOINT

t MAIN PLATE

tSPLICE

P

P/2 tSPLICE

P/2

Rivet under Double Shear



LAP JOINT

t P P

3. PUNCHING SHEAR

4. INDUCED SHEAR BEARING STRESS Contact pressure exerted by one body upon another body. Also know as end stress. The force acts perpendicular to the area.

LAP JOINT

t


P P

P P

P

P

P P

SHEARED AREA

SHEARED AREA



P P

DIAMETER

t=25mm

130mm

PROBLEMS ON SHEARING STRESS AND BEARING STRESS

8. A circular hole is to be punched out of a plate that has a shear strength of 40ksi The working compressive stress in the punch is 50ksi. Compute the maximum thickness of a plate in which a hole 2.5in in diameter can be punched. B) If the plate is 0.25in thick, determine the smallest diameter that can be punched.

9. The lap joint is connected by three 20mm diameter rivets. Assuming that the axial load P = 50KN is distributed equally among the three rivets, find a) the shearing stress in each rivet; b) bearing stress between the plate and a rivet and c) the maximum average tensile stress in each plate.

10. Determine the maximum force P that the top chord can carry if the allow shearing stress is 50MPa, bearing stress is 60MPa and tensile stress is 85MPa of the connection shown

a 200mm b 120mm c 40mm d 75mm q 30


ENLARGEMENT OF THE RIVET HOLES

t OF THE PLATE

b

a c

q

P

d



11. A 25.6 mm diameter bolt having a diameter at the root of the threads of 21.7mm is used to faster two timbers as shown. The nut is tightened to cause a tensile force of 34KN in the bolt. Determine the shearing stress developed in the head of the bolt and the threads. Also determine the outside diameter of the washer if the inside diameter is 30mm and the bearing stress is limited to 0.9MPa.

12. The bracket is supported by inch diameter pins at A and B (the pin at B fits in the 45 slot in the bracket). Neglecting friction, determine the shear stresses in the pins, assuming single shear.

13. Compute the maximum force P

that can be applied to the foot pedal. The inch diameter pin at B is in single shear, and its working shear stress is 40000psi. The cable attached at C has a diameter of 1/8 inch and a working normal stress of 20,000psi.

12in

45 A

B

36in

200lb

P

10

A B C

T

2

6 2

15mm

17mm



600mm

400mm

14. The figure shows a roof truss and the detail of the connection at joint B. Members BC and BE are angle sections with thickness shown in the figure. The working stresses are 70MPa for shear in rivet and 140MPa for bearing stress due to the rivets. How many 19-mm diameter rivets are required to fasten the said members to the gusset plate

THIN WALLED CYLINDERS PROBLEMS ON THIN WALLED CYLINDERS

15. A cylindrical tank 8m in diameter is 12m high. If the tank is completely filled with water, determine the required thickness of tank plating if the allowable stress is 40MPa.

16. The tank shown in the figure is fabricated from steel plate. Determine the minimum thickness of the tank plating if the internal pressure is 1.5MPa and the allowable stress is 40MPa.

96KN 200KN

96KN

A

D

C E

B

H G

F

4m 4m 4m 4m

6m

DETAIL OF JOINT B

P BC

P BE

10mm GUSSET PLATE

75X75X6

75X75X13

LONGITUDINAL JOINT

CIRCUMFERENTIAL JOINT

tD

S NTIALCIRCUMFERE

=2r

tD

S ALLONGITUDIN

=4r



17. A large pipe called a penstock in hydraulic work is 1.5m in diameter. Here it is composed of wooden staves bound together by steel hoops, each 300mm2 in cross-sectional area, and is used to conduct water from a reservoir to a power house. If the maximum tensile stress permitted in the hoops is 130MPa, what is the spacing between hoops under a head of water of 30m?

