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CHAPTER 1
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125 kN
PROBLEM 1.1
B
C
Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Knowing that
= 30 mm and d 2 = 50 mm,
find the average normal stress at the
midsection of (a) rod AB, (b) rod BC.
60 kN
125 kN
0.9 m
1.2 m
SOLUTION
(a)
Rod AB'.
Force:
P = 60 x 103 N
Area:
A = —d? = —(30 x 10"3 ) 2 = 706.86 x 10"6 tn 2
4'4
Normal stress:
(b)
tension
crAB = — = — 6 0 X 1 0 6, = 84.882 x 106 Pa
A 706.86 x IO’
o AB = 84.9 MPa ◄
Rod 5C:
Force:
P = 60 x 103 - (2)(125 x 103 ) = -190 x 103 N
Area:
A = - d i = — (50 x 10-3 )2 = 1.96350 x 10“3 m2
4 4
Normal stress:
a BC = — = ~ 1 9 0 x l ° 3 = -96.766 x 10s Pa
A 1.96350 x 10“
a BC = -96.8 MPa ◄
PROPRIETOR Y Af. ITER L4L. Copyright £:■ 2015 McGraw-Hill Education. This is p r o p r i e t a r y m a t e r i a ] solely for a u t h o r i z e d i n s t r u c t o r use.
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PROBLEM 1.2
125 kN
C
Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Knowing that the average normal stress must
not exceed 150 MPa in either rod, determine
the smallest allowable values of the diameters
d\ and d 2 .
60 kN
125 kN
0.9 m
1.2 m
SOLUTION
(a)
Rod AB:
Force:
P= 60xl0 3 N
Stress:
a AB = 150 x 10 s Pa
Area:
A = -d?
4
a
P
AB ~ ~
■’
A = -------°AB
4 u AB
'
,2
4r
"i =--710
AB
(4)(60 x 103 )
< 2
— ---------6/ = 509.30 x 10“6 m 2
■(150 xlO )
d } = 22.568 x 10 - J m
(b)
d { = 22.6 mm ◄
Rod BC\
Force:
P = 60 x 103 - (2)(125 x 103 ) = -190 x 103 N
Stress:
a BC = -150 x 106 Pa
Area:
A = —di2
4
P
4P
a
BC - ™ - — 7 2
A
ffd 2
BC
7T(-150 X 106 )
d 2 = 40.159 x 10’3 m
d 2 - 40.2 mm ◄
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PROBLEM 1.3
Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Knowing that P = 10 kips, find the average normal stress at the
midsection of (a) rod AB, (b) rod BC.
30 in.
B
25 in.
SOLUTION
(а)
Rod AB\
p = 12 + 10 = 22 kips
A = - d l = — (1.25)2 = 1.22718 in 2
4
4
P
22
<yAB
4B - — = ---------- - 17.927 ksi
A 1.22718
(б)
cr 4B - 17.93 ksi ◄
Rod BC.
P = 10 kips
A=
4 =
-(0.75)2 = 0.44179 in 2
P
10
<jAB
4B - — = ---------- = 22.635 ksi
A
0.44179
crAB
4B = 22.6 ksi ◄
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PROBLEM 1.4
Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Determine the magnitude of the force P for which the tensile stresses in
rods AB and BC are equal.
30 in.
— 1.25 in.
12 kips
25 in.
0.75 in.
C
SOLUTION
(a)
Rod AB\
P = P + 12 kips
Tid
. T
A = - — = - ( 1 . 2 5 m.)2
4
4
A = 1.22718 in 2
P 4- 12 kips
“ 1.22718 in 2
(b)
Rod BC\
P= P
A = —d 1 = —(0.75 in.)2
4
4
A = 0.44179 in 2
_ P
aBC
a
“ 0.44179 in 2
AB =
a
BC
P + 12 kips _ P
1.22718 in 2
0.44179 in 2
5.3015 = 0.78539P
P = 6.75 kips ◄
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1200 N
PROBLEM 1.5
A strain gage located at C on the surface of bone AB indicates that the average normal stress
in the bone is 3.80 MPa when the bone is subjected to two 1200-N forces as shown.
Assuming the cross section of the bone at C to be annular and knowing that its outer diameter
is 25 mm, determine the inner diameter of the bone’s cross section at C.
1200 N
SOLUTION
P
a = —
A
Geometry:
A=
p
.
A= —
<5
-(d? - df)
,2
j2 4/1
4P
2
d 2 = di ------- = d } ------71
77(5
rf 2 = (25 x IO- 3 ) 2 -
w g 200 >
-(3.80 xlO 6 )
= 222.92 x I0“ 6 m 2
d 2 = 14.93 x 10- 3 m
d 2 = 14.93 mm ◄
PROPR/EEdRE .VE1TERL4L. Copyright £:■ 2015 McGraw-Hill Education. This is p r o p r i e t a r y material solely for a u t h o r i z e d i n s t r u c t o r use.
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PROBLEM 1.6
Two brass rods AB and BC, each of uniform diameter, will be brazed together
at B to form a nonuniform rod of total length 100 m, which will be suspended
from a support at J as shown. Knowing that the density of brass is 8470 kg/m 3,
determine (a) the length of rod AB for which the maximum normal stress in
ABC is minimum, (b) the corresponding value of the maximum normal stress.
SOLUTION
Areas:
A AB =
-(15 mm)2 = 176.715 mm 2 = 176.715 x 10“6 m2
ABC =
(10 mm) 2 = 78.54 mm 2 = 78.54 x 10 -6 m 2
From geometry,
Weights:
b= 100 - a
W AB = pg A AB f. AB = (8470)(9.81)(176.715 x 10 )a = 14.683a
W BC = pg ABC f. BC = (8470)(9.81)(78.54 x 10~*)(100 - a) = 652.59 - 6.526a
Normal stresses:
At J,
PZi.4 = W-4R
+ WjDLR r = 652.59 + 8.157a
ID
(1)
v
z
= 3.6930 x 106 + 46.160 x 103a
cr. =
AB
At B,
PB = WBC = 652.59 - 6.526a
<yB =
(2)
= 8.3090 x 106 - 83.090 x 103a
BC
(а)
Length of rod AB. The maximum stress in ABC is minimum when <J A = a B or
4.6160 x 106 - 129.25 x 103a = 0
f. AB = a = 35.7 m ◄
a = 35.71m
(б)
Maximum normal stress.
a A = 3.6930 x 106 + (46.160 x 103 )(35.71)
crfi = 8.3090 x 106 - (83.090 x 10 3 )(35.71)
ff A = (T B = 5.34 x 10s Pa
cr = 5.34 MPa ◄
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PROBLEM 1.7
0.25 m
0.2 m
Each of the four vertical links has an 8 x 36-mm uniform rectangular
cross section and each of the four pins has a 16-mm diameter. Determine
the maximum value of the average normal stress in the links connecting
(a) points B and D, (Z?) points C and E.
20 kN
SOLUTION
Use bar ABC as a free body.
XM C = 0 :
SO HF
(0.040) F BD - (0.025 + 0.040)(20 x 10 3 ) = 0
F BD = 32.5 x 103 N
Link BD is in tension.
3
1M B = 0 : -(0.040) FCE - (0.025)(20 x 10 ) = 0
FCf = -12.5 x 103 N
Link CE is in compression.
Net area of one link for tension = (0.008)(0.036 - 0.016)= 160 x 10“6 m 2
For two parallel links,
(a)
a BD
=
BD
A net = 320 x 10 6 nr
SD= — - 1()3s.. = 101.563 x 106
An a
320x 10“
<rBD = 101.6 MPa ◄
Area for one link in compression = (0.008)(0.036) = 288 x 10- 6 m2
For two parallel links,
a
CE
A = 576 x l0“ 6 m 2
FcE _ -12.5 xlO 3 = -21.701 x 10“6
A ~ 576 xlO - 6
erC£ = -21.7 MPa ◄
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PROBLEM 1.8
Link AC has a uniform rectangular cross section | in. thick and 1 in. wide.
Determine the normal stress in the central portion of the link.
