Stresses in Beams-Basic Topics

Longitudinal Strains in Beams

Problem 5.4-1 Determine the maximum normal strain �max produced ina steel wire of diameter d � 1/16 in. when it is bent around a cylindricaldrum of radius R � 24 in. (see figure).

Solution 5.4-1 Steel wire

5Stresses in Beams(Basic Topics)

285

d

R

R � 24 in.

From Eq. (5-4):

Substitute numerical values:

emax �1�16 in.

2(24 in.) � 1�16 in.� 1300 � 10�6

�d�2

R � d�2�

d

2R � d

emax �y

r

d �1

16 in.

dR

Cylinder

Problem 5.4-2 A copper wire having diameter d � 3 mm is bent into a circle and held with the ends just touching (see figure). If the maximumpermissible strain in the copper is �max � 0.0024, what is the shortestlength L of wire that can be used?

Solution 5.4-2 Copper wire

d = diameter

L = length

d � 3 mm �max � 0.0024

From Eq. (5-4):

Lmin ��demax

�� (3 mm)

0.0024� 3.93 m

emax �y

r�

d�2L�2�

��d

L

L � 2�r�r�L

2�d

�

Problem 5.4-3 A 4.5 in. outside diameter polyethylene pipe designed tocarry chemical wastes is placed in a trench and bent around a quarter-circular 90° bend (see figure). The bent section of the pipe is 46 ft long.

Determine the maximum compressive strain �max in the pipe.

Solution 5.4-3 Polyethylene pipe

286 CHAPTER 5 Stresses in Beams (Basic Topics)

90°

Angle equals 90º or �/2 radians, r � � � radius of curvature

emax ��d

4L�

�

4¢ 4.5 in.

552 in.≤� 6400 � 10�6

r�L

��2�

2L�� emax �

y

r�

d�22L��

r � radius

r

d

L

Problem 5.4-4 A cantilever beam AB is loaded by a couple M0 at its free end (see figure). The length of the beam is L � 1.5 m and thelongitudinal normal strain at the top surface is 0.001. The distance from the top surface of the beam to the neutral surface is 75 mm.

Calculate the radius of curvature �, the curvature �, and the verticaldeflection � at the end of the beam.

Solution 5.4-4 Cantilever beam

A

BM0

�

L

L � length of beamL � 1.5 m �max � 0.001

k�1r

� 0.01333 m�1

∴ r�y

emax�

75 mm

0.001� 75 m

y � 75 mm �emax �y

r

A

C BM0

�

L

��

�

0′

L � length of 90º bendL � 46 ft� 552 in.

d � 4.5 in.

L �2�r

4�

�r

2

Assume that the deflection curve is nearly flat. Then the distance BC is the same as the length Lof the beam.

� � arcsin 0.02 � 0.02 rad� � � (1 � cos �) � (75 m)(1 � cos (0.02 rad))

� 15.0 mm

NOTE: which confirms that the deflection

curve is nearly flat.

L

�� 100,

∴ sin u�Lr

�1.5 m

75 m� 0.02

SECTION 5.4 Longitudinal Strains in Beams 287

Problem 5.4-5 A thin strip of steel of length L � 20 in. and thickness t � 0.2 in. is bent by couples M0 (see figure). The deflection � at themidpoint of the strip (measured from a line joining its end points) is found to be 0.25 in.

Determine the longitudinal normal strain � at the top surface of the strip.

Solution 5.4-5 Thin strip of steel

t

M0M0

�

L2— L

2—

L � 20 in. t � 0.2 in.� � 0.25 in.

The deflection curve is very flat (note that L/� � 80)and therefore � is a very small angle.

For small angles, (� is in radians)

� � � � � cos � � �(1 � cos �)

Substitute numerical values (� � inches):

Solve numerically: � � 200.0 in.

NORMAL STRAIN

(Shortening at the top surface)

e�y

r�

t�2r

�0.1 in.

200 in.� 500 � 10�6

0.25 � r ¢1 � cos 10r≤

� r ¢1 � cos L

2r≤

u� sin u�L�2r

sin u�L�2r

M0M0

�

L2— L

2—

��

0′

� �

Problem 5.4-6 A bar of rectangular cross section is loaded andsupported as shown in the figure. The distance between supports is L � 1.2 m and the height of the bar is h � 100 mm. The deflection � at the midpoint is measured as 3.6 mm.

