Strength of Materials Final

STRENGTH OF MATERIALS (Review Notes) Simple Stress a.

Shearing Deformation

δ S =

Normal Stress (σ) (applied force over cross sectional area)

σ  =

c.

 AS G

Where:

P

δS = shearing deformation V = shearing force acting over A s As = shearing area L = length G = modulus of rigidity (shear modulus)

 A

b. Tangential Stress or Shear Stress ( τ) (shearing force over sheared area)

τ  =

VL

V  Rotational Deformation

 AS

Bearing Stress (σb) (bearing force over bearing area or  contact area)

σ b =

θ  =

TL  JG

Where: J = polar moment of inertia T = torque

Pb  Ab

Stress – Strain Diagram NOTE:

1 Pa 1 kPa 1 MPa 1 GPa

2

= 1 N/m 3 2 = 1x10 N/m 6 2 = 1x10 N/m 9 2 = 1x10 N/m

Actual Strength Ultimate Strength

Simple Strain Rupture Strength

Axial Deformation

Stress, Elastic limit σ=P/A

Hookes Law:

 E  = P  A P  A

σ 

σ  = E ε 

;

ε 

=  E ε  =  E 

;

ε  =

;

δ  =

δ 

 L

δ   L PL  AE 

Where: δ = total deformation P = applied axial load A = constant cross sectional area L = Length E = modulus of elasticity Note: the stress should should not exceed the the proportional  limit.

Stiffness of Rod

 AE   L K  =

= P

δ 

δ  (stiffness)

Flexibility of Rod

δ  P δ  P

=

Proportional limit

0

Strain, ε=S/L

Proportional limit – the stress strain diagram is a straight line. All theories involving the behavior of elastic bodies is based upon a stress-strain proportionality. It is an indication that the proportional limit and not the ultimate strength is the maximum stress to which a material may be subjected. The stress is proportional to the strain:

 E  =

σ  ε 

(Hookes Law)

E = modulus of elasticity σ = stress ε = strain

P

;

Yield Point

Elastic Limit – the stress beyond which the material will not return to its original shape when unloaded but will retain a permanent deformation called permanent set.

 L  AE 

= (  flexibilit y )

Yield Point  –  – the point in which there is an appreciable elongation or yielding of the material without any corresponding increase of load, the load might decrease while the yielding occurs. Ultimate Strength  –  – the highest ordinate on the stress-strain curve. Rupture Strength  –  – the stress at failure

Poisson’s Ratio ( μ)

Torsion

μ = ration of unit lateral deformation to the unit longitudinal deformation.

Shear Stress

1. Uni-axial stress

μ  = −

ε  y

ε  z

=−

ε  x

τ  =

T  ρ   J 

Where: T = torque J = polar moment of inertia ρ = radial distance

ε  x

2. Biaxial Stress

ε  x =

ε  y =

σ  x

− μ 

 E 

σ  y

−

 E 

1. Solid Shaft

 E 

− μ 

 E  σ  y

ε  z = −

σ  y

 J  =

σ  x

 E  σ  x  E 

ε  x = ε  y = ε  z =

 E  1  E  1  E 

32

τ max =

16T  π  D 3

2. Hollow Shaft

3. Tri-axial Stress

1

π  D 4

 J  =

π ( D 4 − d 4 ) 32

[σ  x − μ (σ  x + σ  z )]

[σ 

− μ (σ  z + σ  x )]

[σ 

− μ (σ  x + σ  y ) ]

 y

 z

τ max =

16TD π ( D 4 − d 4 )

Where: D = outside diameter  d = inside diameter 

Where: μ = poisson’s ratio ε = strain σ = stress (P/A) E = modulus of elasticity

Rotational or angle of twist

θ  =

TL  JG

Shear modulus (G) Where:

G=

SS = shearing stress T = torsion θ = angle of twist in radians L = length of shaft G = shear modulus J = polar moment of inertia

E 

2(1 + μ )

where: G = shear modulus E = modulus of elasticity Μ = Poisson’s ratio Power  Bulk Modulus or  Modulus of Compression (k)

k  =

E 

P = T(ω) where: ω = angular speed in radians per unit time

3(1 − 2 μ ) Helical Spring

Thermal Stress

1. Shear stress

δt = αL ΔT δt = linear deformation α = coefficient of linear expansion expressed in units of meters per meter per degree of  temperature change  ΔT = temperature change L = length