120 lb
SOLUTION
Use the plate together with two pulleys as a free body. Note that the cable tension causes at 1200 Ib-in.
clockwise couple to act on the body.
i— >
1*2.0 i b
(20 l b
= 0: - (12 + 4)(F 4C cos30°) + (10)(F<c sin30°) - 1200 lb = 0
F Ac~-
12001b
= ------------------------------- = -135.500 lb
16 cos30° - 10 sin30°
Area of link AC:
1
9
A = 1 in. x - i n . - 0 . 1 2 5 in 2
8
Stress in link AC:
,, cr =
A
=
- 1 3 5 '5 0 = 1084 psi = 1.084 ksi
0.125
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0.100 m
PROBLEM 1.9
Three forces, each of magnitude P = 4 kN, are applied to the mechanism
shown. Determine the cross-sectional area of the uniform portion of rod
BE for which the normal stress in that portion is + 100 MPa.
B
0.150 m
0.300 m
0.250 m
SOLUTION
Draw free body diagrams of AC and CD.
£
fO.|4O
D
Free Body CD:
+y£.M D = 0: 0. 15OP - 0.250C = 0
C = 0.6P
Free Body JC:
+)M A =0:
F£ £ =
Required area of BE:
a
BE
=
_
O.15OF BE - 0.350P - 0.450/’ - 0.450C = 0
1 07
F = 7 1 3 3 3 p = (7.133)(4 kN) = 28.533 kN
BE
BE
_ F£ £ , 28.533 xlO 3
ABE ~—
A
—
crBE
_
100 xlO 6
2
—
zN L vz
111
ABE = 285 mm2 ◄
PROPR/EEdRY AE1TERL4L. Copyright £:■ 2015 McGraw-Hill Education. This is p r o p r i e t a r y m a t e r i a ] solely for a u t h o r i z e d i n s t r u c t o r use.
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PROBLEM 1.10
4 kips
Link BD consists of a single bar 1 in. wide and
k in. thick. Knowing that each pin has a | -in.
diameter, determine the maximum value of the
average normal stress in link BD if (a) 0 = 0,
(b) O = 90°.
SOLUTION
Use bar ABC as a free body.
8
(Y
+ .M A = 0: (18 sin30°)(4) - (12 cos30°)FSD = 0
F BD = 3.4641 kips
(tension)
Area for tension loading:
(
3V 1 \
= (b - d)t = 1 1 - - 1 - 1 = 0.31250 in 2
F BD
Stress:
3.464 1 kips
A ~ 0.31250 in 2
a = 11.09 ksi ◄
e = 90°.
+
M A =0:
-(18 cos30°)(4) - (12 cos30°)FBD = 0
F BD = -6 kips i.e. compression.
Area for compression loading:
A = hi
Stress:
BD _ ~6 k i PS
A
0.5 in 2
cr = 12.00 ksi ◄
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PROBLEM 1.11
7
12 ft
4
E
9 ft
■9 ft-
9ft
80 kips
80 kips
For the Pratt bridge truss and loading shown, determine the
average normal stress in member BE, knowing that the crosssectional area of that member is 5.87 in2 .
-9 ft-
80 kips
SOLUTION
Use entire truss as free body.
F
+)£M H = 0: (9)(80) + (18)(80) + (27)(80) - 36 v. = 0
A y = 120 kips
12 ft
Use portion of truss to the left of a section cutting members
BD, BE, and CE.
Atv
™
*
80 kips
’
SO kips
1
+fEF r =0:
12
120-80- —
= 0
F BE = 50 kips
SO kips
B
BE
FgE = 50 kips
A
5.87 in 2
crBE = 8.52 ksi ◄
PROPRIETARY MATERIAL. Copyright D 2015 McGraw-Hill Education. This is p r o p r i e t a r y material solely for a u t h o r i z e d i n s t r u c t o r use.
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►h-------- 30 in.
I*
o
I
40 in.
PROBLEM 1.12
The frame shown consists of four wooden members, ABC,
DEF, BE, and CF. Knowing that each member has a 2 x 4-in.
rectangular cross section and that each pin has a y-in.
diameter, determine the maximum value of the average
normal stress (a) in member BE, (b) in member CF.
~X+/
L
|E
|F
--------30 in. —------*j
SOLUTION
Add support reactions to figure as shown.
4
J3
F*
Using entire frame as free body,
= 0: 40D v - (45 + 30)(480) = 0
Dx = 900 lb
Use member DEF as free body.
A
Reaction at D must be parallel to FBE and FCF .
D,.
y = — Dx = 12001b
3
LM F = 0 :
- ( 3 0 ) P F fl J - ( 3 0
+ 15)Z»K = 0
FBE = -2250 lb
f4
A
SM £ = 0: (30)1 - F C J - (15)fy = 0
ft.
FC£ = 750 lb
Stress in compression member BE:
Area:
A = 2 in. x 4 in. = 8 in 2
(")
B£ =
= 2g50
°BE = ~ 2 8 1 PS' ◄
Minimum section area occurs at pin.
4nin = (2)(4.0 - 0.5) = 7.0 in 2
Stress in tension member CF\
W
o’er = “ y = ™
a
CF =1071 psi ◄
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PROBLEM 1.13
Dimensions in mm
1
ZZIOO
450
250
850
.500
675
825
An aircraft tow bar is positioned by means of a single
hydraulic cylinder connected by a 25-mm-diameter steel
rod to two identical arm-and-wheel units DEF. The mass
of the entire tow bar is 200 kg, and its center of gravity
is located at G. For the position shown, determine the
normal stress in the rod.
SOLUTION
FREE BODY
ENTIRE TOW BAR:
W = (200 kg)(9.81 m/s 2 ) = 1962.00 N
+ )SM,=0:
8507? - 1150(1962.00 N) = 0
R = 2654.5 N
4
too
I
450 nw)
J.
FREE BODY
BOTH ARM & WHEEL UNITS:
100
tan « = ----675
a = 8.4270°
+)ZM E = 0: (FCD cos«)(550) - F(500) = 0
500
(2654.5 N)
550 cos 8.4270°
= 2439.5 N
S OO
Col 5
m
Fc o
ACD ~
(comp.)
2439.5 N
#(0.0125 m) 2
= -4.9697 x 10 6 Pa
crCD = -4.97 MPa ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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PROBLEM 1.14
150 mm
SOO N
-300 mm -s
A
150 mm
Two hydraulic cylinders are used to control the position
of the robotic arm ABC. Knowing that the control rods
attached at A and D each have a 20-mm diameter and
happen to be parallel in the position shown, determine the
average normal stress in (a) member AE, (h) member DG.
200 mm
SOLUTION
Use member ABC as free body.
Use
B
ABC
Free body-
0OO bi
0: (0. 150) - FAE - (0.600)(800) = 0
FAE = 4 x l O 3 N
Area of rod in member AE is
Stress in rod AE:
A = —d 2
4
a
<T
AE -~
.
A
( 2 0 x l 0 ‘3) 2 = 314.16 x lO m 2
4 x 103
----------------r= 12.7324 x IO6 Pa
314.16 x 10-f>
(a)
a AE = 12.73 MPa ◄
Use combined members ABC and BFD as free body.
+)XM F = 0: (0.150)
- ( 1 .050 - 0.350)(800) = 0
7
DG = -1500 N
Area of rod DG:
= —d 2 = — (20 x 10’ 3 )2 = 314.16 x IO-6 nr
4
4
Stress in rod DG:
■nc =
A
= — ~ 1 5 °° -6 = -4.7746 x 106 Pa
3.1416 xl0“
(6)
a
DG = -4.77 MPa ◄
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PROBLEM 1.15
Determine the diameter of the largest circular hole that can be punched into a sheet of polystyrene 6 mm
thick, knowing that the force exerted by the punch is 45 kN and that a 55-MPa average shearing stress is
required to cause the material to fail.
SOLUTION
For cylindrical failure surface:
Shearing stress:
Therefore,
Finally,
A = 7idt
P
r - — or
A
p
— = 7ldt
T
P
A= —
T
TTtT
_
45xlO3 N
~ >r(0.006 m)(55 x 106 Pa)
= 43.406 x IO- 3 m
d = 43.4 mm ◄
PROPRIETARY MATERIAL. Copyright £ 2015 McGraw-Hill Education. This is p r o p r i e t a r y m a t e r i a ] solely for a u t h o r i z e d i n s t r u c t o r use.
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PROBLEM 1.16
Two wooden planks, each f in. thick and 9 in.
wide, are joined by the dry mortise joint shown.