What is the maximum normal strain � at the top and bottom ofthe bar?

Ph

P

a a

�

L2— L

2—


Solution 5.4-6 Bar of rectangular cross section

L � 1.2 m h � 100 mm � � 3.6 mm

Note that the deflection curve is nearly flat (L/� � 333) and � is a very small angle.

� � r (1 � cos u) � r ¢1 � cos L

2r≤

u�L�2r

(radians)

sin u�L�2r

Substitute numerical values (� � meters):

Solve numerically: � � 50.00 m

NORMAL STRAIN

(Elongation on top; shortening on bottom)

e�y

r�

h�2r

�50 mm

50,000 mm� 1000 � 10�6

0.0036 � r ¢1 � cos 0.6r≤

Ph

P

a a

�

L2— L

2—

� ��

0′

Normal Stresses in Beams

Problem 5.5-1 A thin strip of hard copper (E � 16,400 ksi) havinglength L � 80 in. and thickness t � 3/32 in. is bent into a circle and held with the ends just touching (see figure).

(a) Calculate the maximum bending stress �max in the strip. (b) Does the stress increase or decrease if the thickness of

the strip is increased?

Solution 5.5-1 Copper strip bent into a circle

332

t = — in.

E � 16,400 ksi L � 80 in. t � 3/32 in.

(a) MAXIMUM BENDING STRESS

From Eq. (5-7):

smax �2�E(t�2)

L�

�Et

L

s�Ey

r�

2�Ey

L

L � 2�r � 2�r�r�L

2�


(b) CHANGE IN STRESS

If the thickness t is increased, the stress �max increases.

smax �� (16,400 ksi)(3�32 in.)

80 in.� 60.4 ksi

Problem 5.5-2 A steel wire (E � 200 GPa) of diameter d � 1.0 mm is bent around a pulley of radius R0 � 400 mm (see figure).

(a) What is the maximum stress �max in the wire?(b) Does the stress increase or decrease if the radius of the pulley

is increased?

Solution 5.5-2 Steel wire bent around a pulley

SECTION 5.5 Normal Stresses in Beams 289

d

R0

E � 200 GPa d � 1.0 mm R0 � 400 mm

(a) MAXIMUM STRESS IN THE WIRE

y �d

2� 0.5 mm

r� R0 �d

2� 400 mm � 0.5 mm � 400.5 mm

From Eq. (5-7):


If the radius is increased, the stress �max

decreases.

smax �Ey

r�

(200 GPa) (0.5 mm)

400.5 mm� 250 MPa

Problem 5.5-3 A thin, high-strength steel rule (E � 30 � 106 psi)having thickness t � 0.15 in. and length L � 40 in. is bent by couples M0 into a circular arc subtending a central angle � 45° (see figure).

(a) What is the maximum bending stress �max in the rule? (b) Does the stress increase or decrease if the central angle is

increased?

Solution 5.5-3 Thin steel rule bent into an arc

L = length

M0M0

t

E � 30 � 106 psit � 0.15 in.

L � 40 in. � 45º � 0.78540 rad

(a) MAXIMUM BENDING STRESS

smax �Ey

r�

E(t�2)

L��

Et

2L

L � r �r�L� � radians


� 44,200 psi � 44.2 ksi


If the angle is increased, the stress �max

increases.

smax �(30 � 106 psi) (0.15 in.) (0.78540 rad)

2 (40 in.)

L

�

Problem 5.5-4 A simply supported wood beam AB with span length L � 3.5 m carries a uniform load of intensity q � 6.4 kN/m (see figure).

Calculate the maximum bending stress �max due to the load q if thebeam has a rectangular cross section with width b � 140 mm and heighth � 240 mm.

Solution 5.5-4 Simple beam with uniform load


A

L

B

q

h

b

L � 3.5 m q � 6.4 kN/mb � 140 mm h � 240 mm

smax �Mmax

S�

3qL2

4bh2

Mmax �qL2

8�S �

bh2

6


smax �3(6.4 kN�m)(3.5 m)2

4(140 mm)(240 mm)2 � 7.29 MPa

Problem 5.5-5 Each girder of the lift bridge (see figure) is 180 ft long and simply supported at the ends. The design load for each girder is a uniform load of intensity 1.6 k/ft. The girders are fabricated by welding three steel plates so as to form an I-shaped cross section (see figure) having section modulus S � 3600 in3.

