mechanics of materials - tistory

Problem 6.8-3 A beam of wide-flange shape has the cross sectionshown in the figure. The dimensions are b � 5.25 in., h � 7.9 in., tw � 0.25 in., and tf � 0.4 in. The loads on the beam produce a shearforce V � 6.0 k at the cross section under consideration.

(a) Using centerline dimensions, calculate the maximum shear stress in the web of the beam.

(b) Using the more exact analysis of Section 5.10 in Chapter 5, calculate the maximum shear stress in the web of the beam and compare it with the stress obtained in part (a).

Solution 6.8-3 Wide-flange beam

SECTION 6.8 Beams with Inclined Loads 403

b � 5.25 in. h � 7.9 in. tw � 0.25 in.tf � 0.4 in. V � 6.0 k

(a) CALCULATIONS BASED ON CENTERLINE DIMENSIONS

(SECTION 6.8)

Moment of inertia (Eq. 6-59):

Iz � 10.272 � 65.531 � 75.803 in.4

Maximum shear stress in the web (Eq. 6-54):

� 3244 psi

tmax � ¢btf

tw

�h

4≤ Vh

2Iz

� (10.375 in.) (312.65 lb�in.3)

Iz �tw h

3

12�

btf h2

2

(b) CALCULATIONS BASED ON MORE EXACT ANALYSIS

(SECTION 5.10)

See Figure 5-38. Replace h by h2 and t by tw.

h2 � h � tf � 8.3 in. h1 � h � tf � 7.5 in.


Maximum shear stress in the web (Eq. 5-48a):

� (40.336 lb/in.5)(80.422 in.3)

� 3244 psi

NOTE: Within the accuracy of the calculations, themaximum shear stresses are the same.

tmax �V

8Itw

(bh 22 � bh 1

2 � twh 12)

I �1

12 (892.51 in.4) � 74.376 in.4

I �1

12 (bh 2

3 � bh 13 � twh 1

3)

b

hz

y

C

tf

tf

tw

b

h1 h h2z

y

C

tf

tf

tw

Probs. 6.8-3 and 6.8-4

Problem 6.8-4 Solve the preceding problem for the following data: b � 145 mm, h � 250 mm, tw � 8.0 mm, tf � 14.0 mm, and V � 30 kN.


404 CHAPTER 6 Stresses in Beams

b � 145 mmh � 250 mm

tw � 8.0 mmtf � 14.0 mm

V � 30 kN

(a) CALCULATIONS BASED ON CENTERLINE DIMENSIONS

(SECTION 6.8)


Iz � 10.417 � 106 mm4 � 63.438 � 106 mm4

� 73.855 � 106 mm4

Maximum shear stress in the web (Eq. 6-54):

� (316.25 mm) (0.050775 N/mm3)

� 16.06 MPa

tmax � ¢btf

tw

�h

4≤ Vh

2Iz

Iz �tw h

3

12�

btf h2

2

(b) CALCULATIONS BASED ON MORE EXACT ANALYSIS

(SECTION 5.10)

See Figure 5-38. Replace h by h2 and t by tw.

h2 � h � tf � 264 mm h1 � h � tf � 236 mm


Maximum shear stress in the web (Eq. 5-48a):

� (6.4864 � 10�6 N/mm5)(2.4756 � 106 mm2)

� 16.06 MPa

NOTE: Within the accuracy of the calculations, themaximum shear stresses are the same.

tmax �V

8Itw

(bh 22 � bh 1

2 � twh 12)

I �1

12 (867.20 � 106 mm4) � 72.267 � 106 mm4

I �1

12 (bh 2

3 � bh 13 � twh 1

2)

b

h1 h h2z

y

C

tf

tf

tw

Shear Centers of Thin-Walled Open Sections

When locating the shear centers in the problems for Section 6.9, assumethat the cross sections are thin-walled and use centerline dimensions forall calculations and derivations.

Problem 6.9-1 Calculate the distance e from the centerline of the web of a C12 � 20.7 channel section to the shear center S (see figure).

(Note: For purposes of analysis, consider the flanges to be rectangles withthickness tf equal to the average flange thickness given in Table E-3, Appendix E.)

Solution 6.9-1 Channel section

z

y

CS

e

C 12 � 20.7 d � 12.00 in. tw � 0.282 in.bf � 2.942 in. tf � average flange thicknesstf � 0.501 in. b � bf � tw�2 � 2.801 in.h � d � tf � 11.499 in.

Eq. (6-65): c �3b2tf

htw � 6btf

� 1.011 in.

z

y

CS

e

tw

bf

b

d2

d2

h2

h2

SECTION 6.9 Shear Centers of Thin-Walled Open Sections 405

Problem 6.9-2 Calculate the distance e from the centerline of the web of a C 8 � 18.75 channel section to the shear center S (see figure).

(Note: For purposes of analysis, consider the flanges to be rectangles with thickness tf equal to the average flange thickness given in Table E-3, Appendix E.)

