Solution manual mechanics of materials 8th edition hibbeler chapter 09

Determine a the principal stress and b the maximum in-plane shear stress and average normal stress at the point.. Determine a the principal stress and b the maximum in-plane shear stress

Stress Transformation Equations: Applying Eqs 9-1 and 9-3 of the text.

(Q.E.D.) sx¿ + sy¿ = sx + sy

9–1. Prove that the sum of the normal stresses

is constant See Figs 9–2a and 9–2b.

sx + sy = sx¿ + sy¿

Referring to Fig a, if we assume that the areas of the inclined plane AB is , then

the area of the horizontal and vertical of the triangular element are and

respectively The forces act acting on these two faces indicated on the

FBD of the triangular element, Fig b.

¢Asin 60°

¢Acos 60°

¢A

9–2. The state of stress at a point in a member is shown on

the element Determine the stress components acting on

the inclined plane AB Solve the problem using the method

of equilibrium described in Sec 9.1

The negative sign indicates that sx¿, is a compressive stress

tx¿y¿ = lim¢A:0

Referring to Fig a, if we assume that the area of the inclined plane AB is , then

the areas of the horizontal and vertical surfaces of the triangular element are

and respectively The force acting on these two faces are

indicated on the FBD of the triangular element, Fig b

¢Acos 60°

¢Asin 60°

¢A

9–3. The state of stress at a point in a member is shown on

the element Determine the stress components acting on

the inclined plane AB Solve the problem using the method

of equilibrium described in Sec 9.1

The negative sign indicates that sx¿, is a compressive stress

tx¿y¿ = lim¢A:0

•9–5. Solve Prob 9–4 using the stress-transformation

equations developed in Sec 9.2

Ans.

Ans.

The negative sign indicates that the sense of sx¿, is opposite to that shown on FBD

sx¿y¿ = lim¢A:0

a + ©Fy¿ = 0 ¢Fy¿ - 650(¢Asin 60°)sin 30° - 400(¢ A cos 60°)sin 60° = 0

¢Fx¿= -387.5¢A

Q + ©Fx¿ = 0 ¢Fx¿ - 400(¢Acos 60°)cos 60° + 650(¢ A sin 60°)cos 30° = 0

*9–4. The state of stress at a point in a member is shown

on the element Determine the stress components acting on

the inclined plane AB Solve the problem using the method

of equilibrium described in Sec 9.1

Ans.

The negative sign indicates t acts in y¿direction

= -a90 - 502 bsin(-300°) + (-35) cos (-300°) = -34.8 MPa

tx¿y¿ =

-sx - sy

2 sin 2u + txy cos 2u = 49.7 MPa

sx = 90 MPa sy = 50 MPa txy = -35 MPa u = - 150°

9–7. Solve Prob 9–6 using the stress-transformation

equations developed in Sec 9.2 Show the result on a sketch

-90¢A cos 30° cos 30° + 35¢A cos 30° cos 60° = 0

b + ©Fx¿= 0 ¢Fx¿ - 50¢A sin 30° sin 30° + 35¢A sin 30° sin 60°

¢Fy¿ = -34.82¢A90¢A cos 30° sin 30° + 35¢A cos 30° sin 60° = 0

R + ©Fy¿= 0 ¢Fy¿ - 50¢A sin 30° cos 30° - 35¢A sin 30° cos 60° +

9–6. The state of stress at a point in a member is shown on

the element Determine the stress components acting on

the inclined plane AB Solve the problem using the method

of equilibrium described in Sec 9.1

Force Equllibrium: Referring to Fig a, if we assume that the area of the inclined

plane AB is , then the area of the vertical and horizontal faces of the triangular

sectioned element are and , respectively The forces acting on

the free-body diagram of the triangular sectioned element, Fig b, are

¢A cos45°

¢A sin45°

¢A

*9–8. Determine the normal stress and shear stress acting

on the inclined plane AB Solve the problem using the

method of equilibrium described in Sec 9.1

The negative sign indicates that sx¿is a compressive stress

tx¿y¿ = lim¢A:0

¢Fx¿ = -5A106B¢A

-c80A106B¢A sin 45°dcos 45° = 0 ©Fx¿ = 0; ¢Fx¿ + c45A106B¢Asin 45°dcos 45° + c45A106B¢Acos 45°dsin 45°

Stress Transformation Equations:

we obtain,

Ans.

Ans.

The negative sign indicates that is a compressive stress These results are

indicated on the triangular element shown in Fig b.

u = + 135° (Fig a) sx = 80 MPa sy = 0 txy = 45 MPa

•9–9. Determine the normal stress and shear stress acting

on the inclined plane AB Solve the problem using the

stress transformation equations Show the result on the

45 MPa

A

B

45⬚

Normal and Shear Stress: In accordance with the established sign convention,

Stress Transformation Equations: Applying Eqs 9-1 and 9-2.