What is the maximum bending stress �max in a girder due to the uniform load?

Solution 5.5-5 Bridge girder

L � 180 ft q � 1.6 k/ft

S � 3600 in.3

smax �(1.6 k �ft) (180 ft)2(12 in.�ft)

8(3600 in.3)� 21.6 ksi

smax �Mmax

S�

qL2

8S

Mmax �qL2

8L

q

Problem 5.5-6 A freight-car axle AB is loaded approximately as shownin the figure, with the forces P representing the car loads (transmitted tothe axle through the axle boxes) and the forces R representing the railloads (transmitted to the axle through the wheels). The diameter of theaxle is d � 80 mm, the distance between centers of the rails is L, and the distance between the forces P and R is b � 200 mm.

Calculate the maximum bending stress �max in the axle if P � 47 kN.

Solution 5.5-6 Freight-car axle


dd

b b

A

P P

RR

L

B

Diameter d � 80 mmDistance b � 200 mmLoad P � 47 kN

Mmax � Pb�S ��d3

32

MAXIMUM BENDING STRESS


smax �32(47 kN)(200 mm)

�(80 mm)3 � 187 MPa

smax �Mmax

S�

32Pb

�d 3

Problem 5.5-7 A seesaw weighing 3 lb/ft of length is occupied by twochildren, each weighing 90 lb (see figure). The center of gravity of eachchild is 8 ft from the fulcrum. The board is 19 ft long, 8 in. wide, and 1.5 in. thick.

What is the maximum bending stress in the board?

Solution 5.5-7 Seesaw

b � 8 in. h � 1.5 in.q � 3 lb/ft P � 90 lb d � 8.0 ft L � 9.5 ft

� 855.4 lb-ft � 10,264 lb-in.

smax �M

S�

10,264 lb-in.

3.0 in.3� 3420 psi

S �bh2

6� 3.0 in3.

Mmax � Pd �qL2

2� 720 lb-ft � 135.4 lb-ft

h

L

d d

qPP

L

b

Problem 5.5-8 During construction of a highway bridge, themain girders are cantilevered outward from one pier toward the next (see figure). Each girder has a cantilever length of 46 mand an I-shaped cross section with dimensions as shown in thefigure. The load on each girder (during construction) is assumedto be 11.0 kN/m, which includes the weight of the girder.

Determine the maximum bending stress in a girder due tothis load.

Solution 5.5-8 Bridge girder


25 mm

50 mm

600 mm

2400 mm

L � 46 mq � 11.0 kN/mb � 600 mm h � 2400 mm

tf � 50 mm tw � 25 mmh1 � h � 2tf � 2300 mmb1 � b � tw � 575 mm

� 129 MPa

smax �Mmax c

I�

(11,638 kN � ˇm)(1.2 m)

0.1082 m4

� 0.6912 m4 � 0.5830 m4 � 0.1082 m4

�1

12 (0.6 m)(2.4 m)3 �

1

12 (0.575 m)(2.3 m)3

I �bh3

12�

b1h31

12

smax �Mmax c

I�c �

h

2� 1200 mm

Mmax �qL2

2�

1

2 (11.0 kN�m)(46 m)2 � 11,638 kN � ˇm

b

tf

h1 h2tw

L

q

Problem 5.5-9 The horizontal beam ABC of an oil-well pump has the cross section shown in the figure. If the vertical pumpingforce acting at end C is 8.8 k, and if the distance from the line of action of that force to point B is 14 ft, what is the maximum bending stress in the beam due to the pumping force?

A B C

0.625in.

0.875 in.

8.0 in.

20.0in.

Solution 5.5-9 Beam in an oil-well pump


L � 14 ftP � 8.8 kb � 8.0 in. h � 20.0 in.tf � 0.875 in. tw � 0.625 in.

h1 � h � 2tf � 18.25 in.b1 � b � tw � 7.375 in.

Mmax � PL � (8.8 k)(14 ft)� 123,200 lb-ft � 1,478,400 lb-in.