Solution 6.9-2 Channel section

z

y

CS

e

tw

bf

b

d2

d2

h2

h2

C 8 � 18.75 d � 8.00 in. tw � 0.487 in.bf � 2.527 in. tf � average flange thicknesstf � 0.390 in. b � bf � tw/2 � 2.284 in.h � d � tf � 7.610 in.

Eq. (6-65): c �3b2tf

htw � 6btf

� 0.674 in.

Problem 6.9-3 The cross section of an unbalanced wide-flange beam is shown in the figure. Derive the following formula for the distance h1 from the centerline of one flange to the shear center S:

h1 �

Also, check the formula for the special cases of a T-beam (b2 � t2 � 0) anda balanced wide-flange beam (t2 � t1 and b2 � b1).

Solution 6.9-3 Unbalanced wide-flange beam

t2b32h

��t1b3

1 � t2b32

z

y

CS

t2t1b1

h1 h2

b2

h

FLANGE 1:

F1 �2

3(t1) (b1) (t1) �

Vt1b 13

12Iz

t1 �Vb1

2

8Iz

Q � (b1 �2)(t1) (b1 �4) �t1b 1

2

8

t1 �VQ

Izt1

z

y

CS

t2t1b1

h1 h2

b2

h1

2

FLANGE 2:

Shear force V acts through the shear center S.

�

or (1)

h1 � h2 � h (2)

Solve Eqs. (1) and (2):

T-BEAM

b2 � t2 � 0;� h1 � 0

WIDE-FLANGE BEAM

t2 � t1 and b2 � b1;� h1 � h�2

h1 �t2b 2

3h

t1b 13 � t2b 2

3

(t1b13)h1 � (t2b2

3)h2

aMs � F1h1 � F2h2 � 0

F2 �Vt2b2

3

12Iz

S�2�1

h1 h2

F1 F2


Problem 6.9-4 The cross section of an unbalanced wide-flange beam is shown in the figure. Derive the following formula for the dis-tance e from the centerline of the web to the shear center S:

e �

Also, check the formula for the special cases of a channel section (b1 � 0 and b2 � b) and a doubly symmetric beam (b1 � b2 � b/2).


3tf (b22 � b2

1)��htw � 6tf (b1 � b2)

z

y

CS

tf

tf

tw

b1 b2

e

h2—

h2—

Shear force V acts through the shear center S.� Ms � �F3e � F1h � F2h � 0

e �3tf (62

2 � 621)

htw � 6tf (b1 � b2)

�h2

12[htw � 6tf (b1 � b2) ]

Iz �tw h

3

12� 2 (b1 � b2) (tf)¢h2≤

2

e �F2h � F1h

F3�

h2tf

4Iz

(b22 � b2

1)

a

F3 � VF2 �b2

2htfV

4Iz

F1 �b1t1tf

2�

b21htfV

4Iz

t2 �b2hV

2Iz

t1 �VQ

Itf

�b1hV

2Iz

CHANNEL SECTION (b1 � 0, b2 � b)

(Eq. 6-65)

DOUBLY SYMMETRIC BEAM (b1 � b2 � b/2)e � 0 (Shear center coincides with the centroid)

e �3b2tf

htw � 6btf

�2

�2

�1

�1

z S

tf tw

b1 b2

e

h2—

h2—

F2

F2

F3

F1

F1

C

Problem 6.9-5 The cross section of a channel beam with double flangesand constant thickness throughout the section is shown in the figure.

Derive the following formula for the distance e from the centerline ofthe web to the shear center S:

e �3b2(h2

1 � h22)

��h3

2 � 6b(h21 � h2

2)z

y

h2h1CS

e

b

Solution 6.9-5 Channel beam with double flanges

SECTION 6.9 Shear Centers of Thin-Walled Open Section 407

t � thickness

F3 � V

F2 �b2h1tV

4Iz

tB �bh1V

2Iz

F1 �1

2 tAbt �

b2h2tV

4Iz

tA �VQA

Izt�

V(bt)¢h2

2≤

Izt�

bh2V

2Iz

z h2h1F2S

e

b

F1

F1

F2

F2

B

A Shear force V acts through the shear center S.

� Ms � �F3e � F1h2 � F2h1 � 0

e �3b2(h2

1 � h22)

h32 � 6b(h2

1 � h22)

�t

12 [h3

2 � 6b(h21 � h2

2) ]

Iz �th3

2

12� 2 [bt(h2�2)2 � bt(h1�2)2]

e �F2h1 � F1h2

F3�

b2t

4Iz

(h12 � h2

2)

a

Problem 6.9-6 The cross section of a slit circular tube of constantthickness is shown in the figure. Show that the distance e from the center ofthe circle to the shear center S is equal to 2r.

Solution 6.9-6 Slit circular tube

z

e

r

y

CS

QA � ydA

� r2t(1 � cos �)r � radiust � thickness

Iz � �r3t

tA �VQA

Izt�

Vr 2(1 � cos u)

Iz

� �u

o

(r sin f)rtdf

� At point A: dA � rtd�TC � moment of shear stresses about center C.

Shear force V acts through the shear center S.Moment of the shear force V about any pointmust be equal to the moment of the shearstresses about that same point.