18. A spiral-riveted penstock 1.5m in diameter is made of steel plate 10mm thick. The pitch of the spiral or helix is 3m. the spiral seam is a single-riveted lap joint consisting of 20-mm diameter rivets. Using SS=70Mpa and Sb=140MPa, determine the spacing of the rivets along the seam from a water pressure of 1.25MPa. Neglect end thrust. What is the circumferential stress? SIMPLE STRAIN STRESS- STRAIN DIAGRAM

STRESS

STRAIN

Proportional limit

Elastic limit

Yield Point

Ultimate Strength

Rupture Strength

Actual Rupture Strength



The strength of the material is not only the criterion that must be considered in designing a structure. The stiffness of a material is frequently of equal importance. Hookes Law states that up to the proportional limit, the stress is proportional to strain. The constant of proportionality based from experiment is the modulus of elasticity.

PROBLEMS ON SIMPLE STRAIN

19. During a stress-strain test, the unit deformation at a stress of 35MPa was observed to be 167x10-6 m/m and at a stress of 140MPa it was 667x10-6. If the proportional limit was 200MPa, what is the modulus of elasticity? Would these results be valid if the proportional limit were 150MPa? Explain.

20. The compound bar containing steel bronze and aluminum segments carries the

axial loads shown in the figure. The properties of the segments and the working stresses are listed in the table. Determine the maximum allowable value of P if the change in length of the entire bar is limited to 0.08in and the working stresses are not to be exceeded.

A (in2) E (psi) S (psi) Steel 0.75 30 x 106 20000 Bronze 1.00 12 x 106 18000 Aluminm 0.50 10 x 106 12000

AEPL

AXIAL =d

STRAINSTRESS da

da S AP

S = L

de =

2ft

3m

2ft

Bronze E=12x10 6 psi L=4ft A= 0.25 in2

A

C

Aluminum E=10x10 6 psi L=3ft A= 0.75 in2

D

P

2ft B



21. The rigid bars AB and CD are supported by pins at A and D. The vertical rods are made of aluminum and bronze. Determine the vertical displacement of the point where the force P =10kips is applied. Neglect the weight so the member.

22. Two steel bars AB ad BC

support a load P=30KN as shown. Area of AB=300mm2 and BC= 500mm2. If E = 200GPA, compute the horizontal and vertical components of the movement of B.

dDELP

=p

d 4 23. A round bar of length L tapers uniformly from a diameter D at one end to a

smaller diameter d at the other end. Determine the elongation caused by an axial tensile load P if E is its modulus of elasticity.

24. The rigid bars shown are separated by a roller at C and pinned at A and D. A

steel rod at B helps support the load of 50KN. Compute the vertical displacement of the roller at C. Answer: 2.82mm

Bronze 4 ft

Steel 2 ft

Aluminum 3 ft

P 3P 4P 2P

STEEL E=200x10 6 N/m2 L=3m A= 300 mm2 D

A

P= 30KN

B

C

q=30

A

C

P= 30KN

L=5m

B L=4m



25. The rigid bars AB and CD are supported by pins at A and D. The vertical rods

are made of aluminum and bronze. Determine the vertical displacement of the point where the force P=10kips is applied. Neglect the weights of the member. Answer: 0.115 in

3ft

2ft

2 ft

2ft

Bronze L=4ft A=0.25in2 E=12x106psi

P

Aluminum L=3ft A=0.75in2 E=10x106psi



STATICALLY INDETERMINATE MEMBERS Static indeterminacy does not imply that the problem cannot be solved; it simply means that the solution cannot be obtained from the equilibrium equations alone. A statically indeterminate problem always has geometric restrictions imposed on its deformation. The mathematical expressions of these restrictions, known as the compatibility equations, provide us with the additional equations needed to solve the problem (the term compatibility refers to the geometric compatibility between deformation and the imposed constraints). Because the source of the compatibility equations is deformation, these equations contain as unknowns either strains or elongations. We can, however, use Hookes law to express the deformation measures in terms of stresses or forces. The equations of equilibrium and compatibility can then be solved for the unknown forces.

