Solution of Chapter- 05 - stresses in beam - Strength of Materials by Singer

Stresses in Beams
Forces and couples acting on the beam cause bending (flexural stresses) and shearing
stresses on any cross section of the beam and deflection perpendicular to the
longitudinal axis of the beam. If couples are applied to the ends of the beam and no
forces act on it, the bending is said to be pure bending. If forces produce the bending,
the bending is called ordinary bending.
ASSUMPTIONS
In using the following formulas for flexural and shearing stresses, it is assumed that a
plane section of the beam normal to its longitudinal axis prior to loading remains plane
after the forces and couples have been applied, and that the beam is initially straight
and of uniform cross section and that the moduli of elasticity in tension and
compression are equal.
Flexure Formula
Stresses caused by the bending moment are known as flexural or bending stresses.
Consider a beam to be loaded as shown.
Consider a fiber at a distance y from the neutral axis, because of the beam’s curvature,
as the effect of bending moment, the fiber is stretched by an amount of cd. Since the
curvature of the beam is very small, bcd and Oba are considered as similar triangles.
The strain on this fiber is
By Hooke’s law, ε = σ / E, then
which means that the stress is proportional to the distance y from the neutral axis.

Considering a differential area dA at a distance y from N.A., the force acting over the
area is
The resultant of all the elemental moment about N.A. must be equal to the bending
moment on the section.
but then
substituting ρ = Ey / fb
then
and
The bending stress due to beams curvature is

The beam curvature is:
where ρ is the radius of curvature of the beam in mm (in), M is the bending moment in
N·mm (lb·in), fb is the flexural stress in MPa (psi), I is the centroidal moment of inertia
in mm4
(in4
), and c is the distance from the neutral axis to the outermost fiber in mm
(in).
SECTION MODULUS
In the formula
the ratio I/c is called the section modulus and is usually denoted by S with units of mm3
(in3
). The maximum bending stress may then be written as
This form is convenient because the values of S are available in handbooks for a wide
range of standard structural shapes.
Solved Problems in Flexure Formula
Problem 503
A cantilever beam, 50 mm wide by 150 mm high and 6 m long, carries a load that
varies uniformly from zero at the free end to 1000 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 20 mm from the top of the beam at a section 2 m
from the free end.

Problem 504
A simply supported beam, 2 in wide by 4 in high and 12 ft long is subjected to a
concentrated load of 2000 lb at a point 3 ft from one of the supports. Determine the
maximum fiber stress and the stress in a fiber located 0.5 in from the top of the beam
at midspan.
Solution 504

Problem 505
A high strength steel band saw, 20 mm wide by 0.80 mm thick, runs over pulleys 600
mm in diameter. What maximum flexural stress is developed? What minimum diameter
pulleys can be used without exceeding a flexural stress of 400 MPa? Assume E = 200
GPa.
Solution 505
Problem 506
A flat steel bar, 1 inch wide by ¼ inch thick and 40 inches long, is bent by couples
applied at the ends so that the midpoint deflection is 1.0 inch. Compute the stress in
the bar and the magnitude of the couples. Use E = 29 × 106
psi.
Solution 506

Problem 507
In a laboratory test of a beam loaded by end couples, the fibers at layer AB in Fig. P-
507 are found to increase 60 × 10–3
mm whereas those at CD decrease 100 × 10–3
mm
in the 200-mm-gage length. Using E = 70 GPa, determine the flexural stress in the top
and bottom fibers.
Solution 507

Problem 508
Determine the minimum height h of the beam shown in Fig. P-508 if the flexural stress
is not to exceed 20 MPa.

Solution 508
Problem 509
A section used in aircraft is constructed of tubes connected by thin webs as shown in
Fig. P-509. Each tube has a cross-sectional area of 0.20 in2. If the average stress in the
tubes is no to exceed 10 ksi, determine the total uniformly distributed load that can be
supported in a simple span 12 ft long. Neglect the effect of the webs.

Solution 509
Problem 510
A 50-mm diameter bar is used as a simply supported beam 3 m long. Determine the
largest uniformly distributed load that can be applied over the right two-thirds of the
beam if the flexural stress is limited to 50 MPa.