� MC � Ve � TC e �TC

V� 2ra

TC � �tArdA � �2�

o

Vr�

(1 � cos u) du� 2Vr

tA �V(1 � cos u)

�rt

z

eC

S

A

d��

�

Problem 6.9-7 The cross section of a slit square tube of constantthickness is shown in the figure. Derive the following formula for thedistance e from the corner of the cross section to the shear center S:

e � �2 �

b

2��

Solution 6.9-7 Slit square tube


ze

y

CS

b

b � length of each sidet � thickness

FROM A TO B:

Q �tS2

212

t�VQ

Iz t

At A: Q � 0 �A � 0

At B:

FROM B TO D:

At B: At D:


� Ms � 0

Substitute for F1 and F2 and solve for e:

e �b

212

2(F1�12)(b12 � e) � 2(F2�12)(e) � 0a

F2 � tB bt �2

3 (tD � tB) bt �

5tb3V

612 Iz

tB �b2V12 Iz

tB �b2V

212 Iz

t�VQ

Izt�

V

Iz

B b2

212�

S

212 (2b � S) R

�tb2

212�

tS

212 (2b � S)

Q � bt ¢ b

212≤� St ¢ b12

�S

212≤

F1 �tB bt

3�

b3tV

612 Iz

tB �b2V

212 Iz

Q �tb2

212

ze

y

CS

s

s

A

slit

D

B

F2

F2 F1

F1

�B

�D


Problem 6.9-8 The cross section of a slit rectangular tube of constantthickness is shown in the figure. Derive the following formula for thedistance e from the centerline of the wall of the tube to the shear center S:

e � �b

2

(

(

2

h

h

�

�

3

3

b

b

)

)�

Solution 6.9-8 Slit rectangular tube

z

y

CS

e

h2—

h2—

b2— b

2—

t � thickness

FROM A TO B:

�A � 0

FROM B TO C:

FVERT � V F3 � 2F1 � V

F3 � V ¢1 �th3

24 Iz

≤a

F2 �1

2 (tB � tC) bt �

bht(h � 2b)V

8 Iz

tC �h (h � 4b)V

8 Iz

QC �th

2 ¢h

4≤� bt ¢h

2≤�

th

8 (h � 4b)

tB �h2V

8 Iz

F1 �tB t

3 ¢h

2≤�

th3V

48Iz

tB �h2V

8 Iz

t�VQ

Iz t�

S2V

2 Iz

Q �tS2

2


� Ms � 0 �F3e � F2h � 2F1(b � e) � 0

Substitute for F3, F2 and F1 and solve for e:

e �b (2h � 3b)

2(h � 3b)

Iz � 2 B 1

12 th3 � bt ¢h

2≤

2R �th2

6 (h � 3b)

e �bh2t(2h � 3b)

12 Iz

a

C B

DAG

E F

�D

�B�C

sz

y

CS

e

h2—

h2—

F3

F2

F2

F1

F1

b

Problem 6.9-9 A U-shaped cross section of constant thickness is shownin the figure. Derive the following formula for the distance e from thecenter of the semicircle to the shear center S:

e ��2(2r2

4�

b �

b2

�

�

r�br)

�

Also, plot a graph showing how the distance e (expressed as thenondimensional ratio e/r) varies as a function of the ratio b/r. (Let b/rrange from 0 to 2.)

Solution 6.9-9 U-shaped cross section


z

r

e

b

y

CS

O

r � radius F1 � force in AEt � thickness F2 � force in EFT0 � moment in BDE

FROM A TO B: �A � 0

FROM B TO E:

� r2t sin �QB � btrQe � QB � Q1 � btr � r2t sin �

At angle �: dA � rtd�

�Vr 3t

Iz

(�b � 2r)

� ��

0

Vr 3t (b � r sin u)du

Iz

T0 � �trdA � ��

b

tr 2tdu

te �VQB

Iz t�

Vr (b � r sin u)

Iz

Q1 � �ydA � �u

0

(r cos f) rtdf

F1 �bttB

2�

Vb2rt

2Iz

tB �VQ

Iz t�

V(btr)

Iz t�

Vbr

Iz

Shear force V acts through the shear center S.Moment of the shear force V about any point must beequal to the moment of the shear stresses about thatsame point.

� M0 � Ve � T0 � F1(2r)

GRAPH

NOTE: When b/r � 0,

(Eq. 6-73)e�r �4�

1.27

3

2

1

0

��er

1.72

2.20

2b/r

er

�2(2 � b2�r2 � � b�r)

4 b�r � �

e �2(2r2 � b2 � �br)

4b � �rIz �

�r 3t

2� 2(btr 2)

e �T0 � 2F1r

V�

r 2t

Iz

(�br � 2r 2 � b2)

a

D

B

CO�D

�B

A

E F

z

e

CS O

b

F1

F2

To

E

A

F

D

B

D

B

O

d

Problem 6.9-10 Derive the following formula for the distance e fromthe centerline of the wall to the shear center S for the C-section ofconstant thickness shown in the figure:

e �

Also, check the formula for the special cases of a channel section (a � 0)and a slit rectangular tube (a � h/2).