PROBLEMS ON STATICALLY INDETERMINATE MEMBERS

26. The figure shows a copper rod that is placed in an aluminum sleeve. The rod is 0.005 inch longer than the sleeve. Find the maximum safe load P that can be applied to the bearing plate, using the following data: Answer: 60.3kips

27. A reinforced concrete column 250mm in diameter is designed to carry an axial compressive load of 400KN. Using the allowable stress in concrete of Sconc =6MPa and S steel = 120MPa, determine the required area of reinforcing steel. Assume Econc=14GPa and Esteel = 200GPa. Answer: 1320mm

2

28. A rigid block of mass M is supported by three symmetrically spaced rods as shown. Each copper rod has an area of 900mm2 Ecopper =120GPa and the allowable stress is 70MPa. The steel rod has an area of 1200mm2, Esteel=200GPa and allowable stress of 140MPa. Determine the largest mass M which can be supported.

Answer: 22.3x103Kg

COPPER ALUMINUM Area (in2) 2 3 E (psi) 17x106 10x106 S (ksi) 20 10

0.005

copper

alum

Bearing plate

10 in

Steel 240mm

Copper 160mm

Copper 160mm



29. Before the 400KN load is applied,

the rigid platform rests on two steel bars each of cross-sectional area of 1200mm2, as shown. The cross-sectional area of copper is 2400mm2. . Compute the stress in each rod after the 400KN load is applied. Neglect the weight of the platform. Esteel =200GPa Ealuminum =70GPa.

30. The composite bar is firmly attached to unyielding supports. Compute the stress

in each material caused by the application of the axial load P=50kips. b) If the maximum allowable stress in each material is Salum=22psi and Ssteel=40psi, find the maximum P that the structure can support.

31. The rigid beam is supported by the two bars shown in a horizontal position before the load P is applied. If P=200KN, determine the stress in each rod after its application. B) Find the vertical movement of P. c) If the allowable stress in aluminum is 80MPa and steel is 120MPa, find the maximum load P that the system can carry

Aluminum Steel A in mm2 600 800 E in GPa 70 200

400KN

steel aluminum steel 250mm

0.1mm

Aluminum A=3.25 in2 E=10x106psi

Steel A=5.5 in2 E= 29x106psi

12in 25in

P

P

3m 3m 3 m

L=4.5 m

L=6 m steel aluminum



THERMAL STRESS It is well known that changes in temperature cause dimensional changes in a body. An increase in temperature results in expansion, whereas a temperature decrease produces contraction. The thermal deformation is: The thermal stress is:

Where a = coefficient of thermal expansion and DT= change in temperature.

PROBLEMS ON THERMAL STRESS AND THERMAL DEFORMATION

33. A steel rod with a cross-sectional area of 150mm2 is stretched between two fixed points. The tensile load at 20C is 5000N. What will be the stress at -20C? At what temperature will the stress be zero? asteel = 11.7x10-6/C and E=200GPa. Answer=127MPa ; T= 34.2C

34. Two identical steel bars 500mm long support the rigid beam shown. An aluminum bar is placed exactly in between them, that is 0.1mm shorter. a) If the rigid beam is weightless determine the change in temperature for the middle bar to just touch the beam. b) If the beam weighs 300KN, find the stress in each bar. c) If the beam weighs 250KN, determine the stress in each bar when the temperature raises 35C d) If the beam weighs 320KN, determine the stress in each bar when the temperature drops 25C.

Aluminum Steel A in mm2 2400 1200 a in x 10-6 /C 23 11.7 E in GPa 70 200

TL D= ad TETS D= a

steel aluminum steel 500mm

0.1mm



35. The composite bar is firmly attached

to unyielding supports. The initial temperature is 80F when the load P = 20kips is applied, compute the stress in when the temperature is 150F and when the temperature is 5F.

36. The rigid beam is supported by the two bars shown in a horizontal position before the load P is applied. If P=200KN, determine the stress in each rod after an increase in temperature of 40C B) drop of 65C.