Knowing that the wood used shears off along its
grain when the average shearing stress reaches
1. 20 ksi, determine the magnitude P of the axial
load that will cause the joint to fail.
SOLUTION
Six areas must be sheared off when the joint fails. Each of these areas has dimensions | in. x 1 in., its area
being
5 1
5 o
?
A=-x= — in 2 = 0.3125 in 2
8 2 16
At failure, the force carried by each area is
F = rA = (1.20 ksi)(0.3125 in 2 ) = 0.375 kips
Since there are six failure areas,
P = 6F = (6)(0.375)
P = 2.25 kips ◄
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PROBLEM 1.17
0.6 in.
When the force P reached 1600 lb, the wooden specimen shown failed
in shear along the surface indicated by the dashed line. Determine the
average shearing stress along that surface at the time of failure.
/
Steel
SOLUTION
Area being sheared:
A = 3 in. x 0.6 in. = 1.8 in 2
Force:
P = 1600 lb
Shearing stress:
r = — - 1 6 0 0 - = 8.8889 x 102 psi
A
L8 in
r = 889 psi ◄
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■40 mm
PROBLEM 1.18
10 mm
S mm
12 mm
A load P is applied to a steel rod supported as shown by an aluminum
plate into which a 12-mm-diameter hole has been drilled. Knowing that
the shearing stress must not exceed 180 MPa in the steel rod and 70 MPa
in the aluminum plate, determine the largest load P that can be applied to
the rod.
SOLUTION
For steel:
4 = ndt =
(0.012 m)(0.010 m)
= 376.99 x 10- 6 m 2
r, = —
A
P = 4 q = (376.99 x 10- 6 m 2 )(l80 x 10 s Pa)
= 67.858 x 103 N
For aluminum:
A2 = ndt =
(0.040 m)(0.008 m) = 1.00531 x 10- 3 m 2
r2 = —
■■■ P = A2 r2 = (1.00531 x 10“3 m 2 )(70 x 106 Pa) = 70.372 x 10 3 N
A
i
P = 67.9 kN ◄
Limiting value of P is the smaller value, so
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PROBLEM 1.19
The axial force in the column supporting the timber beam shown is
P = 20 kips. Determine the smallest allowable length L of the bearing
plate if the bearing stress in the timber is not to exceed 400 psi.
SOLUTION
Bearing area: Ah = Lw
6
P
Ah
P
Lw
P
2 0 x l 0 3 lb
T
£ = ------ = ------------------- = 8.33 in.
ah w
(400 psi)(6in.)
L = 8.33 in.
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PROBLEM 1.20
Three wooden planks are fastened together by a series of bolts to form
a column. The diameter of each bolt is 12 mm and the inner diameter
of each washer is 16 mm, which is slightly larger than the diameter of
the holes in the planks. Determine the smallest allowable outer
diameter d of the washers, knowing that the average normal stress in
the bolts is 36 MPa and that the bearing stress between the washers
and the planks must not exceed 8.5 MPa.
12 mm
SOLUTION
Bolt:
Tensile force in bolt:
Boit -
—
■ = *(0.012 m)2 = 1 1 3 0 9 7 x l 0
4
-4 m 2
P
<j = — => P = a A
A
= (36 x IO6 Pa)(l.13097 x 10
m2 )
= 4.0715 x 10 3 N
Bearing area for washer:
and
4.- —
&BRG
Therefore, equating the two expressions for
2
gives
77(7
<2
+42
=-
BRG
4 (4.0715 x 103 N) , „ „ 1Z x2
----------------7---------F
(0.016 m)
il ( 8 . 5 x l 06 Pa)
d* = 8.6588 x 10 m 2
d o = 29.426 x 10‘3 m
d o = 29.4 mm ◄
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PROBLEM 1.21
P = 40 kN
120 mm
A 40-kN axial load is applied to a short wooden post that is
supported by a concrete footing resting on undisturbed soil.
Determine (a) the maximum bearing stress on the concrete
footing, (6) the size of the footing for which the average bearing
stress in the soil is 145 kPa.
100 m m
SOLUTION
(#)
Bearing stress on concrete footing.
p = 40 kN = 40 x 103 N
A = (100)(120) = 12 x I 0 W
= 12 x 10“3 m2
P
40 x 103
----. „6 j,
er = — = ----------3— = 3.3333 x 10 Pa
A 12 x 10“
Footing area. P - 40 x 103 N
3.33 MPa ◄
cr = 145 kPa = 45 x 103 Pa
A = — = 4 0 X 1 0 3 = 0.27586 m 2
cr 145 x 10
Since the area is square, A = b2
b = 4 A = V0.27586 = 0.525 m
b = 525 mm ◄
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PROBLEM 1.22
An axial load P is supported by a short W8 x 40 column of crosssectional area A = 11.7 in 2 and is distributed to a concrete foundation
by a square plate as shown. Knowing that the average normal stress in
the column must not exceed 30 ksi and that the bearing stress on the
concrete foundation must not exceed 3.0 ksi, determine the side a of
the plate that will provide the most economical and safe design.
SOLUTION
p
For the column, a = — or
A
P = aA = (30)(l 1 .7) = 351 kips
For the a x a plate, (J = 3.0 ksi
Since the plate is square, A = a 2
a = y]A = VI 17
a = 10.82 in. ◄
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PROBLEM 1.23
Link AB, of width b = 2 in. and thickness t = k in., is used to support the end of a
horizontal beam. Knowing that the average normal stress in the link is -20 ksi and
that the average shearing stress in each of the two pins is 12 ksi, determine (a) the
diameter cl of the pins, (b) the average bearing stress in the link.
SOLUTION
Rod AB is in compression.
bt
where
-<jA
b = 2 in. and
-20)(2)
t = — in.
10 kips
Pin:
and
4AP
n
(4X1
°? = 1.03006 in.
*(12)
1.030 in.
10
ax = — = --------—--------- = 38.833 ksi
dt
(1.03006)(0.25)
= 38.8 ksi
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PROBLEM 1.24
p
Determine the largest load P which may be applied at A when
0= 60°, knowing that the average shearing stress in the 10-mmdiameter pin at B must not exceed 120 MPa and that the average
bearing stress in member AB and in the bracket at B must not
exceed 90 MPa.
750 inni
750 m m
50 mm
12 mm
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
© = 6 0 ° \lao °
3t>'
C
Use joints as a free body.
P
P
120*
3<r
Force
Ac
Law of sines applied to force triangle:
?
AB
AC
sin 30°
sin 120°
sin 30°
P = F
sm30
sin 120°
° = 0.57735/ ABg
F 4 C sin30° _
sin 30°
r — ------------------------------- — F j r
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PROBLEM 1.24 (Continued)
If shearing stress in pin at B is critical,
A = —d 2 = — (0.010)2 = 78.54 x 10“6 m 2
4
4
FAB = 2Ar = (2)(78.54 x 10~6 )(l 20 x 106 ) = 18.850 x 103 N
If bearing stress in member AB at bracket at A is critical,
Ah = td = (0.016)(0.010) = 160 x 10- 6 m 2
= (1 6 0 x 10- S )(90 x 106 ) = 14.40 x 103 N
If bearing stress in the bracket at B is critical,
A„ = ltd = (2)(0.012)(0.010) = 240 x lO m 2
FAB = Aha h = (240 x lO XW x 106 ) = 21.6 x 103 N
Allowable F AB is the smallest, i.e., 14.40 x 103 N
Then from statics,
PaNow = (0.57735)(14.40 x 103 )
= 8 . 3 1 x l 03 N
8.31 kN ◄
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PROBLEM 1.25
Knowing that 6 - 40° and P = 9 kN, determine (a) the smallest
allowable diameter of the pin at B if the average shearing stress in
the pin is not to exceed 120 MPa, (b) the corresponding average
bearing stress in member AB at B, (c) the corresponding average
bearing stress in each of the support brackets at B.
16 mm
750 mm
750 m m
50 mm
12 mm
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
6C
8
Use joint A as a free body.
P
v||o'
Law of sines applied to force triangle:
_
AB
_
AC
sin20°
sinll0°
sin50°
_ P s i n l l O0
AB
~ sin 20°
24.727 kN
sin 20°
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PROBLEM 1.25 (Continued)
(a)
Allowable pin diameter.