τ  =

16 PR ⎛  d   ⎞ 1 + ⎜ ⎟ π d 3 ⎝  4 R ⎠

Where: τ = max. torsional shearing stress P = axial load R = mean radius of helix D = diameter of wire

Note:  a. Max. Moment occurs where shear is zero b. Point of inflection occurs where moment is zero

2. RAM Wahl Formula (accurate)

τ  =

16 PR ⎡ 4m − 1

π d 3 ⎢⎣ 4m − 4

+

0.615 ⎤ m ⎥⎦

Where:

Moving Loads

m=

 D d 

2 R

=

1.

d 

D = mean diameter  d = diameter of the wire

The bending moment under a particular load is a maximum when the center of the beam is midway between that load and the resultant of all loads then on the span: R = P1 + P2 (resultant of two moving loads)

3. Spring deflection 3

δ  =

64 PR n 4

Gd 

R

P1

Where:

x/2

CL

x/2

P2

δ = spring deflection n = number of coils

occurs here

4. Spring constant

k  =

δ 

1.

Shear and Moment Diagram 2.

Shear Diagram Diagram Increasing

 R

∑ MR

⎛  L1  x ⎞ − ⎟ 2 2 ⎠ ⎝  ⎛  L −  x ⎞  R⎜ ⎟ 2  ⎠ ⎝   R2 =  R2 L =  R1 ⎜

Moment

Decreasing Decreasing

 L

Load (increasing)

Load (decreasing) w

L

P1 S

1

Increasing

R1

R2

Rx = P1S + P2(0) ; S-spacing b/w loads

 x =

How to draw the curves for shear and moment diagrams

L/2-x/2 L/2

L/2

R1

P

Load Diagram

Max. moment

3.

⎛  L  x ⎞ − ⎟ ⎝  2 2 ⎠

 Max. M  =  R2 ⎜

w

2.

L

R2

R1

R2

Maximum shearing force occurs at, and is equal to, the maximum reaction. Maximum reaction is when the load is directly above the reaction, thus, the left most load is above the left reaction; and the right most load is above the reaction.

•

Position at which R2 is maximum P1

L-S

Relation of Load, Shear and Moment Diagrams

L

R1

Load Diagram

S

P2

R2

A1 R1

x2

x1

R2  ΔV=A1

Shear  Diagram

V1

V2

Mmax

•

V1 – V2 =A1 (area of  load diag.)

Position at which R1 is maximum P1

L-S

S

P2

 ΔM=A1

Moment Diagram

M1

M2

M2 – M1 =A1 (area of  shear diag.)

R1

L

R2

b – width Q – statical moment of area Q = Ay

Stresses in Beams Radius of curvature of beams

 ρ  =

 EI   M 

1.

ρ = radius of curvature of beam M = bending moment I = moment of inertia

b

Rectangular section

τ  =

3V 

d/4

2bd 

d/2

Critical section for shear 

2.

ρ

f b

c

τ  =

NA

3. Flexural (or Bending) Stress

 M  =  f b =  I   I  C   I 

S= 1.

 f b =

2.

σ S = nσ W 

n=

 Mc  I 

c=

;

( 2)

d 

2

;

 M  d 

 I  =

3

12 6 M  2

bd 

12

σ  =

Torsional loading;

τ  =

Flexural loading

 f b =

T  ρ   J   Mc  I 

Axial and flexure

π d 

C L

C L P

P

e

C L M

32 =

π  4 R

( R

4

+

− r 4 ) M=Pe

 f  = ±

4. Triangle 3

bd 

24

Shearing Stress

τ  =

 A

; section modulus

6

R = outer radius r = inside radius

S=

; modular ratio

P

Axial loading;

2

bd 

 E W 

bd 

3

bd 

 E S

Combined Stresses

Hollow tube

S=

 AW  = nAS

;

Where:

3

3.