Aluminum Steel A in mm2 600 800 a in x 10-6 /C 23 11.7 E in GPa 70 200

Aluminum A=3.25 in2 E=10x106psi a= 12.8x10-6/F

Steel A=5.5 in2 E= 29x106psi a= 12.8x10-6/F

12in 25in

P

P

3m 3m 3 m

L=4.5 m

L=6 m steel aluminum



BEAMS It is a structure usually horizontal acted upon by transverse loads (forces that acts perpendicular to the plane containing the longitudinal axis of the beam)

axis beam

I. Statically Determinate Beams supported such that the number of reacting forces equals the number of available equations static equilibrium conditions,

a. Simply Supported

b. Cantilever Beams

c. Beams with Overhang

P

P W N/m

W N/m W N/m P

P W N/m



II. Statically Indeterminate Beams

Beams Supported such that the number of reacting forces exceeds the number of equations of static. The analysis requires the use of elastic deformations a. Propped Beam

b. Continuous Beams

c. Restrained Beams

P W N/m

P P W N/m

P W N/m

P W N/m



TYPES OF LOADS 1. Concentrated Loads

2. Distributed Loads

a. Uniformly Distributed

b. Uniformly Varying Triangular

Trapezoidal

c. Parabolic Loads

d. Moving Loads

W N/m

W2 N/m W1 N/m

W N/m

W N/m

P F



The fundamental definitions of shear and bending moments are expressed by

( )lefty

FV = and ( ) ( ) == rightleft MMM in which upward acting forces or loads cause positive effects. The shearing force V should be computed only in terms of the forces to the left of the section being considered: the bending moment M may be computed in terms of the forces to either the left or the right of the section depending on which requires less arithmetical work. Relations between load, shear moment are given by:

dxdv

w = = slope of the shear diagram

dx

dMv = = slope of the moment diagram

These relations are amplified to provide a semi graphical method of computing shear and moment which supplements the equations

( )lefty

FV = and ( ) ( ) == rightleft MMM diagramloadAVV -+= 12 daigramShearAMM -+= 12 A summary of the principles presented suggests the following procedure for the construction of shear and moment diagrams 1. Compute the reactions 2. Compute the values of shear at the change of load points using diagramloadAVV -+= 12 3. Sketch the shear diagram, drawing the correct shape and concavity of the shear diagram. 4. Locate the points of zero shear. 5. Compute values of bending moment at the change of load points and the points of zero shear using daigramShearAMM -+= 12 6. Sketch the moment diagram



PROPERTIES OF THE SHEAR DIAGRAM 1. At every change of loading we have to investigate our shear. 2. For concentrated loads or reactions the left and right portion of the point where they are acting must be investigated 3. Whenever we have a concentrated load or reaction there will always be a vertical line in the shear diagram 4. The shear diagram is one degree higher than the load diagram 5. Reference of the Concavity of the Shear Diagram Note: When M=0 ( point of inflection) there is a change of the concavity of the elastic curve of beam.

PROPERTIES OF THE MOMENT DIAGRAM 1. For every change in shear diagram, the moment must be investigate 2. Consider only the moment at any point and not the left and the right portion of the point except when we have a moment load or reaction 3. Analyze the point where the shear intersects the reference line (V=0) since when shear is zero moment is maximum or minimum 4. Vertical line will only be observed in the moment diagram whenever we have a moment load or reaction 5. The concavity of the moment diagram depends upon the load: if the load is downward, moment diagram is downward 6. The moment diagram is one degree higher than the shear diagram

Negatively increasing load

Negatively decreasing load

Positively decreasing load

Positively increasing load

Concavity of the circle corresponds to the concavity of the shear diagram



SHEAR AND MOMENT DIAGRAMS

Determine the shear and moment diagrams of the following beams loaded as shown

1.

2.

3.

4.

5.

6.