Solution 510
Problem 511
A simply supported rectangular beam, 2 in wide by 4 in deep, carries a uniformly
distributed load of 80 lb/ft over its entire length. What is the maximum length of the
beam if the flexural stress is limited to 3000 psi?
Solution 511

Problem 512
The circular bar 1 inch in diameter shown in Fig. P-512 is bent into a semicircle with a
mean radius of 2 ft. If P = 400 lb and F = 200 lb, compute the maximum flexural stress
developed in section a-a. Neglect the deformation of the bar.
Solution 512
Problem 513
A rectangular steel beam, 2 in wide by 3 in deep, is loaded as shown in Fig. P-513.
Determine the magnitude and the location of the maximum flexural stress.

Problem 514
The right-angled frame shown in Fig. P-514 carries a uniformly distributed loading
equivalent to 200 N for each horizontal projected meter of the frame; that is, the total
load is 1000 N. Compute the maximum flexural stress at section a-a if the cross-section
is 50 mm square.
Solution 514

Problem 515
Repeat Prob. 524 to find the maximum flexural stress at section b-b.
Solution 515
Problem 516
A timber beam AB, 6 in wide by 10 in deep and 10 ft long, is supported by a guy wire
AC in the position shown in Fig. P-516. The beam carries a load, including its own
weight, of 500 lb for each foot of its length. Compute the maximum flexural stress at
the middle of the beam.

Solution 516
Problem 517
A rectangular steel bar, 15 mm wide by 30 mm high and 6 m long, is simply supported
at its ends. If the density of steel is 7850 kg/m3
, determine the maximum bending
stress caused by the weight of the bar.
Solution 517

Problem 518
A cantilever beam 4 m long is composed of two C200 × 28 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 120 MPa if (a) the webs are vertical and
(b) the webs are horizontal? Refer to Appendix B of text book for channel properties.
Solution 518

Problem 519
A 30-ft beam, simply supported at 6 ft from either end carries a uniformly distributed
load of intensity wo over its entire length. The beam is made by welding two S18 × 70
(see appendix B of text book) sections along their flanges to form the section shown in
Fig. P-519. Calculate the maximum value of wo if the flexural stress is limited to 20 ksi.
Be sure to include the weight of the beam.
Solution 519

Problem 520
A beam with an S310 × 74 section (see Appendix B of textbook) is used as a simply
supported beam 6 m long. Find the maximum uniformly distributed load that 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 120 MPa.
Solution 520

Problem 521
A beam made by bolting two C10 × 30 channels back to back, is simply supported at its
ends. The beam supports a central concentrated load of 12 kips and a uniformly
distributed load of 1200 lb/ft, including the weight of the beam. Compute the maximum
length of the beam if the flexural stress is not to exceed 20 ksi.
Solution 521

Problem 522
A box beam is composed of four planks, each 2 inches by 8 inches, securely spiked
together to form the section shown in Fig. P-522. Show that INA = 981.3 in4
. If wo = 300
lb/ft, find P to cause a maximum flexural stress of 1400 psi.
Solution 522
Problem 523
Solve Prob. 522 if wo = 600 lb/ft.

Problem 524
A beam with an S380 &times 74 section carries a total uniformly distributed load of 3W
and a concentrated load W, as shown in Fig. P-524. Determine W if the
flexural stress is limited to 120 MPa.
Solution 524

Problem 525
A square timber beam used as a railroad tie is supported by a uniformly distributed
loads and carries two uniformly distributed loads each totaling 48 kN as shown in Fig. P-
525. Determine the size of the section if the maximum stress is limited to 8 MPa.
Solution 525
Problem 526
A wood beam 6 in wide by 12 in deep is loaded as shown in Fig. P-526. If the maximum
flexural stress is 1200 psi, find the maximum values of wo and P which can be applied
simultaneously?