2 cos2u + txy sin 2u

u = + 60° sx = -3 ksi sy = 2 ksi txy = -4 ksi

9–11. Solve Prob 9–10 using the stress-transformation

equations developed in Sec 9.2 Show the result on a sketch

Force Equllibrium: For the sectioned element,

Normal and Shear Stress: For the inclined plane.

Ans.

Ans.

Negative sign indicates that the sense of sx¿, is opposite to that shown on FBD

tx¿y¿ = lim¢A:0

-2(¢A cos 30°)cos 30° + 4(¢A cos 30°) cos 60° = 0

Q + ©Fx¿= 0; ¢Fx¿ + 3(¢A sin 30°)cos 60° + 4(¢ A sin 30°)cos 30°

¢Fy¿= 4.165 ¢A

-2(¢A cos 30°)sin 30° - 4(¢A cos 30°) sin 60° = 0

a + ©Fy¿= 0; ¢Fy¿- 3(¢A sin 30°)sin 60° + 4(¢ A sin 30°)sin 30°

9–10. The state of stress at a point in a member is shown

on the element Determine the stress components acting on

the inclined plane AB Solve the problem using the method

of equilibrium described in Sec 9.1

2 bsin 100° + ( - 16)cos 100° = 7.70 ksi

tx¿y¿ = -asx -2 sybsin 2u + txycos 2u

*9–12. Determine the equivalent state of stress on an

element if it is oriented 50° counterclockwise from the

element shown Use the stress-transformation equations

16 ksi

10 ksi

In accordance to the established sign covention,

Applying Eqs 9-1, 9-2 and 9-3,

Ans.

Ans.

Ans.

Negative sign indicates that is a compressive stress These result, can be

represented by the element shown in Fig b.

2 cos 2u + txy sin 2u

u = - 60° (Fig a) sx = 200 psi sy = -350 psi txy = 75 psi

•9–13. Determine the equivalent state of stress on an

element if the element is oriented 60° clockwise from the

element shown Show the result on a sketch

200 psi

350 psi

75 psi

Ans.

Ans.

Orientation of principal stress:

Use Eq 9-1 to determine the principal plane of and

By observation, in order to preserve equllibrium along AB, has to act in the

direction shown in the figure

-12( - 30 - 0)>2 = 0.8

9–14. The state of stress at a point is shown on the element

Determine (a) the principal stress and (b) the maximum

in-plane shear stress and average normal stress at the point

Specify the orientation of the element in each case Show

the results on each element

30 ksi

12 ksi

In accordance to the established sign convention,

[ - 60 - ( - 80)]>2 = 5

s1 = -19.0 MPa s2 = -121 MPa = - 70 ; 22600

sx = -60 MPa sy = -80 MPa txy = 50 MPa

9–15. The state of stress at a point is shown on the element

Determine (a) the principal stress and (b) the maximum

in-plane shear stress and average normal stress at the point

Specify the orientation of the element in each case Show

the results on each element

80 MPa

60 MPa

50 MPa

9–15 Continued

Ans.

Ans.

Orientation of principal stress:

Use Eq 9-1 to determine the principal plane of and :

sx = 45 MPa sy = -60 MPa txy = 30 MPa

*9–16. The state of stress at a point is shown on the

element Determine (a) the principal stress and (b) the

maximum in-plane shear stress and average normal stress at

the point Specify the orientation of the element in each case

Sketch the results on each element

60 MPa

45 MPa

30 MPa

Normal and Shear Stress:

In - Plane Principal Stresses:

The element that represents the state of principal stress is shown in Fig a.

Maximum In - Plane Shear Stress:

sx = 125 MPa sy = -75 MPa txy = -50 MPa

•9–17. Determine the equivalent state of stress on an

element at the same point which represents (a) the principal

stress, and (b) the maximum in-plane shear stress and the

associated average normal stress Also, for each case,

determine the corresponding orientation of the element

with respect to the element shown Sketch the results on

each element

50 MPa

125 MPa

75 MPa

Stress Transformation Equations: Applying Eqs 9-1, 9-2, and 9-3

and

Combining the stress components of two elements yields

9–18. A point on a thin plate is subjected to the two

successive states of stress shown Determine the resultant

state of stress represented on the element oriented as

shown on the right

In accordance to the established sign Convention,

-120(0 - 160)>2 = 1.5

s1 = 224 MPa s2= -64.2 MPa = 80 ; 220800

9–19. The state of stress at a point is shown on the element

Determine (a) the principal stress and (b) the maximum

in-plane shear stress and average normal stress at the point

Specify the orientation of the element in each case Sketch

the results on each element

120 MPa

160 MPa

9–19 Continued

Stress Transformation Equations: Applying Eqs 9-2 and 9-1 with ,

*9–20. The stress acting on two planes at a point is

indicated Determine the normal stress and the principal

stresses at the point

ta= -asx -2 sybsin 2u + txy cos u

•9–21. The stress acting on two planes at a point is

indicated Determine the shear stress on plane a–a and the

principal stresses at the point

a b

ta

The location of the centroid c of the T cross-section, Fig a, is

The shear stress is contributed by the transverse shear stress only Thus,

The state of stress of point A can be represented by the element shown in Fig c.