Solution 6.9-10 C-section of constant thickness

3bh2(b � 2a) � 8ba3��h2(h � 6b � 6a) � 4a2(2a � 3h)


z

e

b

a

a

y

CS

h2—

h2—

C B

D

A

GE F

�D

�B�Cs

z

y

CSe

F3

F2

F2

F1

F1

b

h

a

a

t � thickness

FROM A TO B:

�A � 0

�a2t (3h � 4a)V

12 Iz

F1 � �a

0

ttdS �tV

Iz�

a

0

S ¢h2

� a �S

2≤ dS

tB �a

2 (h � a)

V

Iz

t�VQ

Iz t� S ¢h

2� a �

S

2≤

V

Iz

Q � St ¢h2

� a �S

2≤

FROM B TO C:

FROM C TO E:



� MS � 0 �F3(e) � F2h � 2F1(b � e) � 0

Substitute for F1, F2, and F3 and solve for e:

CHANNEL SECTION (a � 0)

(agrees with Eq. 6-65 when tf � tw)

SLIT RECTANGULAR TUBE (a � h�2)

(agrees with the result of Prob. 6.9-8)e �b(2h � 3b)

2(h � 3b)

e �3b2

h � 6b

e �3bh2(b � 2a) � 8ba3

h2(h � 6b � 6a) � 4a2(2a � 3h)

�t

12[h2(h � 6b � 6a) � 4a2(2a � 3h) ]

Iz � 2 ¢ 1

12 th3≤� 2bt ¢h

2≤

2

�6

12 (h � 2a)3

e �bt[3h2(b � 2a) � 8a3 ]

12 Iz

a

F3 � VB1 �a2t (3h � 4a)

6 Iz

Ra

F2 �1

2 (tB � tC) bt �

bt

4 [2a(h � a) � bh ]

V

Iz

tC � B a

2 (h � a) �

bh

2R

V

Iz

�at

2 (h � a) �

bht

2

QC � at ¢h2

�a

2≤� bt ¢h

2≤tB �

a

2 (h � a)

V

Iz

Problem 6.9-11 Derive the following formula for the distance e fromthe centerline of the wall to the shear center S for the hat section ofconstant thickness shown in the figure:

e �

Also, check the formula for the special case of a channel section (a � 0).

Solution 6.9-11 Hat section of constant thickness

3bh2(b � 2a) � 8ba3��h2(h � 6b � 6a) � 4a2(2a � 3h)


z

a

a

b

y

CS

e

h2—

h2—

�C

�B

�B

�D

C B

A

D

E F

G

z

y

CSe

F3

F2

F2

F1

F1

b

h

a

a

t � thickness

FROM A TO B:

�A � 0

FROM B TO C:

F2 �1

2 (tB � tC) bt �

bt

4 [2a(h � a) � bh ]

V

Iz

tC � B a

2 (h � a) �

bh

2R

V

Iz

QC � at ¢h2

�a

2≤� bt ¢h

2≤�

at

2 (h � a) �

bht

2

tB �a

2 (h � a)

V

Iz

�a2t (3h � 4a)V

12 Iz

F1 � �a

0

ttdS �tV

Iz�

a

0

S ¢h2

� a �S

2≤ dS

tB �a

2 (h � a)

V

Iz

t�VQ

Iz t� S ¢h

2� a �

S

2≤

V

Iz

Q � St ¢h2

� a �S

2≤

FROM C TO E:



� MS � 0 �F3e � F2h � 2F1(b � e) � 0

Substitute for F1, F2, and F3 and solve for e:

CHANNEL SECTION (a � 0)

(agrees with Eq. 6-65 when tf � tw)e �3b2

h � 6b

e �3bh2(b � 2a) � 8ba3

h2(h � 6b � 6a) � 4a2(2a � 3h)

�t

12 [h2(h � 6b � 6a) � 4a2(2a � 3h) ]

Iz �1

12 th3 � 2bt ¢h

2≤

2

�t

12 (h � 2a)3 �

1

12 th3

e �bt[3h2(b � 2a) � 8a3 ]

12 Iz

a

F3 � VB1 �a2t (3h � 4a)

6 Iz

Ra

Problem 6.9-12 A cross section in the shape of a circular arc of constantthickness is shown in the figure. Derive the following formula for thedistance e from the center of the arc to the shear center S:

e �

in which � is in radians. Also, plot a graph showing how the distance evaries as � varies from 0 to �.

Solution 6.9-12 Circular arc

2r(sin � � � cos �)��

� � sin � cos �


z

e

y

r

CS

O

�

�

t � thicknessr � radiusAt angle �:

Q � ydA

� r2t(cos � � cos �)

� r3t(� � sin � cos �)

T0 � moment of shear stresses

At angle �, dA � rtd�

�2Vr(sin b� b cos b)

b� sin b cos b

T0 � �trdA � �b

�b

V(cos u� cos b)

t(b� sin b cos b) rtdu

t�V(cos u� cos b)

rt(b� sin b cos b)

Iz � �y 2dA � �b

�b

(r sin f)2rtdf

t�VQ

Iz t�

Vr 2(cos u� cos b)

Iz

� �b

u

(r sin f)rtdf

�

z

e

d

CS O�

�

Shear force V acts through the shear center S.Moment of the shear force V about any point must beequal to the moment of the shear stresses about thatsame point.