Beams of Different Materials (Reinforced Beam)

 M 

Circular section with diameter, d

S=

3 A section

For rectangular sections:

S=

r 

C 

 f b =

h/2

section

4V 

S  I 

 f b =

b/2

Critical

bh

Critical

; section modulus, S

 f b =

h/2

h/3

3V 

Circular section

τ  =

 Mc

C 

Triangular section

VQ

P  A

 I 

Mohr’s Circle 1. Normal Stress (σ) Tension (+) Compression (-)

τ  XY  = −τ YX 

2. Shear Stress (τ)

C  =

 Ib

Clockwise (+)

Where V – shear force I – moment of inertia

±

 Mc

3. Rotation Counterclockwise (+)

σ  X  + σ Y  2

Stability & Determinacy of Structures Beams 1. 2. 3.

r < c+3 r = c+3 r > c+3

1. 2. 3.

3b+r < 3j+c ; unstable 3b+r = 3j+c ; stable & determinate 3b+r > 3j+c ; stable & indeterminate

where:

; unstable ; stable and determinate ; stable and indeterminate

r = unknown reaction elements c = no. of equations of condition = 1.0 hinge, internal connection = 2.0 roller, internal connection = 0 without internal connection

Frames Where: b = no. of bars  j = no. of joints

Trusses 1. 2. 3.

b+r < 2j b+r = 2j b+r > 2j

; unstable ; stable & determinate ; stable & indeterminate

Three Moment Equation

⎡  L1 ⎡ h1 h3 ⎤ 6 A2 a2  L2 ⎤  M 3 L3 6 A1a1 + 2 M 2 ⎢ + ⎥ +  E  I  +  L E  I  +  L  E  I  = 6⎢ L +  L ⎥  E 1 I 1  E   I   E   I  ⎣ 11 2 2⎦ 2 2 1 1 1 2 2 2 ⎣ 1 2⎦

 M 1 L1

When EI is constant

 M 1 L1 + 2 M 2 [ L1 +  L2 ] +  M 3 L2 +

6 A1 a1  L1

+

6 A2 b2  L2

⎡h h ⎤ = 6 EI ⎢ 1 + 3 ⎥ ⎣ L1  L2 ⎦

Where: h1 & h2 – is positive when measured upward

6 Aa

- moment of area of M-diag. resulting from carrying the applied loads on a simple span o f the

 L

length as equivalent beam segment

Types of Loading P a

Pa

b

 L

L

6 Aa

6 Ab

 L

 L

( L

2

w N/m

− a2 )

3

L

Pa  L

( L

2

− a2 )

3

wL

wL

4

4

w N/m 3

L

3

wL

wL

4

4

Moment Distribution Method (Hardy Cross Method)

Absolute stiffness:

Relative Stiffness:

4 EI 

k  =

4 EI 

k  =

 L

when E is constant

 L •

let I = LCM of the span

Distribution Factor:

 DF  =

k 

∑

;

k 

DF= 1.0 for hinge or roller  DF = 0 for fixed end

Fixed End Moment P Ma

M b a

A

b

B

L w N/m

Ma

L 2

 M  A = +

wL

12

 M  A = +

 M  A = +

wL

 M  A = +

wL

2

2

 L

;

 M  A = +

Pab

 M  A = +

Pab

2

2

 L

M b B

A

 M  A = +

Pab

Pab

2

12

;

2

2

 L

w N/m Ma

M b B

A

L

Ma

2

30

;

2

2

 L

M b L

A

B

Δ

 M  A = +

6 EI Δ  L2

;

Beam Deflection Double Integration Method

 EI   EI 

d 2 y dx dy dx

 EIy =

=  M 

2

= ∫  Mdx + C 1

∫∫  Mdx(dx) + C  ( x ) + C  1

2

;

Slope equation

;

Deflection equation

Where: dy/dx

-

y EI C1 & C2 M -

slope/tangent on the elastic curve deflection flexural rigidity constant of integration moment equation

 M  A = +

Pab 2  L2

Area – Moment Method Theorem I: The change of 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 moment diagram between these two points  A and B 

θ  AB =

1  EI 

[areaMD] AB

Theorem II: The deviation of any point B relative to a tangent drawn to the elastic curve at any 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 

t  B /  A =

1  EI 

[areaMD] AB ⋅ x B

Tangent @ A

Tangent @B

Any Loading C

θA

A

B

θB

δC tC/A

tB/A θAB

Elastic curve

Sign convention

1. Slope A

B

A

θAB (-) B

θAB (+)

2. Deviation A

B

tB/A (-)

A B

tB/A (+)

Other 1. 2. 3. 4.