Ans.

s1 = 4.93 MPa s2 = -111 MPa = - 53.10 ; 58.02

9–22. The T-beam is subjected to the distributed loading

that is applied along its centerline Determine the principal

stress at point A and show the results on an element located

( - 106.19 - 0)>2 = -0.4406

9–22 Continued

•9–23. The wood beam is subjected to a load of 12 kN If a

grain of wood in the beam at point A makes an angle of 25°

with the horizontal as shown, determine the normal and

shear stress that act perpendicular and parallel to the grain

due to the loading

-0.1286(2.2857 - 0)>2

*9–24. The wood beam is subjected to a load of 12 kN

Determine the principal stress at point A and specify the

orientation of the element

Using the method of sections and consider the FBD of the rod’s left cut segment,

Fig a.

a

The normal stress developed is the combination of axial and bending stress Thus,

Since no torque and transverse shear acting on the cross - section,

The state of stress at point A can be represented by the element shown in Fig b

100(0.01)2.5(10- 9)p

-y = C = 0.01 m

s = N

A ;

MyI

I = p

4 (0.01

4) = 2.5(10- 9)p m4

A = p(0.012) = 0.1(10- 3) p m2+ ©MC = 0; 400(0.25) - M = 0 M = 100 N#m

:+

©Fx = 0; N - 400 = 0 N = 400 N

•9–25. The bent rod has a diameter of 20 mm and is

subjected to the force of 400 N Determine the principal

stress and the maximum in-plane shear stress that is

developed at point A Show the results on a properly

oriented element located at this point

Internal Loadings: Consider the equilibrium of the free - body diagram from the

bracket’s left cut segment, Fig a.

Normal and Shear Stresses: The normal stress is the combination of axial and

bending stress Thus,

The cross - sectional area and the moment of inertia about the z axis of the bracket’s

cross section is

For point A, Then

Since no shear force is acting on the section,

The state of stress at point A can be represented on the element shown in Fig b.

In - Plane Principal Stress: , , and Since no shear

stress acts on the element,

Ans.

The state of principal stresses can also be represented by the elements shown in Fig b

Maximum In - Plane Shear Stress:

( - 12)(1)0.45573 = 29.76 ksi

©MO = 0; 3(4) - M = 0 M = 12 kip#in

:+

©Fx = 0; N - 3 = 0 N = 3 kip

9–26. The bracket is subjected to the force of 3 kip

Determine the principal stress and maximum in-plane

shear stress at point A on the cross section at section a–a.

Specify the orientation of this state of stress and show the

Substituting into

This indicates that is directed in the positive sense of the axes on the ace

of the element defined by

Average Normal Stress:

Internal Loadings: Consider the equilibrium of the free - body diagram of the

bracket’s left cut segment, Fig a.

Normal and Shear Stresses: The normal stress is the combination of axial and

bending stress Thus,

The cross - sectional area and the moment of inertia about the z axis of the bracket’s

cross section is

For point B, Then

Since no shear force is acting on the section,

The state of stress at point A can be represented on the element shown in Fig b.

In - Plane Principal Stress: , , and Since no shear

stress acts on the element,

Ans.

The state of principal stresses can also be represented by the elements shown in Fig b.

Maximum In - Plane Shear Stress:

( - 12)( - 1)0.45573 = -22.90 ksi

©MO = 0;3(4) - M = 0 M = 12 kip#in

:+

©Fx = 0; N - 3 = 0 N = 3kip

9–27. The bracket is subjected to the force of 3 kip

Determine the principal stress and maximum in-plane

shear stress at point B on the cross section at section a–a.

Specify the orientation of this state of stress and show the

= 11.5 ksi = t

max in-plane

Internal Forces and Moment: As shown on FBD(a).

Section Properties:

Normal Stress:

Shear Stress: Applying the shear formula

point A Applying Eq 9-5.

Ans.

, , and for point B Applying Eq 9-5.