T=
nd 2
where F, B = 24.727 x 103 N
2A P
2 d2
m
(2X24.727 xltf) _
jr(120 x 106 )
*
d = 11.4534 x 10”3 m
(b)
11.45 mm ◄
Bearing stress in J # at J.
Ah = td = (0.016)(11.4534x 10”3 ) = 183.254 x 10”6 m 2
cr. = F '-B- = 24.727 xlO 3 = 1 3 4 9 3 3 x 1 0 6 p a
Af,
183.254x 10”“
(e)
1 3 4 .9 MPa ◄
Bearing stress in support brackets at 5.
A = td = (0.012)(11.4534 x 10“3 ) = 137.441 x 10”6 m 2
A
137.441 x 10- 6
,55x10
,
,o.o M P a <
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PROBLEM 1.26
The hydraulic cylinder CF, which partially controls the position of rod
DE, has been locked in the position shown. Member BD is 15 mm
thick and is connected at C to the vertical rod by a 9-mm-diameter
bolt. Knowing that P = 2 kN and 0 = 75°, determine (a) the average
shearing stress in the bolt, (b) the bearing stress at C in member BD.
200 mm
45 m m
SOLUTION
Free Body: Member BD.
7J*
x
+ y M c = 0:
40
—F
9
(100 cos20°) - - F 4 5 (100 sin 20°)
-(2 kN) cos 75°(l 75 sin 20°) - (2 kN) sin 75°(115 cos 20°) = 0
(40 cos 20° - 9sin20°) = (2 kN)(175)sin(75° + 20°)
F
AB = 4-1424 kN
9
-±-SFx = 0: C x - —(4.1424 kN) + (2 kN)cos75° = 0
C x = 0.39167 kN
»
+J
40
= 0: C y - —(4.1424 kN) - (2 kN)sin75° = 0
C y = 5.9732 kN
C = 5.9860 kN
c
(«)
r
ave ~
~
c
(*)
r
*
=
u
=
86.2°
3
8 6 0 x l 0 N - 94
„ . 1, io
l n 6 op = 94.
n „ ,1 S
4D
_5 . 9--------—
MPa
x
a
(0.0045 m)2
5.9860 x l 0 3 N
------------------------- = 44.3 x 10s Pa = 44.3 MPa
(0.015 m)(0.009m)
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PROBLEM 1.27
For the assembly and loading of Prob. 1.7, determine (a) the average
shearing stress in the pin at B, (b) the average bearing stress at B in
member BD, (c) the average bearing stress at B in member ABC,
knowing that this member has a 10 x 50-mm uniform rectangular cross
section.
0.25 m
0.2 ni
E
20 kN
PROBLEM 1.7 Each of the four vertical links has an 8 x 36-mm
uniform rectangular cross section and each of the four pins has a 16-mm
diameter. Determine the maximum value of the average normal stress in
the links connecting (a) points B and D, (b) points C and E.
SOLUTION
Usebar/t/?C as a free body.
C
BD
3
+)SW C = 0 : (0.040)Ffl£) - (0.025 + 0.040)(20 x 10 ) = 0
FBD = 32.5 x 103 N
(")
Shear pin at B.
r =
where
A = — d 2 = — (0.01 6) 2 = 201.06 x 10“6 m:
2A
for double shear
r = -----3 2 ~5 * 1()3 6 = 80.822 x IO6 Pa
(2)(201.06x 10“ )
80.8 MPa
dt = (0.016)(0.008) = 128 x 10“6 m
Bearing: link BD.
1 , _ <0.5)02.3 xltf).
A
| 2 6 .9 5 x | Q
128 xlO - 6
. p,
a h = 127.0 MPa
dt = (0.016)(0.010) = 160x 10“6 m
Bearing in ABC at B.
ak =
A
= 3 2 - 5 X 1 - 6 = 203.12 x 106 Pa
160 xlO
a h = 203 MPa
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PROBLEM 1.28
12 in.
___L
12m.
i
150011) f
I 15 in.
E
Two identical linkage-and-hydraulic-cylinder systems control the
position of the forks of a fork-lift truck The load supported by the one
system shown is 1500 lb. Knowing that the thickness of member BD is
| in., determine (a) the average shearing stress in the y-in.-diameter
pin at B, (b) the bearing stress at B in member BD.
I J
t
20 in.
SOLUTION
Use one fork as a free body.
+y.M B = 0: 24E - (20)(l 500) = 0
E = 1250 lb —
+,SF =0:
E + Bx = 0
= -E
Bx = 12501b-—
1_
+1SF
i y =0:
B=
(a)
By - 1500 = 0
+ 5 2 =
B.y = 15001b
2
1250
+ 15002 = 1952.56 lb
Shearing stress in pin at B.
=°- 196350 in2
r =
B
pin
(6)
1952.56
= 9.94 x 103 psi
" 0.196350 ’
r = 9,94 ksi ◄
Bearing stress at B.
B
1952.56
.
cr = 6.25 ksi ◄
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PROBLEM 1.29
1 150 mm
Two wooden members of uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that P = 1 1 kN,
determine the normal and shearing stresses in the glued splice.
SOLUTION
e = 90° - 45° == 45°
p = 11 kN = 11 x L03 N
A)
=
(j =
T=
(150)(75) =: 11.25 x 103 mm 2 = 11.25 x 10“3 m 2
Feos 2 6 _
A)
11.25 x IO’3
Psin26> _
<n ’< 1 ° ! X " 9-<3 P .
(2)(11.25 x 10 )
2 o
cr = 489 kPa ◄
4 8 9 x 1 tfp.
r = 489 kPa ◄
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PROBLEM 1.30
150 mm
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
0 = 90° - 45° = 45°
Ao = (150)(75) = 11.25 x 10 3 mm 2 =11.25 x 10"3 m 2
r = 620 kPa = 620 x 103 Pa
P sin 20
T = ---------2 0
p
a
2-l (,r
(2)(11.25x 10~3 ) ( 6 2 0 x l 03 )
~ sin20 ”
sin 90°
= 13.95 x 103 N
-
Pcos261
Ao
~
P = 13.95 kN ◄
(13.95 x 103 )(cos45°)2
11.25 x IO- 3
= 620 x 103 Pa
cr = 620 kPa ◄
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PROBLEM 1.31
5.0 in.
3.0 in.
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.
60°
P'
SOLUTION
P = 1400 lb
e = 90° - 60° = 30°
Ao = (5.0)(3.0) = 15 in 2
(1400)(cos30°)2
Feos 2 0
& = ---------- =
15
0
<T = 70.0 psi ◄
P sin 20
(1400) sin 60°
T = ---------- =
(2)(15)
2 o
r = 40.4 psi ◄
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PROBLEM 1.32
5.0
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, (6) the corresponding
shearing stress in the splice.
SOLUTION
Ao = (5.0)(3.0) = 15 in 2
0 = 90° - 60° = 30°
Pcos 2 #
(7 = --------------0
= ( 7 5 2 1 5 ) . = 1500 lb
cos 30°
(a)
P=
(ft)
Psin20
(1500)sin60°
r = ---------- = -— ---------2Jo
(2)(15)
cos2 8
P = 1.500 kips ◄
T = 43.3 psi ◄
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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, (</ ) the
maximum value of the normal stress in the block.
SOLUTION
Ao = (6)(6) = 36 in 2
max =
2
-5
ksi
0 = 45° for plane of r |naK
(a)
M
24
W
sin 20 = 1 20 = 90°
(c)
45
(d)
max
=
P _
—cos 2 45° =
24) “
0
P
-180
"T ”
p -= 180.0 kips ◄
Iri = 24)r max = (2)(36)(2.5)
6 = 45.0° ◄
180
(2X36)
= -2.50 ksi ◄
max = -5.00 ksi ◄
36
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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 (fl) the normal stress, (b) the shearing stress. Specify the
orientation of the plane on which each of these maximum values occurs.
SOLUTION
J o = (6 )(6 ) = 3 6 i n 2
a = — cos 2 # =
4)
(fl)
2
2
36 cos # = -6.67 cos #
max tensile stress = 0 at # = 90.0°
◄
max. compressive stress = 6.67 ksi at # = 0°
P = 240
=
r,nax
2Jo
(2)(36)
= 3-33 ksi ◄
at 0 = 45°
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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.