Methods (for beam deflection) Method of superposition Virtual work method or Unit Load Method (beams, frames and trusses) Slope Deflection Method Castiqlianos Theorem

Problems 1. Determine the outside diameter of a hollow steel tube that will carry a tensile load of 500 kN at a stress of  2 150 MN/m . Assume the wall thickness to be one-tenth of the outside diameter. (ans. 109mm) 2. An aluminum tube is rigidly fastened between a bronze rod and a steel rod as shown. 20kN Determine the stress in each material. (ans. σ b=28.57MPa; σa=5MPa; σs=12.5MPa )

3. The following data were recorded during a tensile test of a 14.0 mm diameter mild steel rod. The gage length was 50 mm. Find the (a) proportional limit, (b) modulus of  elasticity, (c) elastic limit, (d) yield point, (e) ultimate strength and (f) rapture strength. (ans. a. 246.2MPa; b. 205167 MPa; c. 260.49MPa;d. 270.24MPa; e. 448.23MPa; f. 399.51 MPa)

.

Aluminum 1000mm2

Bronze 700mm2

15kN

15kN 0.50m

0.60m

Elongation (mm)

σ  =

10kN

0.70m

Stress Load (N)

Steel 800mm2

Strain

P

ε  =

 A

0

δ   L

0

0

0

6310

0.010

12600

0.020

18800

0.030

25100

0.040

31300

0.050

37900

0.060

40100

0.163

41600

0.433

270.24

0.0087

46200

1.25

300.12

0.025

52400

2.50

340.40

0.05

58500

4.50

380.02

0.09

68000

7.50

441.74

0.15

69000

12.50

448.23

0.25

67800

15.50

440.44

0.31

65000

20.00

422.25

0.40

61500

fracture

399.51

fracture

40.99 40.99 40.86 81.85 40.28122.13 40.92163.05 40.28203.33 42.87246.20 14.27260.49

0.0002 0.0004 0.0006 0.0008 0.001 0.0012 0.0033

4. Determine what force is required to punch a 15mm hole in a mild steel pipe 5mm thick, when an ultimate shear stress of the plate is 500 MPa? What will then be the compressive stress in the punch? (ans. V = 118kN; σ = 667MPa) -6

5. During a stress-strain test the unit deformation at a stress of 62 MPa was observed to be 160x10 m/m -6 and at a stress of 150 MPa the unit deformation was observed to be 600x10 m/m. Find the value of its -6 modulus of elasticity and the strain corresponding to a stress of 80 MPa. (ans. E=200GPa; ε=250x10 ) 6. A short concrete column 300 mm square in cross-section is reinforced with six symmetrically placed steel 2 bars each has an area of 400 mm . If the column carries a load 1000kN and Es=200GPa (steel); Ec=14GPa (concrete), find the stress of concrete and steel. (ans. σs=117MPa; σc=8MPa) 7. The beam shown is supported by dissimilar  members. The bar is rigid but is not constrained to remain horizontal. What are the reactions in Aa=300mm2 the vertical members? (ans. Pc=554kN; Ea=400GPa Pb=614kN; Pa=169kN)

A b=200mm2 E =300GPa

1000kN

Ac=100mm2 Ec=200GPa

2m

4m

6m

8. A steel wire 10 m long hanging vertically supports a tensile load of 2kN. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 140 MPa and the total elongation is not to exceed 5mm. Use E = 200 GPa. (ans. 5.05mm) 9. A steel block with dimensions shown is subjected to the tri-axial forces indicated. Using E=200GPa and μ=0.30, determine (a) εx; (b) εy; -5 (c) εz; (d) change in volume. (ans. a. 1.85x10 ; b. -5 -6 3 -1.4x10 ; c. 5.5x10 ; d. 150mm )

100kN

150kN 120kN

500mm

100kN

300mm

100mm 120kN

150kN o

10. A steel rod is stretched between two rigid walls and carries a tensile load of 600N at 20 C, using

-6 o

α=11.7x10 / C and E=200GPa. Find the required minimum diameter of the rod if the allowable stress is o not to exceed 140MPa at -25 C. (ans. 15mm)