Problem 527
In Prob. 526, if the load on the overhang is 600 lb/ft and the overhang is x ft long, find
the maximum values of P and x that can be used simultaneously.
Solution 527

Economic Sections
From the flexure formula fb = My / I, it can be seen that the bending stress at the
neutral axis, where y = 0, is zero and increases linearly outwards. This means that for a
rectangular or circular section a large portion of the cross section near the middle
section is understressed.
For steel beams or composite beams, instead of adopting the rectangular shape, the
area may be arranged so as to give more area on the outer fiber and maintaining the
same overall depth, and saving a lot of weight.
When using a wide flange or I-beam section for long beams, the compression flanges
tend to buckle horizontally sidewise. This buckling is a column effect, which may be
prevented by providing lateral support such as a floor system so that the full allowable
stresses may be used, otherwise the stress should be reduced. The reduction of stresses
for these beams will be discussed in steel design. In selecting a structural section to be
used as a beam, the resisting moment must be equal or greater than the applied
bending moment. Note: ( fb )max = M/S.
The equation above indicates that the required section modulus of the beam must be
equal or greater than the ratio of bending moment to the maximum allowable stress. A
check that includes the weight of the selected beam is necessary to complete the
calculation. In checking, the beams resisting moment must be equal or greater than the
sum of the live-load moment caused by the applied loads and the dead-load moment
caused by dead weight of the beam.
Dividing both sides of the above equation by ( fb )max, we obtain the checking equation
Assume that the beams in the following problems are properly braced against lateral
deflection. Be sure to include the weight of the beam itself.

Solved Problems in Economic Sections
Problem 529
A 10-m beam simply supported at the ends carries a uniformly distributed load of 16
kN/m over its entire length. What is the lightest W shape beam that will not exceed a
flexural stress of 120 MPa? What is the actual maximum stress in the beam selected?
Solution 529

Problem 530
Repeat Prob. 529 if the distributed load is 12 kN/m and the length of the beam is 8 m.
Solution 530

Problem 531
A 15-ft beam simply supported at the ends carries a concentrated load of 9000 lb at
midspan. Select the lightest S section that can be employed using an allowable stress of
18 ksi. What is the actual maximum stress in the beam selected?
Solution 531

Problem 532
A beam simply supported at the ends of a 25-ft span carries a uniformly distributed load
of 1000 lb/ft over its entire length. Select the lightest S section that can be used if the
allowable stress is 20 ksi. What is the actual maximum stress in the beam selected?
Solution 532

Problem 533
A beam simply supported on a 36-ft span carries a uniformly distributed load of 2000
lb/ft over the middle 18 ft. Using an allowable stress of 20 ksi, determine the lightest
suitable W shape beam. What is the actual maximum stress in the selected beam?
Solution 533

Problem 534
Repeat Prob. 533 if the uniformly distributed load is changed to 5000 lb/ft.
Solution 534

Problem 535
A simply supported beam 24 ft long carries a uniformly distributed load of 2000 lb/ft
over its entire length and a concentrated load of 12 kips at 8 ft from left end. If the
allowable stress is 18 ksi, select the lightest suitable W shape. What is the actual
maximum stress in the selected beam?
Solution 535

Problem 536
A simply supported beam 10 m long carries a uniformly distributed load of 20 kN/m
over its entire length and a concentrated load of 40 kN at midspan. If the allowable
stress is 120 MPa, determine the lightest W shape beam that can be used.
Solution 536

Floor Framing
In floor framing, the subfloor is supported by light beams called floor joists or simply
joists which in turn supported by heavier beams called girders then girders pass the
load to columns. Typically, joist act as simply supported beam carrying a uniform load
of magnitude p over an area of sL,
where
p = floor load per unit area
L = length (or span) of joist
s = center to center spacing of joists and
wo = sp = intensity of distributed load in joist.

Solved Problems in Floor Framing
Problem 538
Floor joists 50 mm wide by 200 mm high, simply supported on a 4-m span, carry a floor
loaded at 5 kN/m2
. Compute the center-line spacing between joists to develop a
bending stress of 8 MPa. What safe floor load could be carried on a center-line spacing
of 0.40 m?
Solution 538

Problem 539
Timbers 12 inches by 12 inches, spaced 3 feet apart on centers, are driven into the
ground and act as cantilever beams to back-up the sheet piling of a coffer dam. What is
the maximum safe height of water behind the dam if water weighs = 62.5 lb/ft3
and ( fb
)max = 1200 psi?
Solution 539
Problem 540
Timbers 8 inches wide by 12 inches deep and 15 feet long, supported at top and
bottom, back up a dam restraining water 9 feet deep. Water weighs 62.5 lb/ft3
. (a)
Compute the center-line spacing of the timbers to cause fb = 1000 psi. (b) Will this
spacing be safe if the maximum fb, ( fb )max = 1600 psi, and the water reaches its
maximum depth of 15 ft?