� 5,333.3 in.4 � 3,735.7 in.4 � 1,597.7 in.4

� 9250 psi � 9.25 ksi

smax �Mmax c

I�

(1.4784 � 106 lb-in.)(10.0 in.)

1,597.7 in.4

�1

12 (8.0 in.) (20.0 in.)3 �

1

12 (7.375 in.)(18.25 in.)3

I �bh3

12�

b1h13

12

smax �Mmax c

I�c �

h

2� 10.0 in.

b

tf

h1 h2tw

L

P

Problem 5.5-10 A railroad tie (or sleeper) is subjected to two rail loads, each of magnitude P � 175 kN, acting as shown in the figure. The reaction q of the ballast is assumed to be uniformly distributed over the length of the tie, which has cross-sectional dimensions b � 300 mm and h � 250 mm.

Calculate the maximum bending stress �max in the tie due to the loads P, assuming the distance L � 1500 mm and the overhang length a � 500 mm.

Solution 5.5-10 Railroad tie (or sleeper)

L

q

P Pb

h

a a

DATA P � 175 kN b � 300 mm h � 250 mmL � 1500 mm a � 500 mm

BENDING-MOMENT DIAGRAM

�P

4 (2a � L)

�P

L � 2a ¢L

2� a≤

2

�PL

2

M2 �q

2 ¢L

2� a≤

2

�PL

2

M1 �qa2

2�

Pa2

L � 2a

q �2P

L � 2a�S �

bh2

6� 3.125 � 10�3 m3


M1 � 17,500 N � m M2 � �21,875 N � m

Mmax � 21,875 N � m


(Tension on top; compression on bottom)

smax �Mmax

5�

21,875 N � ˇm

3.125 � 10�3 m3 � 7.0 MPa0

M1

M2

M1

Problem 5.5-11 A fiberglass pipe is lifted by a sling, as shown in thefigure. The outer diameter of the pipe is 6.0 in., its thickness is 0.25 in.,and its weight density is 0.053 lb/in.3 The length of the pipe is L � 36 ft and the distance between lifting points is s � 11 ft.

Determine the maximum bending stress in the pipe due to its ownweight.

Solution 5.5-11 Pipe lifted by a sling


s

L

L � 36 ft � 432 in. d2 � 6.0 in. t � 0.25 in.s � 11 ft � 132 in. d1 � d2 � 2t � 5.5 in.

� 0.053 lb/in.3

a � (L � s)/2 � 150 in.

A ��

4 (d 2

2 � d 12) � 4.5160 in.2

q � A � (0.053 lb/in.3)(4.5160 in.2) � 0.23935 lb/in.

I ��

64 (d 2

4 � d 14) � 18.699 in.4

L

q

a as d1

d2

t


Mmax � 2,692.7 lb-in.

M2 � �qL

4 ¢L

2� s≤� �2,171.4 lb-in.

M1 � �qa2

2� �2,692.7 lb-in.


(Tension on top)

smax �(2,692.7 lb-in.)(3.0 in.)

18.699 in.4� 432 psi

smax �Mmax c

I�c �

d2

2� 3.0 in.

0

M1 M1

M2

Problem 5.5-12 A small dam of height h � 2.0 m is constructed ofvertical wood beams AB of thickness t � 120 mm, as shown in the figure.Consider the beams to be simply supported at the top and bottom.

Determine the maximum bending stress �max in the beams, assumingthat the weight density of water is � 9.81 kN/m3.

Solution 5.5-12 Vertical wood beam


h

t

A

B

h � 2.0 mt � 120 mm � 9.81 kN/m3 (water)

Let b � width of beam perpendicular to the plane of the figure

Let q0 � maximum intensity of distributed load

q0 � gbh�S �bt 2

6

MAXIMUM BENDING MOMENT

Substitute into the equation for M:

For the vertical wood beam:

Maximum bending stress

SUBSTITUTE NUMERICAL VALUES:

�max � 2.10 MPa

NOTE: For b � 1.0 m, we obtain q0 � 19,620 N/m, S � 0.0024 m3, Mmax � 5,034.5 N � m, and�max � Mmax/S � 2.10 MPa

smax �Mmax

S�

2q0ˇˇ h2

3�3 bt 2�

2gh3

3�3 t 2

L � h; Mmax �q0ˇˇ h2

9�3

Mmax �q0ˇˇ L

6 ¢ L

�3≤�

q0ˇ

6L ¢ L3

3�3≤�

q0ˇˇ L2

9�3

x � L� �3

dM

dx�

q0ˇˇ L

6�

q0ˇˇ x2

2L� 0�x �

L

�3

�q0ˇˇ Lx

6�

q0ˇˇ x3

6L

M � RAx �q0ˇ x3

6L

RA �q0 ˇL

6

h

A

Bq0

t

L

A B

q0

RA

x

(��)xq � q0 L

Problem 5.5-13 Determine the maximum tensile stress �t (due to purebending by positive bending moments M) for beams having cross sectionsas follows (see figure): (a) a semicircle of diameter d, and (b) an isoscelestrapezoid with bases b1 � b and b2 � 4b/3, and altitude h.

Solution 5.5-13 Maximum tensile stress


C C h

(a) (b)

d b2

b1

(a) SEMICIRCLE

From Appendix D, Case 10:

st �Mc

IC

�768M

(9�2 � 64)d3 � 30.93 M

d3

c �4r

3��

2d

3�

IC �(9� 2 � 64)r4

72��

(9� 2 � 64)d 4

1152�

(b) TRAPEZOID

From Appendix D, Case 8:

st �Mc

IC

�360M

73bh2

c �h(2b1 � b2)

3(b1 � b2)�

10h

21

�73bh3

756

IC �h3(b1

2 � 4b1b2 � b22)

36(b1 � b2)

b1 � b�b2 �4b

3

C

d

c C h

b2

b1

c

Problem 5.5-14 Determine the maximum bending stress �max(due to pure bending by a moment M) for a beam having a cross section in the form of a circular core (see figure). The circle has diameter d and the angle � � 60°. (Hint: Use the formulas given in Appendix D, Cases 9 and 15.)

C

d

�

�

Solution 5.5-14 Circular core


From Appendix D, Cases 9 and 15:

� � radians � radians a � r sin � b � r cos �

�d 4

128 (4b� sin 4b)

��d 4

64�

d 4

32 ¢�

2� b�

1

4 sin 4b≤

��d 4

64�

d 4

32 ¢�

2� b� ¢1

2 sin 2b≤(�cos 2b)≤

��d 4

64�

d 4

32 ¢�

2� b� (sin b cos b)(1 � 2 cos2b)≤

Iy ��d 4

64�

d 4

32 ¢�

2� b� sin b cos b� 2 sin b cos3 b≤

r �d

2� �

�

2� b

Iy ��r4

4�

r4

2 ¢ �

ab

r 2�

2ab3

r4 ≤MAXIMUM BENDING STRESS

For � � 60º � �/3 rad:

smax �576M

(8��3 � 9)d 3� 10.96

M

d 3

smax �64M sin b

d3(4b� sin 4b)

smax �Mc

Iy

�c � r sin b�d

2 sin b

C

d

�

�y y

Problem 5.5-15 A simple beam AB of span length L � 24 ft is subjected to two wheel loads acting at distance d � 5 ft apart (see figure). Each wheel transmits a load P � 3.0 k, and the carriage may occupy any position on the beam.

Determine the maximum bending stress �max due to the wheel loads if the beam is an I-beam having section modulus S � 16.2 in.3

Solution 5.5-15 Wheel loads on a beam

A B C

dP P

L

L � 24 ft � 288 in.d � 5 ft � 60 in.P � 3 kS � 16.2 in.3


dM

dx�

P

L (2L � d � 4x) � 0�x �

L

2�

d

4

M � RAx �P

L (2Lx � dx � 2x2)

RA �P

L (L � x) �

P

L (L � x � d) �

P

L (2L � d � 2x)

Substitute x into the equation for M:



� 21.4 ksi

smax �3k

2(288 in.) (16.2 in.3) (288 in. � 30 in.)2

smax �Mmax

S�

P

2LS ¢L �

d

2≤

2

Mmax �P

2L ¢L �

d

2≤

2

A B

d

P P

LRA

x

Problem 5.5-16 Determine the maximum tensilestress �t and maximum compressive stress �c due to the load P acting on the simple beam AB (see figure).