Ans.

s1 = 1.60 MPa s2 = -143 MPa = - 70.538 ; 72.134

= 15.09 MPa

t = VQIt

sB= 3.608 - 144.685 = - 141.1 MPa

sA= 3.608 + 144.685 = 148.3 MPa

= 21.65(10

3)6.00(10- 3) ;

73.5(103)(0.1)50.8(10- 6)

s = N

A ;

MyI

*9–28. The wide-flange beam is subjected to the loading

shown Determine the principal stress in the beam at point A

and at point B These points are located at the top and

bottom of the web, respectively Although it is not very

accurate, use the shear formula to determine the shear stress

B A

Using the method of sections and consider the FBD of the left cut segment of the

(59.71 - 0)>2 = 0.6122

s1= 64.9 MPa s2 = -5.15 MPa = 29.86 ; 35.01

= 18.28(106)Pa = 18.28 MPa

s = My

I =

39.15(103)(0.075)49.175(10- 6)

•9–29. The wide-flange beam is subjected to the loading

shown Determine the principal stress in the beam at

point A, which is located at the top of the web Although

it is not very accurate, use the shear formula to determine

the shear stress Show the result on an element located at

9–30. The cantilevered rectangular bar is subjected to the

force of 5 kip Determine the principal stress at points A

3

Support Reactions: Referring to the free - body diagram of the entire arm shown

in Fig a,

Internal Loadings: Consider the equilibrium of the free - body diagram of the

arm’s left segment, Fig b.

Section Properties: The cross - sectional area and the moment of inertia about the z

axis of the arm’s cross section are

Referring to Fig b,

Normal and Shear Stress: The normal stress is a combination of axial and bending

stress Thus,

The shear stress is caused by transverse shear stress

The share of stress at point A can be represented on the element shown in Fig d.

It =

583.33C3.1875A10- 6B D0.16367A10- 6B(0.0075)

= 1.515 MPa

= -1876.390.5625A10- 3B +

87.5(0.0175)0.16367A10- 6B = 6.020 MPa

sA = N

A +

MyAI

+ ©MO = 0;

V = 583.33 N

V - 583.33 = 0+ c ©Fy = 0;

N = 1876.39N1876.39 - N = 0

:+

©Fx = 0;

By = 583.33N2166.67 sin 30° - 500 - By = 0

+ c ©Fy = 0;

Bx = 1876.39N

Bx - 2166.67 cos 30° = 0:+

©Fx = 0;

©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0 FCD = 2166.67N

9–31. Determine the principal stress at point A on the

cross section of the arm at section a–a Specify the

orientation of this state of stress and indicate the results on

an element at the point

Section a – a

a a

Support Reactions: Referring to the free - body diagram of the entire arm shown

in Fig a,

Internal Loadings: Considering the equilibrium of the free - body diagram of the

arm’s left cut segment, Fig b,

Section Properties: The cross - sectional area and the moment of inertia about the z

axis of the arm’s cross section are

Referring to Fig b,

Normal and Shear Stress: The normal stress is a combination of axial and bending

stress Thus,

The shear stress is contributed only by transverse shear stress

It =

583.33C3.1875A10- 6B D0.16367A10- 6B(0.0075)

= 1.515 MPa

= -1876.390.5625A10- 3B +

87.5(0.0175)0.16367A10- 6B = 6.020 MPa

sA = N

A +

MyAI

+ ©MO = 0;

V = 583.33 N

V - 583.33 = 0+ c ©Fy = 0;

N = 1876.39 N1876.39 - N = 0

:+

©Fx = 0;

By = 583.33 N2166.67 sin 30° - 500 - By = 0

+ c ©Fy = 0;

Bx = 1876.39 N

Bx - 2166.67 cos 30° = 0:+

©Fx = 0;

©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0 FCD = 2166.67N

*9–32. Determine the maximum in-plane shear stress

developed at point A on the cross section of the arm at

section a–a Specify the orientation of this state of stress and

indicate the results on an element at the point

Section a – a

a a

Orientation of the Plane of Maximum In - Plane Shear Stress:

Ans.

Substituting into

This indicates that is directed in the positive sense of the axis on the face

of the element defined by

Average Normal Stress:

9–32 Continued

Support Reactions: As shown on FBD(a).

Internal Forces and Moment: As shown on FBD(b).

Section Properties:

Normal Stress: Applying the flexure formula

Shear Stress: Applying the shear formula

Since no shear stress acts on the element

= 24.0 MPa

tA =

24.0(103)(0)0.3125(10- 6)(0.03)

= 0

t = VQIt

sB = 2.40(103)(0)0.3125(10- 6)

-= 0

sA= 2.40(103)(0.025)0.3125(10- 6)

-= -192 MPa

s = -MyI

QB= y¿A¿ = 0.0125(0.025)(0.03) = 9.375A10- 6B m3

QA= 0

I = 1

12 (0.03) A0.053B = 0.3125 A10- 6B m4

•9–33. The clamp bears down on the smooth surface at E

by tightening the bolt If the tensile force in the bolt is 40

kN, determine the principal stress at points A and B and

show the results on elements located at each of these

points The cross-sectional area at A and B is shown in the

adjacent figure

100 mm

50 mm

A E