� M0 � Ve � T0 e � T0 �V

GRAPH

SEMICIRCULAR ARC (� � �/2):

(Eq. 6-73)

SLIT CIRCULAR ARC (� � �):

(Prob. 6.9-6)er

� 2

er

�4�

er

�2(sin b� b cos b)

b� sin b cos b

�� 2

��

1.27

0

1

2��er

e �2r(sin b� b cos b)

b� sin b cos b

a

Elastoplastic Bending

The problems for Section 6.10 are to be solved using the assumption thatthe material is elastoplastic with yield stress �Y.

Problem 6.10-1 Determine the shape factor f for a cross section in the shape of a double trapezoid having the dimensions shown in the figure.

Also, check your result for the special cases of a rhombus (b1 � 0)and a rectangle (b1 � b2).

Solution 6.10-1 Double trapezoid


z

y

h2—

h2—

C

b2

b1

b1

Neutral axis passes through the centroid C.

Use case 8, Appendix D.

SECTION MODULUS S

C � h/2

PLASTIC MODULUS Z (EQ. 6-78)

z �A

2 (y1 � y2) �

h2

12 (2b1 � b2)

y1 � y2 �1

3 ¢h

2≤ ¢2b1 � b2

b1 � b2≤

A � 2 ¢h2≤(b1 � b2)�2 �

h

2 (b1 � b2)

S �I

C�

h2

24 (3b1 � b2)

�h3

48 (3b1 � b2)

Iz � 2 ¢h2≤

3

(3b1 � b2)�12

z

y

h2—

h2—

C

b2

b1

SHAPE FACTOR f (EQ. 6.79)

SPECIAL CASE – RHOMBUS

b1 � 0 f � 2

SPECIAL CASE – RECTANGLE

b1 � b2 f �3

2

z

y

z

y

F �z

S�

2(2b1 � b2)

3b1 � b2

Problem 6.10-2 (a) Determine the shape factor f for a hollow circularcross section having inner radius r1 and outer radius r2 (see figure).(b) If the section is very thin, what is the shape factor?

Solution 6.10-2 Hollow circular cross sections

SECTION 6.10 Elastoplastic Bending 415

z

y

C

r1

r2

Neutral axis passes through the centroid C.Use cases 9 and 10, Appendix D.

SECTION MODULUS S

c � r2

PLASTIC MODULUS Z (EQ. 6-78)

For a semicircle,

z �A

2 (y1 � y2) �

4

3 (r 32 � r 31)y1 � y2

�4

3� ¢r 32 � r 31

r 22 � r 21≤

y1 �a yiAi

a Ai

�

¢4r2

3�≤ ¢�r 22

2≤� ¢4r1

3�≤ ¢�r2

1

2≤

��2(r 22 � r 21)

y �4r

3�A � �(r 22 � r 21)

S �Iz

c�

�

4r2 (r2

4 � r14)

Iz ��

4 (r4

2 � r41)

z

y

C

r1

r2

y1�

(a) SHAPE FACTOR f (EQ. 6-79)

(b) THIN SECTION (r1 ➞ r2)

Rewrite the expression for the shape factor f.

Let r1/r2 ➞ 1

SPECIAL CASE OF A SOLID CIRCULAR CROSS SECTION

Let r1 � 0 (Eq. 6-90)f �16

3� ¢1

1≤�

16

3�

f �16

3� ¢3

4≤�

4�

� 1.27

�16

3�B 1 � r1�r2 � (r1�r2)2

(1 � r1�r2) (1 � r 21�r 22)R

f �16r2

3� B r 22 � r1r2 � r 21

(r2 � r1) (r 22 � r 21)R

(r 42 � r 41) � (r2 � r1) (r2 � r1) (r 22 � r 21)

(r 32 � r 31) � (r2 � r1) (r 22 � r1r2 � r 21)

f �z

S�

16r2(r32 � r3

1)

3�(r42 � r4

1)

Problem 6.10-3 A cantilever beam of length L � 54 in. supports a uniform load of intensity q (see figure). The beam is made of steel (�Y � 36 ksi) and has a rectangular cross section of width b � 4.5 in. and height h � 6.0 in.

What load intensity q will produce a fully plastic condition in the beam?

L = 54 in.

q

z

y

C h = 6.0 in.

b = 4.5 in.

Solution 6.10-3 Cantilever beam (rectangular cross section)


MAXIMUM BENDING MOMENT:

PLASTIC MOMENT:

Mmax � MP

qL2

2�sy bh2

4� q �

sy bh2

2L2

MP �sy bh2

4

Mmax �qL2

2SUBSTITUTE NUMERICAL DATA:

L � 54 in. �Y � 36 ksib � 4.5 in. h � 6.0 in.

� q � 1000 lb/in.

L

q

h

b

Problem 6.10-4 A steel beam of rectangular cross section is 50 mm wide and 80 mm high (see figure). The yield stress of the steel is 210 MPa.

