PROBLEM 1.

29

Two wooden members of uniform rectangular cross section are joined

by the simple glued scarf splice shown. Knowing that P  11 kN,
determine the normal and shearing stresses in the glued splice.

SOLUTION

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 PROBLEM 1.30

Two wooden members of uniform rectangular cross section are joined

by the simple glued scarf splice shown. Knowing that the maximum
allowable shearing stress in the glued splice is 620 kPa, determine
(a) the largest load P that can be safely applied, (b) the corresponding
tensile stress in the splice.

SOLUTION

(a)


(b)


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 PROBLEM 1.31

The 1.4-kip load P is supported by two wooden members of uniform cross section
that are joined by the simple glued scarf splice shown. Determine the normal and
shearing stresses in the glued splice.

SOLUTION

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 PROBLEM 1.32

Two wooden members of uniform cross section are joined by the simple scarf splice
shown. Knowing that the maximum allowable tensile stress in the glued splice is 75 psi,
determine (a) the largest load P that can be safely supported, (b) the corresponding
shearing stress in the splice.

SOLUTION

(a) 

(b) 

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 PROBLEM 1.33

A centric load P is applied to the granite block shown. Knowing that the
resulting maximum value of the shearing stress in the block is 2.5 ksi, determine
(a) the magnitude of P, (b) the orientation of the surface on which the maximum
shearing stress occurs, (c) the normal stress exerted on that surface, (d ) the
maximum value of the normal stress in the block.

SOLUTION

(a) 
(b) 

(c) 

(d ) 

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 PROBLEM 1.34

A 240-kip load P is applied to the granite block shown. Determine the resulting
maximum value of (a) the normal stress, (b) the shearing stress. Specify the
orientation of the plane on which each of these maximum values occurs.

SOLUTION

(a) max tensile stress  0 at

max. compressive stress  at 

(b)

at

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 PROBLEM 1.35

A steel pipe of 400-mm outer diameter is fabricated from 10-mm thick

plate by welding along a helix that forms an angle of 20 with a plane
perpendicular to the axis of the pipe. Knowing that a 300-kN axial
force P is applied to the pipe, determine the normal and shearing
stresses in directions respectively normal and tangential to the weld.

SOLUTION

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 PROBLEM 1.36

A steel pipe of 400-mm outer diameter is fabricated from 10-mm

thick plate by welding along a helix that forms an angle of 20° with a
plane perpendicular to the axis of the pipe. Knowing that the
maximum allowable normal and shearing stresses in the directions
respectively normal and tangential to the weld are and
determine the magnitude P of the largest axial force
that can be applied to the pipe.

SOLUTION

Based on

Based on

Smaller value is the allowable value of P. 

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 PROBLEM 1.37 1
4
The horizontal link BC is in. thick, has a width w = 1.25 in., and is
made of a steel with a 65-ksi ultimate strength in tension. What is the
factor of safety if the structure shown is designed to support a load of P
= 10 kips?

SOLUTION

Use member ACD as a free body, and note that member BC is a two-force member.

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 PROBLEM 1.38

Member ABC, which is supported by a pin and bracket at C and a cable

BD, was designed to support the 16-kN load P as shown. Knowing that
the ultimate load for cable BD is 100 kN, determine the factor of safety
with respect to cable failure.

SOLUTION

Use member ABC as a free body, and note that member BD is a two-force member.

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 PROBLEM 1.39

Knowing that the ultimate load for cable BD is 100 kN and that a factor
of safety of 3.2 with respect to cable failure is required, determine the
magnitude of the largest force P that can be safely applied as shown to
member ABC.

SOLUTION

Use member ABC as a free body, and note that member BD is a two-force member.

Allowable value of FBD :

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 PROBLEM 1.40

A 20 mm-diameter rod made of the same material as rods AC and AD in

the truss shown was tested to failure and an ultimate load of 130 kN was
recorded. Using a factor of safety of 3.0, determine the required diameter
(a) of rod AC, (b) of rod AD.

SOLUTION

Forces in AC and AD.

Joint C:

Joint D:

Ultimate stress. From test on rod:

Allowable stress:

(a) Diameter of rod AC.



(b) Diameter of rod AD.

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 PROBLEM 1.41

In the truss shown, members AC and AD consist of rods made of the

same metal alloy. Knowing that AC is of 25-mm diameter and that
the ultimate load for that rod is 345 kN, determine (a) the factor of
safety for AC, (b) the required diameter of AD if it is desired that both
rods have the same factor of safety.

SOLUTION

Forces in AC and AD.

Joint C:

Joint D:

(a) Factor of safety for AC. 

(b) For the same factor of safety in AC and AD,

Required diameter: 

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 PROBLEM 1.42

Link AB is to be made of a steel for which the ultimate normal stress is

65 ksi. Determine the cross-sectional area of AB for which the factor
of safety will be 3.20. Assume that the link will be adequately
reinforced around the pins at A and B.