20°
SOLUTION
d o = 0.400 m
r
o=
- 0.200 m
= ro - t = 0.200 — 0.010 = 0.190 m
Ao =
- rf) =
(0.2002 - 0.190 2 )
= 12.2522 x 10“3 m2
20°
— = cos 2 0 =
4
2J 0
= sin 20 =
-300 x 103 cos 2 20°
= 21.621 x 10 6 Pa
12.2522 x IO- 3
-300 x 103 sin 40°
(2)(12.2522 x KF3 )
= 7.8695 x 106 Pa
a = -21.6 MPa
T = 7.87 MPa
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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 cr = 60 MPa and
T = 36 MPa, determine the magnitude P of the largest axial force that
can be applied to the pipe.
10 mm
SOLUTION
do = 0.400 m
=
~d o - 0.200 m
= ro - t = 0.200 - 0.010 = 0.190 m
Ao = n t f - r?) = zr(0.2002 - 0.190 2 )
= 12.2522 x 10“3 m 2
0 = 20°
Based on
P
i
cr = 60 MPa: tr = — cos~l?
4
P=
Based on
cos 2 0
= (12-2522 x l 02- 3 )(60x 10 ) =
cos 20°
N
| r | = 3 0 M P a : r = — sin2£
24,
p . 2 4 / _ (2)(12J522x IO~3 )(36x IO6 ) _
sin 23
sin 40°
Smaller value is the allowable value of P.
N
P = 833 kN
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PROBLEM 1.37
Q
<— 12 in.
r
9 in.
A steel loop ABCD of length 5 ft and of | -in. diameter is placed as
shown around a 1-in.-diameter aluminum rod AC. Cables BE and DF,
each of y -in. diameter, are used to apply the load Q. Knowing that the
ultimate strength of the steel used for the loop and the cables is 70 ksi,
and that the ultimate strength of the aluminum used for the rod is 38 ksi,
determine the largest load Q that can be applied if an overall factor of
safety of 3 is desired.
B
1 in.
I
t
p
9 in.
L_
l in
D
— T in
Q'
SOLUTION
Using joint B as a free body and considering symmetry,
2-|
fi - g
=o
Q=
FAB
Using joint A as a free body and considering symmetry,
2’
FAB - F AC = 0
8 5
7-Tejo
3
Q = - F AC
4
c=0
Based on strength of cable BE,
Qu = a v A = O u
d 2 = (70)
( i V
|-
= 13.7445 kips
Based on strength of steel loop,
Qu =
6
= p 70 )
=
6
lgl
6
7i
~
2
=9.2775 kips
Based on strength of rod AC,
Qu =
F
AC.u
Actual ultimate load
= | ( 3 8 ) J(1.0) 2 =22.384 kips
=
is the smallest,
Allowable load:
Q L< = 9.2775 kips
Q=
F.S.
3
= 3.0925 kips
g = 3.09 kips ◄
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PROBLEM 1.38
Link BC is 6 mm thick, has a width w = 25 mm, and is made of a steel w ith
a 480-MPa ultimate strength in tension. What was the safety factor used if the
structure shown was designed to support a 16-kN load P?
480 mm
SOLUTION
Use bar ACD as a free body and note that member BC is a two-force member.
= 0:
(480)FflC - (600)P = 0
BC
Ultimate load for member BC:
480
480
FL, = OyA
Fv = (480 x 10s )(0.006)(0.025) = 72 x 103 N
Factor of safety:
F,
77 x 103
FS. = —— = ---------3r
Fs c
20xl0
FS. = 3.60 ◄
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PROBLEM 1.39
Link BC is 6 mm thick and is made of a steel with a 450- MPa ultimate strength in
tension. What should be its width w if the structure shown is being designed to
support a 20-kN load P with a factor of safety of 3?
90°
480 mm
SOLUTION
Use bar ACD as a free body and note that member BC is a two-force member.
♦ Ay
Al
pl
LM A = 0:
(480)Fflc - 600P = 0
60gp = (600X20 x 1(f) = 2 5 x i 0
BC
480
3N
480
For a factor of safety F.S. = 3, the ultimate load of member BC is
Fv = (F.S)(Ffic ) = (3)(25 x 103 ) = 75 x 103 N
But F, = a v A
A=
F„
av
75 x 103
_s ,
= -----------7= 166.667 x 10 5 nr
450 xlO 6
r.
,
A 166.667 x IO”6
„„ „ „
, ,
For a rectangular section, A = wt or w = — = ------------------- = 27.778 x 10 m
t
0.006
w = 27.8 mm ◄
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PROBLEM 1.40
->-0.75 m
-7- h
A _______
0.4 m
__L
B
Members AB and BC of the truss shown are made of the same alloy. It is known
that a 20-mm-square bar of the same alloy was tested to failure and that an
ultimate load of 120 kN was recorded. If a factor of safety of 3.2 is to be
achieved for both bars, determine the required cross-sectional area of (a) bar
AB, (6) bar AC.
SOLUTION
Length of member AB‘.
£AB = Vo.752 + 0.4 2 = 0.85 m
Use entire truss as a free body.
- 0: 1.4X. - (0.75)(28) = 0
Ax = 15 kN
Ay = 28 kN
2$ kN
Use Joint J as free body.
c = 28 _MlZ) = 20kN
For the test bar,
A = (0.020)2 = 400 x 10-6 m 2
Pv = 1 2 O x l O3 N
For the material,
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PROBLEM 1.40 (Continued)
(a)
For member AB:
F.S. =
?U
_
a
U
AB
(*>)
For member AC:
A.A oR = (
&)FAB
F.S. = (jj
_
crL,
AC
AC -
AB
AB
_ (3.2)(17x IO3 ) _: 181.333 x 10“s m 2
300 x 10s
A 4B = 18 L3 mm 2 ◄
AC
(F.S.)FJ C _ (3.2)(20x 103 )
= 213.33 x 10“s m2
”
300 x 10s
A 4C = 213 mm 2 ◄
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PROBLEM 1.41
*-0.75
T
1
k
0 4m
. _L
!B
1.4 m
Members AB and BC of the truss shown are made of the same alloy. It is known
that a 20-mm-square bar of the same alloy was tested to failure and that an
ultimate load of 120 kN was recorded. If bar AB has a cross-sectional area of
225 mm 2, determine (a) the factor of safety for bar AB and (b) the crosssectional area of bar AC if it is to have the same factor of safety as bar AB.
SOLUTION
Length of member AB:
f AB = Vo.752 + 0.4 2 = 0.85 m
Use entire truss as a free body.
'.M e = 0: 1.4J, - (0.75)(28) = 0
Ax = 15 kN
0:
A v - 28
28 kN
Use Joint J as free body.
-+
F t = 0:
0 75
0.85
-------- .48 ~
(0.85)(15)
0.75
AB
+ SF„ = 0:
F
4C-—
F
0 85
AB = 0
Aa
F, c = 2 8 - M ! Z )
0.85
For the test bar,
For the material,
A = (0.020)2 = 400 x 10“6 m 2
=
-=
A
120x
10
400 x IO’6
17kN
= 20kN
Pt/ = 1 2 0 x l 0 3 N
= 300 x 106 Pa
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PROBLEM 1.41 (Continued)
(a)
For bar AB'.
FS
=
=
=
FAB
(300 x 106 )(2 25 x IO"6 )
17xl0 3
F.S. = 3.97 ◄
(b)
For bar AC:
FS = —
=
AC
,
=
crL,
PROPRIETARY MATERIAL. Copyright
AC
(3.97)(20x Itf) _
3 0 0 x l 06
.263
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PROBLEM 1.42
600 Ib/ft
Oj
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.
a
5 kip:
1.4 ft
1.4 ft
L4 ft
SOLUTION
—
P = (4.2)(0.6) = 2.52 kips
+)£M D = 0 :
-(2.8)(F sin35°)
+(0.7)(2.52) + (1.4)(5) = 0
FAB = 5.4570 kips
a
AB _
°~ult
AB
AB
45
_
(F.S,)FAB
(3.20X5.4570 kips)
“
77 i ■
crult
65 ksi
= 0.26854 in 2
AAB = 0.268 in 2 ◄
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PROBLEM 1.43
16 kN
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.