1m

7m 3m

2m

20 KN/m

40KN/m

100 KN 80 KN

1m 3m 1m

40 KN/m 120KN-m 60 KN

1m

5m

3m 40 KN/m

w1 KN/m

2m

w2 KN/m

2m

2m 2m

20 KN/m

40 KN

1m

80 KN

1m w KN/m

20 kN

2m

10kN/m

1 m 20 KN/m 1 m

2 m

25 kN

1m 3 m 2m

Hinge



7.

8.

9.

10. Determine P so that the moment at each support is equal to the moment at midspan.

4m

2m

1m

Hinge 80 KN

R1

20 KN/m 40 KN/m 1m

6m R3 R2

6m 1m

10 KN/m

P KN

1m

P KN

1m 3m 1m

20 KN/m 50KN-m

2m

2 m

120KN-m

2m 1 m

10KN/m



II SHEAR AND MOMENT DIAGRAMS OF BEAMS WITH TRIANGULAR LOADINGS

1.

2.

3.

4.

5.

4m

9m

12 KN/m 12 KN

1m

1m w KN/m

40 KN

2m 3m

60KN/m

1m

40KN

2m 3m

120KN/m

2m

40 KN/m

MAX M=137.5KN-m

1m 3m

80KN/m

1m

20 KN/m

3m 2m

80KN/m

1m 2m

40 KN/m MAX M=80KN-m

MAX M=160KN-m



6.

7.

8. The shear diagram is shown in the figure, determine the moment and load diagrams.

9.

10.

3m

120KN/m

3m 3m 3m

120KN/m

MAX M=-450KN-m

12KN/m

3m 6m

18KN/m

MAX M=49.5KN-m

2m

45KN

1m 1m 2m

-15KN

105KN

45KN

-20KN

2m

2m 1m

-10KN

20KN

-30KN

2m 3m 1m

-10KN

-8KN

-2KN

1m



DESIGN FOR FLEXURAL STRESS 1. A cantilever beam, 75mm wide by 200 mm high and 6m long carries a load that varies uniformly from zero at the free end to 1500 N/m at the wall. (a) Compute the magnitude and location of the maximum flexural stress (b) Determine the type and magnitude of the stress in a fiber 25mm from the top of the beam at section 2.5m from the free end. 2. Determine the minimum width b of the beam shown if the flexural stress is not to exceed 10Mpa.

3. A 2 diameter bronze bar is used as a simply supported beam 8ft long. Determine the

largest uniformly distributed load, which can be, applied over the right half of the beam if the flexural stress is limited to go 10ksi?

4. A simply supported rectangular yacal beam, 75mm wide by 150mm deep, carries a uniformly distributed load of 2250 N / m over its entire length. What is the maximum length of the beam if the flexural stress is limited to 18 Mpa?

5. A simply supported beam 6m long is composed of two C 200x28 channels riveted back to back. What uniformly distributed load can be carried, in addition to the weight of the beam, without exceeding a flexural stress of 125 MN/m2 if (a) the webs are vertical and (b) the webs are horizontal. Refer to Appendix B for channel properties.

6. A beam with a S380 x 74 section is simply supported at the ends. It supports a central concentrated load of 40 kN and a uniformly distributed load of 15 kN/m over its entire length, including the weight of the beam. Determine the maximum length of the beam if the flexural stress is not to exceed 130 Mpa. Refer to Appendix B for properties of S shapes.

7. A beam 15 m long is simply supported 2 m from each end. It is a builtup made of four angle bars 100x75x13, with long legs horizontal (see Table B-6) welded to a flat bar 25mmx300mm as shown in the figure. Determine the total uniformly distributed load that can be carried along its entire length without exceeding a flexural stress of 120 MPa.

8. A beam with a W2360x33section (see Table B-2) is used as a cantilever beam 7.5 m long. Find the maximum uniformly distributed load which can be applied over the entire length of the beam, in addition to the weight of the beam, if the flexural stress is not to exceed 140 MN / m2

9. A 12-m beam simply supported at the ends carries a uniformly distributed load of 20kN / m over its entire length. What is the lightest W shape beam that will not exceed a flexural stress of 120MPa ? What is the actual stress in the beam selected?