11. A steel marine propeller is to transmit 4.5 MW at 3 rev/sec without exceeding a shearing stress of 50 2 o MN/m or twisting through more than 1 in a length of 25 diameter. Compute the diameter if G=83GPa. (ans. 351mm) 12. A hollow steel tube having an outside diameter of 220mm and inside diameter of 200mm, which of the following gives the maximum torque that it could carry if the allowable shearing stress is 75.5 MPa. 13. Three wheel loads roll as a unit across a 10m span. The loads are A=60kN; B=50kN, 3m to the right of A, and C=90kN, 4m to the right of B. Find maximum shear and moment. (ans. V=138kN; M=257.66kN-m) 50kN/m

14. Give the maximum shear and moment of the beam shown. (ans. V=90kN; M=81kN)

2m

20kN/m

2m

2m

15. A simply supported beam, 60mm wide by 100mm high and 4m long is subjected to a concentrated load of  800N at a point 1m from one of the supports. Determine the maximum flexural stress and the stress developed in a fiber located 10mm from the top of the beam at midspan. (ans. 6MPa; 3.20MPa) 16. A timber beam 150mm by 300mm is reinforced on the bottom only, with a steel strip 75mm wide by 10mm thick. Determine the maximum resisting moment of the beam if the allowable stress of steel and wood are 120MPa and 8MPa, respectively. Use n=20. By what amount is the moment increased by the reinforcement? (ans. 25.77 kN-m; 7.77 kN-m)

17. For the beam shown, determine the maximum stress in each material when the section is resisting a bending moment of 70 kN-m. Es=200GPa (steel), Ea=70GPa (aluminum), Ew=10GPa (wood). (ans. σs=249MPa; σa=112MP, σw=10MPa)

80mm Steel

20mm

wood

150mm

aluminum

150mm

18. A cross section of the beam is built-up of two 40mm by 120mmwood planks, securely spiked together to act as a single unit. It is used as a simple span 3m long supporting a load of 200 N/m including its own weight. Calculate the maximum bending stress in the beam. (ans. 1.03MPa)

19. At certain point in a stressed body, the principal stresses are σx=80MPa, and σy=-40MPa. Determine the o normal stress and shear stress on the plane whose normal is at +30 with the x-axis. (ans. σz=50MPa, y τ=51.96MPa) 80MPa 20. If a point is subjected to the state of stresses 50MPa shown , determine the principal stresses and the maximum shearing stress. (ans. σ: 58 & -98MPa; 50MPa τ=78MPa) 40MPa 40MPa x

80MPa

21. A beam has a rectangular cross section 100mm wide by 200 mm deep. What is the maximum safe value of P if the allowable flexural stress is 10 MPa and the allowable shearing stress of 1.5 MPa? (ans. 2kN)

P

P

30kN/m 0.6m

3m

0.6m

P

22. From the given beam, find the value of P that will cause the tangent to the elastic curve over the support B to be horizontal. (ans. 1350kN)

300 N

B

2m

1m A

2m

10 kN/m

B

23. Find the deflection and the rotational deflection at point C. E = 70 GPa and I = 6 4 10.57 x10 mm . (ans. δ=257mm; θ=0.2433rad)

A

C

6m

1m

D 1m

E 2m

20kN

24. Find the reactions of the beam shown. (ans. Ma=83kN-m)

10 kN/m

B

A

C

6m

1m

D 1m

25. Analyze the beam shown. (ans. Rb=18.61kN, Rc=9.03kN, Rd=27.36kN)

E 2m

10 kN/m 5 kN

6 kN/m

5 kN

A B

C

D

4m

1m

1m

2m

E

1m

3m

200kN

26. Determine the deflection at midspan of the given beam. Use any method. (ans. EIy=8107)

20 kN/m

A

B

4m

C 2m

100kN

27. Determine the support reactions. (ans. Rb=126kN, Rc=134kN, Mc=182.5kN-m)

3m 20 kN/m

A 2m

3m C

B

28. Analyze the beam shown. (ans. Ra=0.775, Rb=2.855, Rc=11.17) 900 N/m

A

6 kN

3 kN

B 4m

1m

1m

D 2m

C 1m

800 N/m

D 5m