Data are as follows: P � 5.4 kN, L � 3.0 m, d � 1.2 m, b � 75 mm, t � 25 mm, h � 100 mm, and h1 � 75 mm.

Solution 5.5-16 Simple beam of T-section


d

A B

P

b

h1h

t

L

P � 5.4 kN L � 3.0 m

b � 75 mm t � 25 mm

d � 1.2 m h � 100 mm h1 � 75 mm

PROPERTIES OF THE CROSS SECTION

A � 3750 mm2

c1� 62.5 mm c2 � 37.5 mm

IC� 3.3203 � 106 mm4

REACTIONS OF THE BEAM

RA � 2.16 kN RB � 3.24 kN


Mmax � RA(L � d) � RB(d) � 3888 N � m

MAXIMUM TENSILE STRESS

� 43.9 MPa

MAXIMUM COMPRESSIVE STRESS

� 73.2 MPa

sc �Mmax c1

IC

�(3888 N � ˇm)(0.0625 m)

3.3203 � 106 mm4

st �Mmax c2

IC

�(3888 N � ˇm)(0.0375 m)

3.3203 � 106 mm4

d

A B

P

b

h1 hc1

c2

t

L RBRA

C

Problem 5.5-17 A cantilever beam AB, loaded by a uniform load and a concentrated load (see figure), is constructed of a channel section.

Find the maximum tensile stress �t and maximum compressive stress �c if the cross section has the dimensions indicated and the moment of inertia about the z axis (the neutral axis) is I � 2.81 in.4

(Note: The uniform load represents the weight of the beam.)A B

5.0 ft 3.0 ft

20 lb/ft

z

y

C

0.606 in.

200 lb

2.133 in.

Solution 5.5-17 Cantilever beam (channel section)


Problem 5.5-18 A cantilever beam AB of triangular cross section haslength L � 0.8 m, width b � 80 mm, and height h � 120 mm (see figure).The beam is made of brass weighing 85 kN/m3.

(a) Determine the maximum tensile stress �t and maximumcompressive stress �c due to the beam’s own weight.

(b) If the width b is doubled, what happens to the stresses? (c) If the height h is doubled, what happens to the stresses?

Solution 5.5-18 Triangular beam

I � 2.81 in.4 c1 � 0.606 in. c2 � 2.133 in.

� 1000 lb-ft � 640 lb-ft � 1640 lb-ft

� 19,680 lb-in.

MAXIMUM TENSILE STRESS

� 4,240 psi

MAXIMUM COMPRESSIVE STRESS

� 14,940 psi

sc �Mc2

I�

(19,680 lb-in.) (2.133 in.)

2.81 in.4

st �Mc1

I�

(19,680 lb-in.) (0.606 in.)

2.81 in.4

Mmax � (200 lb)(5.0 ft) � (20 lb�ft) (8.0 ft)¢8.0 ft

2≤

A B

5.0 ft

8.0 ft

3.0 ft

20 lb/ft

z

y

C

0.606 in.

200 lb

2.133 in.

h

A

B

b

L

L

q

h

b

z C

y

h/3

2h3

L � 0.8 m b � 80 mm h � 120 mm � 85 kN/m3

(a) MAXIMUM STRESSES

Tensile stress: st �Mc1

Iz

�3gL2

h

Iz � IC �bh3

36�c1 �

h

3�c2 �

2h

3

q � gA � g ¢bh

2≤�Mmax �

qL2

2�gbhL2

4

Compressive stress: �c � 2�t

Substitute numerical values: �t � 1.36 MPa

�c � 2.72 MPa

(b) WIDTH b IS DOUBLED

No change in stresses.

(c) HEIGHT h IS DOUBLED

Stresses are reduced by half.

Problem 5.5-19 A beam ABC with an overhang from B to C supports auniform load of 160 lb/ft throughout its length (see figure). The beam is a channel section with dimensions as shown in the figure. The moment ofinertia about the z axis (the neutral axis) equals 5.14 in.4

Calculate the maximum tensile stress �t and maximum compressivestress �c due to the uniform load.

Solution 5.5-19 Beam with an overhang


AB

C

10 ft 5 ft

160 lb/ft

z

y

C

0.674 in.

2.496 in.