6 min
125 mm
SOLUTION
=
2.5 MPa = 0 9 Q 9 0 9 M p a
2.75
On one face of the upper contact surface,
. L - 0.006 m ... .
.
A = -------- -------(0.125 m)
Since there are 2 contact surfaces,
P_
r
all
0.90909 x 106
L
2A
16 x 103
(A - 0.006)(0.125)
146.8 mm ◄
0.14680 m
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PROBLEM 1.44
16 kN
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.
125 mm
SOLUTION
Area of one face of upper contact surface:
0. 180 m - 0.006 m
,
------------------------(0.125 m)
A = 10.8750 x 10 -3 m 2
Since there are two surfaces,
P
16xl0 3 N
2A ~ 2(10.8750 x 10“3 m2 )
rall = 0.73563 MPa
FS = h =
rall
2 5 MPa
-
0.73563 MPa
= 3.40
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PROBLEM 1.45
Three -in.-diameter 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 = 24 kips and that
the ultimate shearing stress for the steel used is 52 ksi, determine the factor of
safety for this design.
SOLUTION
For each bolt,
A = —d 2 = — 4
4V4j
= 0.44179 in 2
= Arv = (0.44179)(52)
= 22.973 kips
For the three bolts,
PL, = (3)(22.973) = 68.919 kips
Factor of safety:
r.5. —
P
—
68.919
24
FS-287 4
2 .o. — x.o /
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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,
24
P = y = 8 kips
Required:
Pv = {F.S.)P = (3.25)(8.0) = 26.0 kips
Py _
U
d =
A
PV
d
_ APy
2
nd 2
= R4 )(26 -Q) = Q.79789 in .
y
(52)
J = 0.798 in. ◄
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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
b = 40 mm, c = 55 mm, and d = 12 mm, determine the load P if an
overall factor of safety of 3.2 is desired.
SOLUTION
Based on double shear in pin,
PL, = 2Ar v =
= — (2)(0.012)2 (145 x 106 ) = 32.80 x 103 N
4
Based on tension in wood,
Pfj —
— w(b
d CTfj
= (0.040)(0.040 - 0.012)(60 x 106 )
= 6 7 . 2 x l O3 N
Based on double shear in the wood,
P,j = 2Ar v = 2wcrv = (2)(0.040)(0.055)(7.5 x 106 )
= 3 3 . 0 x l 03 N
Use smallest
Pv = 32.8 x 103 N
Allowable:
P=
FS.
= — - 10
3.2
= 10.25 x 103 N
10.25 kN ◄
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PROBLEM 1.48
For the support of Prob. 1.47, knowing that the diameter of the pin is
d = 16 mm and that the magnitude of the load is P = 20kN, 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 b = 40 mm, c = 55 mm, and d = 12 mm, determine the
load P if an overall factor of safety of 3.2 is desired.
SOLUTION
P = 20 kN = 20 x 103 N
(«)
Pin:
A = —d 2 = — (0.016)2 = 2.01.06 x 10"6 m2
4
4
Double shear:
T = _L
2A
r
U
2A
-s
s
3
Pu ~ lAtu = (2)(201.16 x 10 )(145 x 10 ) = 58.336 x 10 N
FS.
W
Tension in wood:
5036x10=
20xl0 3
P
3
Pu = 58.336 x 10 N
cr,,
=
u
A
for same F.S.
= -----C ---- where
w(b - d)
w = 40 mm = 0.040 m
6 = </ + - - = 0.016 + 5 8 3 3 6 x 1 0
= 4 0 . 3 x l 0 ~3 m
wa v
(0.040)(60 x 10s )
Shear in wood:
b =40.3 mm ◄
Pu = 58.336 x 103 N for same F.S.
Pu _ Pu
2A
2wc
Double shear: each area is A = we
58.336 x 103
(2)(0.040)(7.5 x 10s )
= 97.2 x 10- 3 m
c = 97.2 mm 4
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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 (fl) 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,
(a — d)t
u
‘v ~ u
FS = ?u — a u
a
~
Solving for a,
a
(3.60)(2.5)
(FS.)P
(Tyt
(36)(1)
(a) a = 1.750 in.
Based on shear between plate and concrete slab,
A = perimeter x depth = 2(a + t)b
pu = T(J A = 2rv (a + t)b
Solving for b,
(F5.)P
= 0.300 ksi
F.S. = —
_________________
(3.6)(2.5)
2(a + r)r £/ “ (2)(1.75 + |)(0.300)
(b) b = 7.50 in.
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PROBLEM 1.50
Determine the factor of safety for the cable
anchor in Prob. L49 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 highstrength cable as shown. The diameter of the hole
in the plate is j 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,
A = (a - d)t
(
3V l i
T
= 2 -- = 0.31250 in 2
V
4 JU J
2*5
Pv = VyA
= (36)(0.31250) = 11.2500 kips
Pj
11.2500
A cfx
FS. = — = ---------- = 4.50
P
3.5
Based on shear between plate and concrete slab,
A = perimeter x depth = 2(a + t)b = 21 2 + — [(6.0)
k
4J
.4 = 27.0 in 2
= 0.300 ksi
= T u A = (0.300)(27.0) = 8.10 kips
F.S. =
- = — = 3.240
P
2.5
FJf. = 3.24 ◄
Actual factor of safety is the smaller value.
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PROBLEM 1.51
T
1
Link AC is made of a steel with a 65-ksi ultimate normal stress and has
a
x 1 -in. uniform rectangular cross section. It is connected to a
support at A and to member BCD at C by -j- -in.-diameter pins, while
member BCD is connected to its support at B by a k- -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.
8 in.
8
Xc
Z ---------*—
| — ---- 6 in. -----
4 in.-
SOLUTION
Use free body BCD.
V
8,—
X
P = 0.485
\
|1
I
5,--
’P
(1)
= 0
c
Bx = — F ,ACc = 1.25P
10
+)M C = 0 : - 6 5 v - 45 = 0
By = A _ p
_______
I
B* + B 2 = J1.252 +
B=
i.e .
P = 1.416675
By
=
P\
P = 0.705885
Shear in pins at /I and C.
FAC
= rA oia
=
AC
pln
f —Y±Y2] = 0.84959 kips
-—d 2 =
F.S.4
1 3.25 A 4 A 8 J
Tension on net section of A and C.
FAL
AC = &A net =
“Y
- - - | = 0-625 kips
4 2 8J
-A
=\
F.S. net
3.25
Smaller value of F AC is 0.625 kips.
From (1),
P = (0.48)(0.625) = 0.300 kips
Shear in pin at B.
B=
From (2),
P = (0.70588)(0.58999) = 0.416 kips
r
pin
F.S. 4
Allowable value of P is the smaller value.
d2
25
325
x z5 yx2
vX zfl-y
A16J = 0.58999 kips
P = 0.300 kips
or
P = 300 lb ◄
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PROBLEM 1.52
T
1
Solve Prob. 1.51, assuming that the structure has been redesigned to use
-in-diameter pins at J and C as well as at B and that no other changes
have been made.
8 in.
B
\\
[Z
PROBLEM 1.51 Link AC is made of a steel with a 65-ksi ultimate
normal stress and has a d- x -L - 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.
c
MS
I’* ---- 6 in. ------
in.
SOLUTION
Use free body BCD.
+ > s = 0:
(6)
c j-10P
V FXe
„
=0
p = 0.48 FAC
---------------\ %-------------.
-----4,
il
By I
+|1 EF. = 0 :
>
Ip
(1)
BYx - — F AC = 0
10
Bx = ± F AC = l . 2 5 P
+)M C = 0 : - 6 B y - 4F = 0
2
____
B=
2
I
2
2
2 I
i.e. By = —p\
z 2 \2
B + B = J1.25 + | |
P = 1.41667P
P = 0.10583 B
Shear in pins at J and C.
FAC =
r
f' 2S
fS
~d 2 = —
= 0.58999 kips
F.S. 4
\ 3 . 2 5 7 \ 4 yV 16 )
pi„ = ~
Tension on net section of J and C.
F, c =
F.S.
~
|| - ¥ - - — | = 0.9375 kips
V3.25 J U J U
16j
Smaller value of F Ac is 0.58999 kips.