(a) What percent of the cross-sectional area is occupied by the elasticcore if the beam is subjected to a bending moment of 13.0 kN m actingabout the z axis?

(b) What is the magnitude of the bending moment that will cause 50%of the cross section to yield?

Solution 6.10-4 Rectangular cross section

z

y

C

50 mm

80 mm

b � 50 mmh � 80 mm

�Y � 210 MPa

(a) ELASTIC CORE (M � 13.0 kN � m)

M is between MY and MP.

Eq. (6-85): e � hB1

2 ¢3

2�

M

My

≤� 32.950 mm

MP �sY bh2

4� 16,800 N � ˇm

MY �sY bh2

6� 11,200 N � ˇm

b

h

Percent of cross-sectional area is

� 82.4 %

(b) ELASTIC CORE

Eq. (6-84):

� 15.4 kN � m

M � MY ¢32

�2e2

h2 ≤¢e �

h

4� 20 ≤

2e

h (100) �

65.90 (100)

80

MM

Problem 6.10-5 Calculate the shape factor f for the wide-flange beam shown in the figure if h � 12.0 in., b � 6.0 in., tf � 0.6 in., and tw � 0.4 in.



z

y

C

tf

tftw

h

b

h � 12.0 in. b � 6.0 in. tf � 0.6 in. tw � 0.4 in.

SECTION MODULUS (S � I/c)

� 276.1 in.4

S �Ic

� 96.0 in.3c �h

2� 6.0 in.

I �1

12 bh3 �

1

12 (b � tw)(h � 2tf)

3

PLASTIC MODULUS (EQ. 6-86)

SHAPE FACTOR (EQ. 6-79)

f �z

S� 1.15

z �1

4[bh2 � (b � tw)(h � 2tf)

2 ] � 52.7 in.3

Problem 6.10-6 Solve the preceding problem for a wide-flange beam with h � 400 mm, b � 150 mm, tf � 12 mm, and tw � 8 mm.

Solution 6.10-6 Wide-flange beamh � 400 mm b � 150 mm tf � 12 mm tw � 8 mm


� 171.0 � 106 mm4

S �Ic

� 854.9 � 103 mm3c �h

2� 200 mm

I �1

12 bh3 �

1

12 (b � tw)(h � 2tf)

3


� 981.2 � 103 mm3


f �z

S� 1.15

z �1

4[bh2 � (b � tw)(h � 2tf)

2 ]

Problem 6.10-7 Determine the plastic modulus Z and shape factor f for a W 10 � 30 wide-flange beam. (Note: Obtain the cross-sectional dimensions and section modulus of the beam from Table E-1, Appendix E.)

Solution 6.10-7 Wide-flange beamW10 � 30h � 10.47 in. b � 5.810 in.tf � 0.510 in. tw � 0.300 in. S � 32.4 in.3


� 36.21 in.3Z �1

4[bh2 � (b � tw)(h � 2tf)

2 ]


f �Z

S� 1.12

Probs. 6.10-5 and 6.10-6

Problem 6.10-8 Solve the preceding problem for a W 8 � 28 wide-flange beam.



W 8 � 28h � 8.06 in. b � 6.535 in.tf � 0.465 in. tw � 0.285 in. S � 24.3 in.3


� 26.70 in.3Z �1

4[bh2 � (b � tw)(h � 2tf)

2 ]


f �Z

S� 1.10

Problem 6.10-9 Determine the yield moment MY, plastic moment MP,and shape factor f for a W 16 � 77 wide-flange beam if �Y � 36 ksi.(Note: Obtain the cross-sectional dimensions and section modulus of the beam from Table E-1, Appendix E.)

Solution 6.10-9 Wide-flange beamW 16 � 77 h � 16.52 in. b � 10.295 in.tf � 0.760 in. tw � 0.455 in. �Y � 36 ksiS � 134 in.3

YIELD MOMENT (EQ. 6-74)

MY � �YS � 4820 k-in.


� 148.9 in.3Z �1

4[bh2 � (b � tw)(h � 2tf)

2 ]

PLASTIC MOMENT (EQ. 6-77)

MP � �YZ � 5360 k-in.


f �MP

MY

� 1.11

Problem 6.10-10 Solve the preceding problem for a W 10 � 45 wide-flange beam.


W 10 � 45 h � 10.10 in. b � 8.020 in.tf � 0.620 in. tw � 0.350 in. �Y � 36 ksiS � 49.1 in.3


MY � �YS � 1770 k-in.


� 54.0 in.3

Z �1

4[bh2 � (b � tw)(h � 2tf)

2 ]


MP � �Y z � 1940 k-in.


f �MP

MY

� 1.10

Problem 6.10-11 A hollow box beam with height h � 16 in., width b � 8 in.,and constant wall thickness t � 0.75 in. is shown in the figure. The beam isconstructed of steel with yield stress �Y � 32 ksi.

Determine the yield moment MY, plastic moment MP, and shape factor f.

Solution 6.10-11 Hollow box beam


z

y

C

t

t

h

b

h � 16 in. b � 8 in.t � 0.75 in. �Y � 32 ksi


� 1079 in.4


MY � �YS � 4320 k-in.