L � 10 ft

q � 160 lb/ft

CBA

b � 5 ft

z

y

C

0.674 in.

2.496 in.

0

M1

M23.75 ft

Iz � 5.14 in.4

c1 � 0.674 in. c2 � 2.496 in.RA � 600 lb RB � 1800 lbM1 � 1125 lb-ft � 13,500 lb-in.M2 � 2000 lb-ft � 24,000 lb-in.

AT CROSS SECTION OF MAXIMUM POSITIVE

BENDING MOMENT

AT CROSS SECTION OF MAXIMUM NEGATIVE

BENDING MOMENT

MAXIMUM STRESSES

�t � 6,560 psi �c � 11,650 psi

sc �M2c2

Iz

�(24,000 lb-in.) (2.496 in.)

5.14 in.4� 11,650 psi

st �M2c1

Iz

�(24,000 lb-in.) (0.674 in.)

5.14 in.4� 3,150 psi

sc �M1c1

Iz

�(13,500 lb-in.) (0.674 in.)

5.14 in.4� 1,770 psi

st �M1c2

Iz

�(13,500 lb-in.) (2.496 in.)

5.14 in.4� 6,560 psi

Problem 5.5-20 A frame ABC travels horizontally with an accelerationa0 (see figure). Obtain a formula for the maximum stress �max in thevertical arm AB, which has length L, thickness t, and mass density �.

L

CB

At

a0 = acceleration

Solution 5.5-20 Accelerating frame

SECTION 5.5 Normal Stresses in Beamss 301

L � length of vertical armt � thickness of vertical arm� � mass density

a0 � accelerationLet b � width of arm perpendicular to the plane of the figureLet q � inertia force per unit distance along vertical arm

VERTICAL ARM

S �bt 2

6�smax �

Mmax

S�

3rL2a0

t

q � rbta0�Mmax �qL2

2�rbta0 L2

2

TYPICAL UNITS FOR USE

IN THE PRECEDING EQUATION

SI UNITS: � � kg/m3 � N � s2/m4

L � meters (m)

a0 � m/s2

t � meters (m)

�max � N/m2 (pascals)

USCS UNITS: � � slug/ft3 � lb-s2/ft4

L � ft a0 � ft/s2 t � ft

�max � lb/ft2 (Divide by 144 to obtain psi)

qL

t

Problem 5.5-21 A beam of T-section is supported and loaded as shown in the figure. The cross section has width b �2 1/2 in., height h � 3 in., and thickness t � 1/2 in.

Determine the maximum tensile and compressive stresses in the beam.

Solution 5.5-21 Beam of T-section

q = 80 lb/ftP = 625 lb

h =3 in.

b = 2 in.1—2

L1 = 4 ft

L3 = 5 ftL2 = 8 ft

t = 1—2 in.

t = 1—2 in.

L1 � 4 ft � 48 in. L2 � 8 ft � 96 in. L3 � 5 ft � 60 in.

P � 625 lb q � 80 lb/ft � 6.6667 lb/in.


b � 2.5 in. h � 3.0 in. t � 0.5 in.A � bt � (h � t)t � 2.50 in.2

c1 � 2.0 in. c2 � 1.0 in.

REACTIONS

RA � 187.5 lb (upward)RB � 837.5 lb (upward)

IC �25

12 in.4 � 2.0833 in.4


AT CROSS SECTION OF MAXIMUM POSITIVE

BENDING MOMENT

AT CROSS SECTION OF MAXIMUM NEGATIVE

BENDING MOMENT

MAXIMUM STRESSES

�t � 11,520 psi �c � 8,640 psi

st �M2c1

IC

� 11,520 psi�sc �M2c2

IC

� 5,760 psi

st �M1c2

IC

� 4,320 psi�sc �M1c1

IC

� 8,640 psi

qP

L1

L3L2RA RB

AB

C

h

b

C

t

t

c1

c2

M1 � RA L1 � 9,000 lb – in.

M2 � �qL3

2

� �12,000 lb – in.�� 2

Problem 5.5-22 A cantilever beam AB with a rectangular crosssection has a longitudinal hole drilled throughout its length (seefigure). The beam supports a load P � 600 N. The cross section is 25 mm wide and 50 mm high, and the hole has a diameter of 10 mm.