SOLUTION

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 PROBLEM 1.43

Two wooden members are joined by plywood splice plates that are fully glued on
the contact surfaces. Knowing that the clearance between the ends of the members
is 6 mm and that the ultimate shearing stress in the glued joint is 2.5 MPa,
determine the length L for which the factor of safety is 2.75 for the loading shown.

SOLUTION

On one face of the upper contact surface,

Since there are 2 contact surfaces,

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 PROBLEM 1.44

For the joint and loading of Prob. 1.43, determine the factor of safety when
L = 180 mm.

PROBLEM 1.43 Two wooden members are joined by plywood splice plates that
are fully glued on the contact surfaces. Knowing that the clearance between the
ends of the members is 6 mm and that the ultimate shearing stress in the glued
joint is 2.5 MPa, determine the length L for which the factor of safety is 2.75 for
the loading shown.

SOLUTION

Area of one face of upper contact surface:

Since there are two surfaces,

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 PROBLEM 1.45

1
Two plates, each in. thick, are used to splice a plastic strip as
8
shown. Knowing that the ultimate shearing stress of the bonding
between the surfaces is 130 psi, determine the factor of safety with
respect to shear when P = 325 lb.

SOLUTION

Bond area,

Factor of safety:

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 PROBLEM 1.46

Three steel bolts are to be used to attach the steel plate shown to a wooden beam.
Knowing that the plate will support a load P = 28 kips, that the ultimate shearing
stress for the steel used is 52 ksi, and that a factor of safety of 3.25 is desired,
determine the required diameter of the bolts.

SOLUTION

For each bolt,

Required:

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 PROBLEM 1.47

A load P is supported as shown by a steel pin that has been inserted in a

short wooden member hanging from the ceiling. The ultimate strength of
the wood used is 60 MPa in tension and 7.5 MPa in shear, while the
ultimate strength of the steel is 145 MPa in shear. Knowing that
and determine the load P if an
overall factor of safety of 3.2 is desired.

SOLUTION

Based on double shear in pin,

Based on tension in wood,

Based on double shear in the wood,

Use smallest

Allowable: 

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 PROBLEM 1.48

For the support of Prob. 1.47, knowing that the diameter of the pin is
and that the magnitude of the load is determine
(a) the factor of safety for the pin, (b) the required values of b and c if the
factor of safety for the wooden members is the same as that found in part a
for the pin.

PROBLEM 1.47 A load P is supported as shown by a steel pin that has

been inserted in a short wooden member hanging from the ceiling. The
ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in
shear, while the ultimate strength of the steel is 145 MPa in shear. Knowing
that and determine the load P if an
overall factor of safety of 3.2 is desired.

SOLUTION

(a) Pin:

Double shear:

(b) Tension in wood:

Shear in wood:

Double shear: each area is

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 PROBLEM 1.49

A steel plate in. thick is embedded in a

concrete wall to anchor a high-strength cable
as shown. The diameter of the hole in the plate is
in., the ultimate strength of the steel used is
36 ksi, and the ultimate bonding stress between
plate and concrete is 300 psi. Knowing that a
factor of safety of 3.60 is desired when
P = 2.5 kips, determine (a) the required width a
of the plate, (b) the minimum depth b to which a
plate of that width should be embedded in the
concrete slab. (Neglect the normal stresses
between the concrete and the end of the plate.)

SOLUTION

Based on tension in plate,

Solving for a,

(a) 
Based on shear between plate and concrete slab,

Solving for b,
(b) 

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 PROBLEM 1.50

Determine the factor of safety for the cable

anchor in Prob. 1.49 when P  2.5 kips, knowing
that a  2 in. and b  6 in.

PROBLEM 1.49 A steel plate in. thick is

embedded in a concrete wall to anchor a high-
strength cable as shown. The diameter of the hole
in the plate is in., the ultimate strength of the
steel used is 36 ksi, and the ultimate bonding
stress between plate and concrete is 300 psi.
Knowing that a factor of safety of 3.60 is desired
when P = 2.5 kips, determine (a) the required
width a of the plate, (b) the minimum depth b to
which a plate of that width should be embedded
in the concrete slab. (Neglect the normal stresses
between the concrete and the end of the plate.)

SOLUTION

Based on tension in plate,

Based on shear between plate and concrete slab,

Actual factor of safety is the smaller value. 

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 PROBLEM 1.51

Link AC is made of a steel with a 65-ksi ultimate normal stress and has a

-in. uniform rectangular cross section. It is connected to a support at

A and to member BCD at C by -in.-diameter pins, while member BCD is

connected to its support at B by a -in.-diameter pin. All of the pins are

made of a steel with a 25-ksi ultimate shearing stress and are in single
shear. Knowing that a factor of safety of 3.25 is desired, determine the
largest load P that can be applied at D. Note that link AC is not reinforced
around the pin holes.

SOLUTION

Use free body BCD.

(1)

(2)
Shear in pins at A and C.

Tension on net section of A and C.

Smaller value of FAC is 0.625 kips.