From (1),
P = (0.48)(0.58999) = 0.283 kips
Shear in pin at B.
B
From (2),
P = (0.70588)(0.58999) = 0.416 kips
=
pin
■ -d 2
F.S. 4
Allowable value of P is the smaller value.
25 y Y A) 2
= 0.58999 kips
3.25JI4JI16;
P = 0.283 kips
or
P = 283 lb ◄
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PROBLEM 1.53
400 m m
Each of the two vertical links CF connecting the two horizontal
members AD and EG has a 10 x 40-mm uniform rectangular cross
section and is made of a steel with an ultimate strength in tension of
400 MPa, while each of the pins at C and F has a 20-mm diameter and
are made of a steel with an ultimate strength in shear of 150 MPa.
Determine the overall factor of safety for the links CF and the pins
connecting them to the horizontal members.
24 kN
SOLUTION
Use member GFG
faebofy
CF
O.4o
YM E = 0 : 0.40Fc f - (0.65)(24 x 103 ) = 0
FCF = 39 x 10 3 N
Based on tension in links CF,
A = (b - d)t = (0.040 - 0.02)(0.010) = 200 x 10“° m2
Fv =
6
-6
(one link)
3
= (2)(400 x 10 )(200 x 10 ) = 160.0 x 10 N
Based on double shear in pins,
A = —d 2 = — (0.020)2 = 314.16 x lO m
Fv = 2r y 4 = (2)(150xl06 )(314. 1 6 x 1 0
) = 94.248 xl0 3 N
Actual Fu is smaller value, i.e. Fv = 94.248 x 103 N
Factor of safety:
F
94.248 x 103
c „
r.L>.
— v — ------------,—
FCF
39xl0 3
F.S. = 2.42
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PROBLEM 1.54
400 mm
Solve Prob. 1 .53, assuming that the pins at C and F have been replaced
by pins with a 30-mm diameter.
24 kN
PROBLEM 1.53 Each of the two vertical links CF connecting the two
horizontal members AD and EG has a 10 x 40- mm uniform rectangular
cross section and is made of a steel with an ultimate strength in tension of
400 MPa, while each of the pins at C and F has a 20-mm diameter and
are made of a steel with an ultimate strength in shear of 150 MPa.
Determine the overall factor of safety for the links CF and the pins
connecting them to the horizontal members.
SOLUTION
Use member £FG as free body.
Use member EFG as faebUf.
J, F M
I*
t peF
r
----- (D.4o -------
a
— 6.2S"-— J
24 kN
+
M E = 0 : 0.40FCF - (0.65)(24 x 103 ) = 0
FCF = 39 x 103 N
Based on tension in links CF,
A = ( b - d ) t = (0.040 - 0.030)(0.010) = 100 x 10“6 m 2
6
-6
(one link)
3
Fv = 2a v A = (2)(400 x 10 )(l 00 x 10 ) = 80.0 x 10 N
Based on double shear in pins,
A = —d 2 = — (0.030)2 = 706.86 x 10- 6 m 2
4
4
Fy = 2r v A = (2)(150x 10s )(706.86x 10"6 ) = 212.06 x 103 N
Actual Fy is smaller value, i.e. Fy = 80.0 x 103 N
Factor of safety:
F.S. =
FCF
x
12
39 x 103
F.S. = 2.05 ◄
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PROBLEM 1.55
12 mm
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.
_L
_
c “ I”
1
8 mm < T
B
20 mm
■8 mm
8 mm
JZ?
12 mm —
-*—
Side view
SOLUTION
A
Statics: Use/tl?C as free body.
C
+)SM fl = 0 :
0.20F, - 0.18P = 0
P=
= 0 : 0.20Ffi£) - 0.38P = 0
P=
-F A
<—0.2.0—
V
FBD
OJ«
5.
Based on double shear in pin , A = —d 2 = — (0.008) 2 = 50.266 x 10 6 m2
4
4
F
P W x 10 6 )(50,266 x IQ"6 ) _ 3 3 5 1
FS.
3.0
10
i3
P = — F AA = 3 . 7 2 x l 0 N
9
A
1
Based on double shear in pins at/? and D, A = — d 2 = — (0.012) 2 = 1 13.10 x 10 6 m2
4
4
_ (2|(IOI> x. O‘)(l !3,l0x !0"> _
BD
F.S.
,
3.0
3
P = — F BD
B n = 3.97 x 10 N
19
Based on compression in links BD, for one link, A = (0.020)(0.008) = 160 x 10 -6 m 2
2a v A
(2)(250 x 106 )(160 x 10- 6 ) , ,
lft 3 N
fnn = -------= --------------------------------- = 20. 7/xX
1U IN
F.S.
3.0
p = — F BD = 14.04 x 10 3 N
19
Allowable value of P is smallest,
P = 3.72 x 103 N
P = 3.72 kN ◄
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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.
12 mm
_L
c t
8 mm
B
20 mm Vp
•8 mm
8 mm
12 mm —*—
Side view
F ront view
PROBLEM 1.55 In the structure shown, an 8mm-diameter pin is used at A, and 12-mmdiameter 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 £), determine the allowable load
P if an overall factor of safely of 3.0 is
desired.
SOLUTION
e>
A
c
Statics: Use ABC as free body.
fl
=0:
0.20Fj - 0.18P = 0
■A = 0: 0.20FS D - 0.38F = 0
Based on double shear in pin J,
A=
2
FS.
A
OJ8
P = yF<
P= l
V
FA
FBD
d 2 = - (O.OIO)2 = 78.54 x 10 6 m 2
(2)(IOO x 106 )(78.54 x 10"*) .
3.0
5
N
P = — F.A = 5.82 x 103 N
9
Based on double shear in pins at B and D, A = ~ d 2 =
Fnn
BD
-
2rv A
FS.
(0.0 12)2 = 113.10x10 6 m 2
(2)(100x 106 )(l 13.10 x IO’6 )
3.0
---------- — ---------------------------------------------------- — {.2)4 X 1U
P = —F
19
3
IN
= 3 . 9 7 x l O3 N
Based on compression in links BD, for one link, A = (0.020)(0.008) = 160 x 10 6 m 2
2a b. A
FBD
-~rr-
(2)(250 x 106 )(l 60 x IO’6 )
3
M
-26.7x10 N
3
P = — F„
n = 14.04 x 10 N
19 BD
Allowable value of P is smallest,
P - 3.97 x 103 N
= 3.97 kN ◄
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r
L
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
= 0.90 and load factors
= 1.25 and
= 1.6, determine the largest load that can be safely
placed on the platform. (6) What is the corresponding conventional
factor of safety for rod BC?
1.8 in
A
I----------2.4 m
SOLUTION
3
= 0 : (2.4)-P-2.4
J
For dead loading,
- L2FF2
6
W t = (40)(9.81) = 392.4 N, W 2 = (50)(9.81) = 490.5 N
PD = ( 1 ) (392.4) + | j ( 4 9 0 . 5 ) = 1.0628 x 103 N
For live loading,
W] = mg
From which
3 P
m = -----—
5 g
Design criterion:
=0
PL -
/p/p + yL PL = $Pu
1=
- YpPp
(0-90X12 X 103 ) - (1.25)(1.0628 x 10~3 )
n
‘
i-6
= 5.920 x 103 N
(a)
Allowable load.
3 5.92X 103
5
9.81
m = 362 kg ◄
Conventional factor of safety:
P = PD + PL = 1.0628 x 10 3 + 5.920 x 103 = 6.983 x l 0 3 N
F.s. = — = 1 2 * 1 0 ’ 3
P
6.983 xlO
F.S. = 1.718 ◄
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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
= 0.85 and load factors
y D = 1.2 and yL = 1 .5, determine the required minimum ultimate load of
one cable, (b) What is the corresponding conventional factor of safety
for the selected cables?
SOLUTION
Y o P o + YL P L = <f> p u
(a)
pv = Y
D±nPL
</>
fl
A
(3
(1.2) - x 160 + (1,5) - x 2 x 1 9 5
= -----0.85
A
P„
= 629 lb ◄
u
Conventional factor of safety:
P = PD + PL = J X 160 + 0.75 X 2 X 195 = 372.5 lb
(ft)
F.S. =
-=- ~
P
372.5
F.S. = 1.689 4
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PROBLEM 1.59
15 m
in the marine crane shown, link CD is known to have a
uniform cross section of 50 x 150 mm. For the loading
shown, determine' ' the normal stress in the central portion
of that link.