10. A simply supported steel beam 10m long carries a uniformly distributed load of 18kn/m load over the entire length and a central concentrated load of 25Kn. Determine the lightest Wide flanged section that can be used to support the load. What is the actual resulting stress in the beam selected.

11. A cantilever wooden beam is composed of two segments with rectangular cross sections. The width of each section is 75mm but their depths (150mm and 250mm) are different, as shown in the figure. Determine the maximum bending stess in each beam.

5000N

1m 3m

3m

2000N/m

b

200mm

2m 1.5m

50K



UNSYMMETRICAL SECTION DESIGN FOR BENDING STRESS

The previous discussions are all about beams symmetric with respect to the neutral axis. Because flexural stress vary directly with distance from the neutral axis of symmetric beams, which is the centroidal axis, such beam sections are desirable for materials that are equally strong in tension and compression. However, for materials relatively weak in tension and strong in compression such as cast iron, it is desirable to use beams that are unsymmetrical with respect to the neutral axis. With such a cross section, the stronger fibers can be located at a greater distance from the neutral axis than the weaker fibers. The ideal treatment for such materials is to locate the centroidal or neutral axis in such a position that the ratio of the distances from it to the fivers in tension and in compression is exactly the same as the ratio of the allowable stresses in tension and in compression. The allowable stresses thus reach their permitted values simultaneously. 1. A cast-iron beam carries a uniformly distributed load on a simple span. Compute the

flange width b of the inverted T section so that the allowable stresses fb(tension)=30MPa and fb(compression)=90MPa reach their limits simultaneously.

2. Compute the maximum tensile and compressive stresses developed in the beam that is

loaded and has the cross-sectional properties shown. 3. Determine the maximum safe value of W that can be carried by the beam shown if given

the following allowable stresses fb(tension)=60MPa and fb(compression)=90MPa

yc

yt

120mm

20mm

NA

20mm

b

L

W N/m

30mm

25mm

125mm 4m 1m

10KN/m 8KN

100mm

120mm

6W KN W W

2m 8m 2m

35mm

80mm 20mm 20mm



DESIGN FOR SHEARING STRESS

The vertical shear sets up numerically equal shearing stresses on longitudinal and transverse sections, which are determined from:

IbVQ

fv = ; Q=AB in which A is the partial area of the cross section above a line drawn through the point at which the shearing stress is desired. Q=Ay is the static moment about the NA of this area (or of the area below this line). Maximum shearing stresses occur at the section of maximum V and usually at the NA. For rectangular beams, the maximum shearing stress is

bhV

fMax v 23

. = .

1. Draw the shearing stress distribution for a rectangular beam 75mmx200mm which is simply supported on a 10m beam with a load of 20KN/m over the entire length.

2. Determine the maximum and minimum shearing stress in the web of the wide flange

section if V=120KN.

3. The distributed load shown is supported by a box beam shown. Determine the maximum safe value of w that will not exceed a flexural stress of 10 MPa or a shearing stress of 1MPa.

4. The distributed load shown is supported by a wide-flange section W 360x45 of the given dimensions. Determine the maximum safe w that will not exceed a flexural stress of 140 MPa or a shearing stress of 75 MPa.

3m 1m

2m 4m



DESIGN FOR FLEXURE AND SHEAR

In heavily loaded short beams the design is usually governed by shearing stress; but in longer beams the flexure stress generally governs because the bending moment varies with both the load and length of beam. Shearing is more important in timber beams than in steel beams because of the low shearing strength of wood.

1. A box beam supports the loads shown. Compute the maximum value of P that will not exceed a flexural stress fb(flexure)=8MPa and fv(shearing)=1.2MPa for section between the support.

2. A simply supported beam L meters long carries a uniformly distributed load of 16KN/m over its entire length and has the cross section shown . Find L to cause a maximum flexural stress of 40MPa. What maximum shearing stress is then developed?