S �Ic

� 134.9 in.3c �h

2� 8.0 in.

I �1

12 bh3 �

1

12 (b � 2t)(h � 2t)3

PLASTIC MODULUS

use (Eq. 6-86) with tw � 2t and tf � t:

� 170.3 in.3


Mp � �yz � 5450 k-in.


f �MP

MY

�Z

S� 1.26

Z �1

4[bh2 � (b � 2t)(h � 2t)2 ]

Problem 6.10-12 Solve the preceding problem for a box beam withdimensions h � 0.4 m, b � 0.2 m, and t � 20 mm. The yield stress of the steel is 230 MPa.

Solution 6.10-12 Hollow box beamh � 400 mm b � 200 mmt � 20 mm �Y � 230 MPa


� 444.6 � 106 mm4


MY � �y S � 511 kN � m

S �Ic

� 2.223 � 106 mm3c �h

2� 200 mm

I �1

12 bh3 �

1

12 (b � 2t)(h � 2t)3

PLASTIC MODULUS

use (Eq. 6-86) with tw � 2t and tf � t:

� 2.816 � 106 mm3


MP � �Yz � 648 kN � m

SHAPE FACTOR (Eq. 6-79)

f �Z

S� 1.27

Z �1

4[bh2 � (b � 2t)(h � 2t)2 ]

Probs. 6.10-11 and 6.10-12

Problem 6.10-13 A hollow box beam with height h � 9.0 in., insideheight h1 � 7.5 in., width b � 5.0 in., and inside width b1 � 4.0 in. isshown in the figure.

Assuming that the beam is constructed of steel with yield stress �Y � 33 ksi, calculate the yield moment MY, plastic moment MP, andshape factor f.



z

y

C hh1

b1

b

h � 9.0 in. b � 5.0 in.h1 � 7.5 in. b1 � 4.0 in. �Y � 33 ksi



MY � �YS � 1196 k-in.

S �Ic

� 36.25 in.3c �h

2� 4.5 in.

I �1

12 (bh3 � b1h1

3) � 163.12 in.4

PLASTIC MODULUS

use (Eq. 6-86) with b � tw � b1 and h � 2tf � h1:


MP � �Yz � 1485 k-in.


f �MP

MY

�Z

S� 1.24

Z �1

4 (bh2 � b1h1

2) � 45.0 in.3

Problem 6.10-14 Solve the preceding problem for a box beam with dimensions h � 200 mm, h1 � 160 mm, b � 150 mm, and b1 � 130 mm. Assume that the beam is constructed of steel with yield stress �Y � 220 MPa.


h � 200 mm b � 150 mmh1 � 160 mm b1 � 130 mm �Y � 220 MPa



MY � �YS � 122 kN � m

S �Ic

� 556.3 � 103 mm3c �h

2� 100 mm

I �1

12 (bh3 � b1h1

3) � 55.63 � 106 mm4

PLASTIC MODULUS

use (Eq. 6-86) with b � tw � b1 and h � 2tf � h1:


MP � �YZ � 147 kN � m


f �MP

MY

�Z

S� 1.20

Z �1

4 (bh2 � b1h1

2) � 668.0 � 103 mm3

Probs. 6.10-13 through 6.10-16

Problem 6.10-15 The hollow box beam shown in the figure is subjectedto a bending moment M of such magnitude that the flanges yield but thewebs remain linearly elastic.

(a) Calculate the magnitude of the moment M if the dimensions ofthe cross section are h � 14 in., h1 � 12.5 in., b � 8 in., and b1 � 7 in.Also, the yield stress is �Y � 42 ksi.

(b) What percent of the moment M is produced by the elastic core?



h � 14 in. b � 8 in.h1 � 12.5 in. b1 � 7 in. �Y � 42 ksi(see Figure 6-47, Example 6-9)

ELASTIC CORE

M1 � �Y S1 � 1094 k-in.

PLASTIC FLANGES

F � force in one flange

M2 � F ¢h � h1

2≤� 3339 k-in.

F �sY b ¢12≤(h � h1) � 252.0 k

S1 �1

6 (b � b1)h2

1 � 26.04 in.3

(a) BENDING MOMENT

M � M1 � M2 � 4430 k-in.

(b) PERCENT DUE TO ELASTIC CORE

Percent �M1

M (100) � 25%

Problem 6.10-16 Solve the preceding problem for a box beam withdimensions h � 400 mm, h1 � 360 mm, b � 200 mm, and b1 � 160 mm,and with yield stress �Y � 220 MPa.

Solution 6.10-16 Hollow box beamh � 400 mm b � 200 mmh1 � 360 mm b1 � 160 mm �Y � 220 MPa(see Figure 6-47, Example 6-9)

ELASTIC CORE

M1 � �YS1 � 190.1 kN � m

PLASTIC FLANGES


M2 � F ¢h � h1

2≤� 334.4 kN � ˇm

F �sY b ¢12≤(h � h1) � 880.0 kN

S1 �1

6 (b � b1)h2

1 � 864 � 103 mm3

(a) BENDING MOMENT

M � M1 � M2 � 524 kN � m


Percent �M1

M (100) � 36%

Problem 6.10-17 A W 12 � 50 wide-flange beam is subjected to abending moment M of such magnitude that the flanges yield but the web remains linearly elastic.