From (1),

Shear in pin at B.
From (2),

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Allowable value of P is the smaller value. or 

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 PROBLEM 1.52

Solve Prob. 1.51, assuming that the structure has been redesigned to use
-in-diameter pins at A and C as well as at B and that no other changes
have been made.

PROBLEM 1.51 Link AC is made of a steel with a 65-ksi ultimate

normal stress and has a -in. uniform rectangular cross section. It is
connected to a support at A and to member BCD at C by -in.-diameter
pins, while member BCD is connected to its support at B by a -in.-
diameter pin. All of the pins are made of a steel with a 25-ksi ultimate
shearing stress and are in single shear. Knowing that a factor of safety of
3.25 is desired, determine the largest load P that can be applied at D.
Note that link AC is not reinforced around the pin holes.

SOLUTION

Use free body BCD.

(1)

(2)
Shear in pins at A and C.

Tension on net section of A and C.

Smaller value of FAC is 0.58999 kips.

From (1),

Shear in pin at B.
From (2),

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Allowable value of P is the smaller value. or 

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 PROBLEM 1.53

In the steel structure shown, a 6-mm-diameter pin

is used at C and 10-mm-diameter pins are used at B
and D. The ultimate shearing stress is 150 MPa at
all connections, and the ultimate normal stress is
400 MPa in link BD. Knowing that a factor of
safety of 3.0 is desired, determine the largest load
P that can be applied at A. Note that link BD is not
reinforced around the pin holes.

SOLUTION

Use free body ABC.

(1)

(2)
Tension on net section of link BD.

Shear in pins at B and D.

Smaller value of FBD is

From (1)

Shear in pin at C.

From (2)

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Smaller value of P is allowable value. 

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 PROBLEM 1.54

Solve Prob. 1.53, assuming that the structure has

been redesigned to use 12-mm-diameter pins at B
and D and no other change has been made.

PROBLEM 1.53 In the steel structure shown, a 6-

mm-diameter pin is used at C and 10-mm-diameter
pins are used at B and D. The ultimate shearing
stress is 150 MPa at all connections, and the
ultimate normal stress is 400 MPa in link BD.
Knowing that a factor of safety of 3.0 is desired,
determine the largest load P that can be applied at
A. Note that link BD is not reinforced around the
pin holes.

SOLUTION

Use free body ABC.

(1)

(2)
Tension on net section of link BD.

Shear in pins at B and D.

Smaller value of FBD is

From (1),

Shear in pin at C.

From (2),

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Smaller value of P is the allowable value. 

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 PROBLEM 1.55

In the structure shown, an 8-mm-diameter pin

is used at A, and 12-mm-diameter pins are
used at B and D. Knowing that the ultimate
shearing stress is 100 MPa at all connections
and that the ultimate normal stress is 250 MPa
in each of the two links joining B and D,
determine the allowable load P if an overall
factor of safety of 3.0 is desired.

SOLUTION

Statics: Use ABC as free body.

Based on double shear in pin A,

Based on double shear in pins at B and D,

Based on compression in links BD, for one link,

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Allowable value of P is smallest,  

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 PROBLEM 1.56

In an alternative design for the structure of

Prob. 1.55, a pin of 10-mm-diameter is to be
used at A. Assuming that all other
specifications remain unchanged, determine
the allowable load P if an overall factor of
safety of 3.0 is desired.

PROBLEM 1.55 In the structure shown, an 8-

mm-diameter pin is used at A, and 12-mm-
diameter pins are used at B and D. Knowing
that the ultimate shearing stress is 100 MPa at
all connections and that the ultimate normal
stress is 250 MPa in each of the two links
joining B and D, determine the allowable load
P if an overall factor of safety of 3.0 is desired.

SOLUTION

Statics: Use ABC as free body.

Based on double shear in pin A,

Based on double shear in pins at B and D,

Based on compression in links BD, for one link,

Allowable value of P is smallest,  

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 PROBLEM 1.57

A 40-kg platform is attached to the end B of a 50-kg wooden beam AB,

which is supported as shown by a pin at A and by a slender steel rod BC
with a 12-kN ultimate load. (a) Using the Load and Resistance Factor
Design method with a resistance factor and load factors
and 1.6, determine the largest load that can be safely
placed on the platform. (b) What is the corresponding conventional
factor of safety for rod BC?

SOLUTION

For dead loading,

For live loading,

From which

Design criterion:

(a) Allowable load. 

Conventional factor of safety:

(b) 

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 PROBLEM 1.58

The Load and Resistance Factor Design method is to be used to select

the two cables that will raise and lower a platform supporting two
window washers. The platform weighs 160 lb and each of the window
washers is assumed to weigh 195 lb with equipment. Since these
workers are free to move on the platform, 75% of their total weight and
the weight of their equipment will be used as the design live load of each
cable. (a) Assuming a resistance factor and load factors
1.2 and determine the required minimum ultimate load of one
cable. (b) What is the corresponding conventional factor of safety for the
selected cables?

SOLUTION

(a)


Conventional factor of safety:

(b) 

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