Problem 541
The 18-ft long floor beams in a building are simply supported at their ends and carry a
floor load of 0.6 lb/in2
. If the beams have W10 × 30 sections, determine the center-line
spacing using an allowable flexural stress of 18 ksi.
Solution 541

Problem 542
Select the lightest W shape sections that can be used for the beams and girders in
Illustrative Problem 537 of text book if the allowable flexural stress is 120 MPa. Neglect
the weights of the members.
Solution 542

Problem 543
A portion of the floor plan of a building is shown in Fig. P-543. The total loading
(including live and dead loads) in each bay is as shown. Select the lightest suitable W if
the allowable flexural stress is 120 MPa.
Solution 543

Unsymmetrical Beams
Flexural Stress varies directly linearly with distance from the neutral axis. Thus for a
symmetrical section such as wide flange, the compressive and tensile stresses will be
the same. This will be desirable if the material is both equally strong in tension and
compression. However, there are materials, such as cast iron, which are strong in
compression than in tension. It is therefore desirable to use a beam with unsymmetrical
cross section giving more area in the compression part making the stronger fiber
located at a greater distance from the neutral axis than the weaker fiber. Some of these
sections are shown below.
The proportioning of these sections is such that the ratio of the distance of the neutral
axis from the outermost fibers in tension and in compression is the same as the ratio of
the allowable stresses in tension and in compression. Thus, the allowable stresses are
reached simultaneously.
In this section, the following notation will be use:
fbt = flexure stress of fiber in tension
fbc = flexure stress of fiber in compression
N.A. = neutral axis
yt = distance of fiber in tension from N.A.
yc = distance of fiber in compression from N.A.
Mr = resisting moment
Mc = resisting moment in compression
Mt = resisting moment in tension

Solved Problems in Unsymmetrical Beams
Problem 548
The inverted T section of a 4-m simply supported beam has the properties shown in Fig.
P-548. The beam carries a uniformly distributed load of intensity wo over its entire
length. Determine wo if fbt ≤ 40 MPa and fbc ≤ 80 MPa.
Solution 548

Problem 549
A beam with cross-section shown in Fig. P-549 is loaded in such a way that the
maximum moments are +1.0P lb·ft and -1.5P lb·ft, where P is the applied load in
pounds. Determine the maximum safe value of P if the working stresses are 4 ksi in
tension and 10 ksi in compression.
Solution 549

Problem 550
Resolve Prob. 549 if the maximum moments are +2.5P lb·ft and -5.0P lb·ft.
Solution 550

Problem 551
Find the maximum tensile and compressive flexure stresses for the cantilever beam
shown in Fig. P-551.
Solution 551

Problem 552
A cantilever beam carries the force and couple shown in Fig. P-552. Determine the
maximum tensile and compressive bending stresses developed in the beam.
Solution 552

Problem 553
Determine the maximum tensile and compressive bending stresses developed in the
beam as shown in Fig. P-553.
Solution 553

Problem 554
Determine the maximum tensile and compressive stresses developed in the
overhanging beam shown in Fig. P-554. The cross-section is an inverted T with the
given properties.
Solution 554

Problem 555
A beam carries a concentrated load W and a total uniformly distributed load of 4W as
shown in Fig. P-555. What safe value of W can be applied if fbc ≤ 100 MPa and fbt ≤ 60
MPa? Can a greater load be applied if the section is inverted? Explain.

Problem 556
A T beam supports the three concentrated loads shown in Fig. P-556. Prove that the NA
is 3.5 in. above the bottom and that INA = 97.0 in4
. Then use these values to determine
the maximum value of P so that fbt ≤ 4 ksi and fbc ≤ 10 ksi.
Solution 556

Problem 557
A cast-iron beam 10 m long and supported as shown in Fig. P-557 carries a uniformly
distributed load of intensity wo (including its own weight). The allowable stresses are fbt
≤ 20 MPa and fbc ≤ 80 MPa. Determine the maximum safe value of wo if x = 1.0 m.
Solution 557

Problem 558
In Prob. 557, find the values of x and wo so that wo is a maximum.
Solution 558

Solution of Chapter- 05 - stresses in beam - Strength of Materials by Singer

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