3. The wide flenage beam shown in the fig supports the concentrated load W and a total uniformly distributed load 2W KN. Determine the maximum safe value of W if fb(flexure)=10MPa and fv(shearing)=1.4MPa

20mm 160mm

140mm

20mm

2m 2m 1m

2W W

20mm 220mm

180mm

200mm

2m 2m 2m

4000 N P

200mm

160mm 20mm

160mm

20mm



DEFLECTION IN BEAMS Frequently the design of a beam is determined by its rigidity rather than by its strength. Several methods are available for determining beam deflections. Although based on the same principles, they differ in technique and in their immediate objective. We consider first a variation of the double-integration method that greatly broadens and simplifies its application. Another method, the area-moment method, is thought to be the most direct of any, especially when the deflection at a specific location is desired.

DOUBLE INTEGRATION METHOD A mathematical approach in solving for the deflection at any portion of the beam involve a complete determination of the moment, slope and deflection equation for the entire beam:

Moment Equation Mdxyd

EI =

2

2

Slope Equation +=

12

2

CMdxdx

ydEI

Deflection Equation 212

2

CxCMdxdxdx

ydEI ++=

The constants of integration will become zero if the origin of axes is selected at a position where the slope and deflection are known to be zero, as at a perfectly restrained end or at the center of a symmetrically loaded beam. 1. Determine the midspan deflection of a simply supported beam carrying a

a) uniformly distributed load over the entire length of the beam. b) central concentrated load.

2. Determine the free end deflection of a cantilever beam carrying a a) uniformly distributed load over the entire length of the beam. b) free end concentrated load.

3. Find the value of EIy at the position midway between the supports and at the overhanging end for the beam shown and also determine the location of maximum deflection.

400N/m 600N

1m 3m 2m 2m



AREA-MOMENT METHOD Theorem I: The change in slope between tangents drawn to the elastic curve at any two points A and B is equal to the product of 1/EI multiplied by the area of the moment diagram between these two points.

( )ABAB AREAEI1

=q Theorem II: The deviation of any point B relative to a tangent drawn to the elastic curve at any other point A, in a direction perpendicular to the original position of the beam, is equal to the product of 1/EI multiplied by the moment of area about B of that part of the moment diagram between points A and B.

BBAAB xareaEIt = )(

1/

The product EI is called flexural rigidity. In two theorems, (area)AB is the area of the moment

diagram between points A and B, and x B is the moment arm of this area measured from B. When the area of the moment diagram is composed of several parts, the expression (area)AB includes the moment of the area of all such parts. The moment of the area is always taken about an ordinate through the points at which the deviation is being computed. One rule is very important: The deviation at any point is positive if the point lies above the reference tangent from which the deviation is measured and negative if the point lies below the reference tangent. 1. Determine the midspan deflection of a simply supported beam carrying a

a) uniformly distributed load over the entire length of the beam. b) central concentrated load.

2. Determine the free end deflection of a cantilever beam carrying a a) uniformly distributed load over the entire length of the beam. b) free end concentrated load.

3. Find the value of EIy at the position midway between the supports and at the overhanging end for the beam shown and also determine the location of maximum deflection.

tB/A

B A

Positive deviation: B located above reference tangent

tB/A

B

A

Negative deviation: B located below reference tangent

400N/m 600N

1m 3m 2m 2m



PROBLEM SET

1. Compute the value of EId at midspan for the beam loaded as shown. If E=10GPa, what value of I is required to limit the midspan deflection to 1/360 of the span?

2. Compute the value of EId at midspan for the beam loaded as shown.

3. Compute the value of EId at the right end of the beam loaded as shown.

4. Compute the value of EIy midway between the supports for the overhanging beam shown.

2m 2

300N/m

4m

0.5m 300N/m1.5m

Answer=EId=5000N-m3;

Answer: EId=657N-

4m 1m 400N/m

4001

Answer: EId=195N-

Answer: EId=-4.66kN-

2m

2kN/m

4kN

2

2kN/m

2



TORSION Torsion is a variable shearing stress experienced by solid or hollow circular shafts subjected to moment or torque. This shearing stress varies directly with the radial distance from the center of the cross section and is expressed by

JT

Tr

=~ . The radial distance r becomes r for maximum torsional shearing stress. For

maximum torsional shearing stress in solid shafts of diameter d becomes 3

16~dT

Tp

=

The angular deformation in a length L

is expressed in radians by JGTL

=q .