Find the bending stresses at the top of the beam, at the top ofthe hole, and at the bottom of the beam.

Solution 5.5-22 Rectangular beam with a hole


50 mm12.5 mm

25 mm

10 mm

A B

P = 600 N

L = 0.4 m

37.5 mm


M � PL � (600 N)(0.4 m) � 240 N � m


A1 � area of rectangle

� (25 mm)(50 mm) � 1250 mm2

A2 � area of hole

A � area of cross section

� A1 � A2 � 1171.5 mm2

Using line B-B as reference axis:

∑Aiyi � A1(25 mm) � A2(37.5 mm) � 28,305 mm3

Distances to the centroid C:

c1 � 50 mm � c2 � 25.838 mm

c2 � y � 24.162 mm

y �a Ai yi

A�

28,305 mm3

1171.5 mm2 � 24.162 mm

��

4(10 mm)2 � 78.54 mm2

MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS

(THE z AXIS)

All dimensions in millimeters.

Rectangle: Iz � Ic � Ad 2

Hole:

� 490.87 � 13,972 � 14,460 mm4

Cross-section: I � 261,300 � 14,460 � 246,800 mm4

STRESS AT THE TOP OF THE BEAM

� 25.1 MPa(tension)

STRESS AT THE TOP OF THE HOLE

(tension)

STRESS AT THE BOTTOM OF THE BEAM

� �23.5 MPa(compression)

s3 � �Mc2

I� �

(240 N � ˇm)(24.162 mm)

246,800 mm4

s2 �(240 N � ˇm)(18.338 mm)

246,800 mm4 � 17.8 MPa

s2 �My

I�y � c1 � 7.5 mm � 18.338 mm

s1 �Mc1

I�

(240 N � ˇm)(25.838 mm)

246,800 mm4

Iz � Ic � Ad 2 ��

64(10)4 � (78.54)(37.5 � 24.162)2

� 260,420 � 878 � 261,300 mm4

�1

12(25)(50)3 � (25)(50)(25 � 24.162)2

c1

c2

y

B B

Cy

zC

–

Problem 5.5-23 A small dam of height h � 6 ft is constructed ofvertical wood beams AB, as shown in the figure. The wood beams,which have thickness t � 2.5 in., are simply supported by horizontalsteel beams at A and B.

Construct a graph showing the maximum bending stress �max inthe wood beams versus the depth d of the water above the lowersupport at B. Plot the stress �max (psi) as the ordinate and the depth d (ft) as the abscissa. (Note: The weight density of water equals 62.4 lb/ft3.)

Solution 5.5-23 Vertical wood beam in a dam


Wood beam

Wood beam

Steel beam

Steel beam

Side view Top view

B

A

h

d

t

t

h � 6 ftt � 2.5 in.

� 62.4 lb/ft3

Let b � width of beam(perpendicular to thefigure)Let q0 � intensity ofload at depth dq0 � bd

ANALYSIS OF BEAM

L � h � 6 ft

x0 � dB d

3L

RB �q0 d

6 ¢3 �

d

L≤

RA �q0 d 2

6L


Section modulus:

q0 � bd

SUBSTITUTE NUMERICAL VALUES:d � depth of water (ft) (Max. d � h � 6 ft)L � h � 6 ft � 62.4 lb/ft3 t � 2.5 in.�max � psi

d (ft) �max (psi)

0 01 92 593 1714 3475 5736 830

� 0.1849d 3(54 � 9d � d�2d)

smax �(62.4)d 3

(2.5)2 ¢1 �d

6�

d

9B d

18≤

smax �gd 3

t 2 ¢1 �

d

L�

2d

3LB d

3L≤

smax �Mmax

S�

6

bt 2B q0 d 2

6¢1 �

d

L�

2d

3LB d

3L≤ R

S �1

6 bt 2

A

Bq0

d

h

t

CA

q0

B

d

L

RBRA

V

M

x0

0

0

MC

C

RA

�RB

Mmax

Mmax �q0 d 2

6 ¢1 �

d

L�

2d

3LB d

3L≤

830 psi

0 1 2 3 4 5 6 d (ft)

250

500

750

1000

�max(psi)

MC � RA(L � d) �q0 d 2

6 ¢1 �

d

L≤

Stresses in Beams-Basic Topics

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