(a) Calculate the magnitude of the moment M if the yield stress is �Y � 36 ksi.

(b) What percent of the moment M is produced by the elastic core?



W 12 � 50 h � 12.19 in. b � 8.080 in.tf � 0.640 in. tw � 0.370 in. �Y � 36 ksi

ELASTIC CORE

M1 � �YS1 � 264.2 k-in.

PLASTIC FLANGES


F � �Ybtf � 186.2 k

M2 � F(h � tf) � 2151 k-in.

S1 �1

6 tw (h � 2tf)

2 � 7.340 in.3

(a) BENDING MOMENT

M � M1 � M2 � 2410 k-in.


Percent �M1

M (100) � 11%

Problem 6.10-18 A singly symmetric beam of T-section (see figure) hascross-sectional dimensions b � 140 mm, a � 200 mm, tw � 20 mm, andtf � 25 mm.

Calculate the plastic modulus Z and the shape factor f.

Solution 6.10-18 Beam of T-section

tf

tw

z

y

O

a

b

b � 140 mm a � 200 mmtw � 20 mm tf � 25 mm

tf

tw

z

y

C

a

b

c1

c2

ELASTIC BENDING

� 78.50 mmc1 � a � tf � c2 � 152.50 mm

� 37.14 � 106 mm4

S �Ic1

� 243.5 � 103 mm3

Iz �1

3tw c 1

3 �1

3 bc 2

3 �1

3(b � tw)(c2 � tf)

3

c2 �a yi Ai

a Ai

�

¢ tf

2≤(btf) � ¢a

2� tf≤(atw)

btf � atw

Problem 6.10-19 A wide-flange beam of unbalanced cross section has the dimensions shown in the figure.

Determine the plastic moment MP if �Y � 36 ksi.



PLASTIC BENDING

A � btf � atw � 7500 mm2

h1 � 187.5 mm

h2 � a � tf � h1 � 37.5 mm

h1tw �A

2

h2

O

h1c1y1

c2y2

f �Z

S� 1.81

Z �A

2(y1 � y2) � 441 � 103 mm3

�

1

2 bh 2

2 �1

2 (b � tw)(h2 � tf)

2

A�2� 23.75 mm

y2 �a yi Ai

A�2

y1 �h1

2� 93.75 mm

z

y10 in.

0.5 in.

0.5 in.

0.5 in.

O7 in.

5 in.

�Y � 36 ksi b1 � 10 in. b2 � 5 in.tw � 0.5 in. d � 8 in. d1 � 7 in.tf � 0.5 in. A � b1tf � b2tf � d1tw � 11.0 in.2

NEUTRAL AXIS UNDER FULLY PLASTIC CONDITIONS

from which we get h1 � 1.50 in.h2 � d � h1 � 8.50 in.

A

2� h1tw � (b1 � tw)tf

z

y

tw

tf

tf

Od1 d

b2

b1

h2

h1

PLASTIC MODULUS

� 1.182 in.

� 4.477 in.

PLASTIC MOMENT

MP � �yz � 1120 k-in.

Z �A

2(y1 � y2) � 31.12 in.3

�(h2 �2)(tw)(h2) � (h2 � tf �2)(b2 � tw)(tf)

A�2

y2 �a yi Ai

A�2

�(h1 �2)(tw)(h1) � (h1 � tf �2)(b1 � tw)(tf)

A�2

y1 �a yi Ai

A�2

Problem 6.10-20 Determine the plastic moment MP for a beam havingthe cross section shown in the figure if �Y � 210 MPa.

Solution 6.10-20 Cross section of beam


z

y

O

30 mm

120 mm

150 mm

250 mm

�Y � 210 MPa d2 � 150 mm d1 � 120 mm

NEUTRAL AXIS FOR FULLY PLASTIC CONDITIONS

Cross section is divided into two equal areas.

� (250 mm)(30 mm) � 13,862 mm2

h2 � 231.0 mmh1 � 150 mm � 250 mm � h2 � 169.0 mm

(h2) (30 mm) �A

2� 6931 mm2

A

2� 6931 mm2

A ��

4[ (150 mm)2 � (120 mm)2]

z

y

O

30 mm

120 mm

150 mm

250 mmh2

h1

PLASTIC MODULUS

for upper half of cross section

for lower half of cross section

(Dimensions are in millimeters)

� 598,000 � 5,400 � 600,400

� 1404 � 103 mm3

PLASTIC MOMENT

MP � �PZ � (210 MPa)(1404 � 103 mm3)

� 295 kN � m

� ¢h2

2≤(30)(h2)

� B ¢h1 � 150

2≤(30)(h1 � 150) R

Z � (h1 � 75)¢�4≤(d 2

2 � d 12)

Z �A

2(y1 � y2) � (a yi Ai)upper � (a yi Ai)lower

y2 �a yi Ai

A�2

y1 �a yi Ai

A�2

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