III. Flanged Bolt Coupling Connection A commonly used connection between two shafts is a flanged bolt coupling. It cobnsists of flanges rigidly attached to the ends of the shafts and bolted together. The torque is transmitted by the shearing force P created in the bolts.

T=PRn or SRnd

PRnT4

2p== .

Occasionally a coupling has two concentric rows of bolt. Letting the subscript 1 refer to bolts on the outer circle and subscript 2 refer to bolts on the inner circle, the torque capacity of the coupling is T=P1R1n1+P2R2n2. The relation between P1 and P2 can be determined from the fact that the comparatively rigid flanges cause shear deformations in the bolts which are proportional to their radial distances from the shaft axis. Using Hookes Law for shear,

122

2

2

11

1

1

RGA

P

RGA

P

=

P=SA



TORSION PROBLEMS

1. What is the minimum diameter of a solid steel shaft that will not twist through more than 3 in a 6m length when subjected to a torque of 14KN-m? What maximum shearing stress is developed? Use G=83Gpa.

2. A stepped steel shaft consists of a hollow shaft 2m long, with an outside diameter of 100 mm and an inside diameter of 70mm, rigidly attached to a solid shaft 1.5m long, and 70mm in diameter. Determine the maximum torque that can be applied without exceeding a shearing stress of 70MPa or a twist of 2.5 in a 3.5 m length. Use G=83Gpa

3. A solid steel shaft is loaded as shown. Using G=83Gpa, determine the required diameter of the shaft if the shearing stress is limited to 60MPa and the angle of rotation at the free end is not to exceed 4.

4. A flanged bolt coupling consists of eight steel 20-mm bolts spaced evenly around a bolt

circle of300mm in diameter. Determine the torque capacity of the coupling if the allowable shearing stress in the bolts is 40MPa.

5. A flange bolt coupling consists of 6-10mm diameter steel bolts on a bolt circle 300mm in

diameter and 4-10mm diameter steel on a concentric bolt circle 200mm in diameter. What torque can be applied without exceeding a shearing stress of 60MPa in the bolts? B) Determine the number of 10mm steel bolts that must be used on the 300mm bolt circle of the coupling to increase the torque capacity to 8KN-m.

a. ECCENTRICALLY LOADED RIVETED CONNECTIONS



6. The connection of 12 rivets shown, the load P = 200kN passes through the center of rivet C and has a slope of 4/3. Determine the resultant load on the most heavily loaded rivet. Y 100 mm X CCC c 100 mm 80 mm 80 mm 80 mm 7. Determine the most and least heavily loaded 16mm diameter rivet in the connection shown. Note that rivets B and C were not properly driven that they do not carry any load.

C

A

c.g.

120 mm

90 mm

90 mm

120 mm

P=50KN I H G

F E D

C B A



8. Rivets 22 mm in diameter are used in the connection shown in the figure. If P = 90 kN, what thickness of plate is required so as not to exceed a bearing stress of 140 MPa?

P

9. In the gusset plate connection shown in the figure, if P = 60 kN, determine the shearing stress in the most heavily loaded of the four 22-mm rivet.

100 mm 100 mm 200 mm 150 mm

100 mm

80 mm

80 mm



10. For the connection shown in the figure, determine the shearing stress in the most heavily loaded of the three 22-mm rivets.

80 mm 80 mm 90 mm

11. A gusset plate is riveted to a larger plate by four 22mm rivets arranged and loaded as shown. Determine the minimum shear stress developed in the rivets.

100 mm

3

4 60 kN

80 mm

80 mm

100 mm

P=40KN

80 mm

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