35 m
■SO Mg
u
15 m
J
D
SOLUTION
IF = (80 Mg)(9.81 m/s 2 ) = 784.8 kN
Weight of loading:
Free Body: Portion JBC.
+)X M A =
F
co( 15 m ) - >D2 8 m) = 0
FCD = — W = —(784.8 kN)
15
15
FCD = +1465 kN
CD
+ 1465 x 103 N
= +195.3 x 10s Pa
(0.050 m)(0.150 m)
<rCD ~ +195.3 MPa ◄
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PROBLEM 1.60
0.5 in.
Two horizontal 5-kip forces are applied to pin B of the assembly shown.
Knowing that a pin of 0.8-in. diameter is used at each connection,
determine the maximum value of the average normal stress (a) in link
AB, (b) in link BC.
5 kips
5 kips
0.5 in.
60°
.
\
45°;
SOLUTION
Use joint B as free body.
|O kr'ps
8c
IO
pores
Law of Sines:
F
AB _ F BC _ 10
sin 45°
sin 60°
sin 95°
FAB = 7.3205 kips
FBC = 8.9658 kips
Link AB is a tension member.
Minimum section at pin:
(a)
Stress in AB :
net
= (1,8 - 0.8)(0.5) = 0.5 in 2
.
FAB
7.3205
AB
»net
0.5
a AB = 14.64 ksi
Link BC is a compression member.
Cross sectional area is A = (1.8)(0.5) = 0.9 in 2
(b)
Stress in BC:
a
BC
~ F sc
-------
-8-9658
0.9
a BC - -9-96 ksi
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PROBLEM 1.61
For the assembly and loading of Prob. 1.60, determine (a) the average
shearing stress in the pin at C, (b) the average bearing stress at C in
member BC, (c) the average bearing stress at B in member BC.
5 kip?
5 kips
PROBLEM 1.60 Two horizontal 5-kip forces are applied to pin B of the
assembly shown. Knowing that a pin of 0.8-in. diameter is used at each
connection, determine the maximum value of the average normal stress
(a) in link AB, (b) in link BC.
0.5 in.
60° \
>
45°;
SOLUTION
Use joint B as free body.
io k/ps
Fac
10 k>ps
Force
Law of Sines:
AB
sin 45°
(a)
_
BC
sin 60°
Shearing stress in pin at C.
_
10
sin 95°
FBC = 8.9658 kips
yr
r = ——
2A„
AFp = - d 2 = — (0.8) 2 = 0.5026 in 2
4
4
r=
(2)(0.5026)
= 8.92
8.92 ksi ◄
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PROBLEM 1.61 (Continued)
F
(b)
Bearing stress at C in member BC.
RC
A
A = td = (0.5)(0.8) = 0.4 in 2
_ 22 4 ksi ◄
_ 8.9658 _ 22
h
(r)
Bearing stress at B in member BC.
h
0.4
b
A
A = 2td = 2(0.5)(0.8) = 0.8 in 2
8.9658
<5b = --------- =11.21
0.8
, .
= 11.2 1 ksi ◄
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PROBLEM 1.62
Two steel plates are to be held together by means of 16-mmdiameter high-strength steel bolts fitting snugly inside cylindrical
brass spacers. Knowing that the average normal stress must not
exceed 200 MPa in the bolts and 130 MPa in the spacers,
determine the outer diameter of the spacers that yields the most
economical and safe design.
SOLUTION
At each bolt location the upper plate is pulled down by the tensile force Ph of the bolt. At the same time, the
spacer pushes that plate upward with a compressive force Ps in order to maintain equilibrium.
P„ = Ps
For the bolt,
= — = — 5-
or
p =
For the spacer,
P
4P
= — = ----- *—54,
-d h )
or
Ps =
» 7°*
- d b)
Equating Pb and Px ,
fit
A
d s z=
1+
\d h = l( 1 + — 1(16)
V
130 )
d = 25.2 mm ◄
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PROBLEM 1.63
A couple M of magnitude 1500 N * m is applied to the crank of an engine. For the
position shown, determine (a) the force P required to hold the engine system in
equilibrium, (b) the average normal stress in the connecting rod BC, which has a
450-mnr uniform cross section.
200 m m
60 mm
SOLUTION
Use piston, rod, and crank together as free body. Add wall reaction H
and bearing reactions Ax and
+)s M A = 0 : (0.280 m)H - 1500 N • m = 0
H = 5.3571 x 103 N
Use piston alone as free body. Note that rod is a two-force member;
hence the direction of force FBc is known. Draw the force triangle
and solve for P and FBB by proportions.
I = >/2002 + 60 2 = 208.81 mm
P
200
P = 17.86 x 103 N
~H ~ ”60~
P = 17.86 kN ◄
(a)
F _=
H
2Q8 81
60
.
F
= 18
6436 x
N
Rod BC is a compression member. Its area is
450 mm 2 = 450x l0“ 6 m2
2t>e>
Stress:
-18.6436 x 103
450 x IO"6
= -41.430 xlO 6 Pa
(6)
a BC = -41.4 MPa ◄
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PROBLEM 1.64
Knowing that the link DE is | in. thick and 1 in. wide, determine
the normal stress in the central portion of that link when
(a) 0 = 0°, (6) 0 = 90°.
60 lb
a
SOLUTION
Use member CEF as a free body
+)LM C = 0 : - 1 2 FDE - (8)(60 sin0) - (16)(60 cos(?) = 0
FDE = -40 sin 0 - 8 0 cos0 lb
DE ~= (1)1 1 ]
a
_
= 0.125 in 2
DE
DE DE
(a)
0 = 0:
FDE = -80 lb
-80
(ft)
0 = 90°:
<rDE ~ “640 psi ◄
FDE = -40 lb
-40
rjr — ~ “
0.125
a DE = -320 psi ◄
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1500 lb
B
PROBLEM 1.65
A |-in.-diameter steel rod AB is fitted to a round hole near end C of
the wooden member CD. For the loading shown, determine (a) the
maximum average normal stress in the wood, (6) the distance b for
which the average shearing stress is 100 psi on the surfaces indicated
by the dashed lines, (c) the average bearing stress on the wood.
SOLUTION
(fl)
Maximum normal stress in the wood.
A na = ( l ) | 4 - | j
= 3.375 in 2
P
1500
= 444 psi
J n a " 3.375
(b)
fT = 444 psi ◄
Distance 6 for r = 100 psi.
For sheared area see dotted lines.
P_
2bt
1500
= 7.50 in.
(2)(l)(100)
b = 7.50 in. ◄
Average bearing stress on the wood.
1500
= 2400 psi
= 2400 psi
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PROBLEM 1.66
D
Front view
6 mm
18 nun
fjp B
160 mm
Side view
120 mm
Top view
In the steel structure shown, a 6-mmdiameter pin is used at C and 10-mmdiameter 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.
C
SOLUTION
Use free body ABC.
r
c
= 0 : 0 . 2 8 0 P - 0.120FSI)
(1)
P = ~ P BD
af B =0:
0 . 1 6 0 P - 0.120C
-C
(2)
Tension on net section of link BD\
400
PB D
G
net
r
c
F.S.
—1° |(6 X 10~ )(l 8 - 10)(10“ ) = 6.40 x 10 N
net
3
3
3
Shear in pins at B and D\
2
FaBD
= f 150X10
D = rJ_p,n
in = - - — d
F.S. 4
3
V —1(10 x 10’3 ) 2 = 3.9270 x 10 3 N
UJ
Smaller value of Fan is 3.9270 x 103 N.
[(3.92 70 x 103 ) = 1.683 x 103 N
From (1),
Shear in pin at C:
From (2),
2r
p,n =
2-Sl-£rf 2 = (2)f 1 5 0 X 1C>6 Y— 3(6 x 10~3 )2 = 2.8274 x l 0 3 N
F.S.4
3
UT
'
[(2.8274 x 103 ) = 2.12 x 10 3 N
Smaller value of P is allowable value.
1.683 x 103 N
1.683 kN
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PROBLEM 1.67
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.
■*------0.8 ni -------
—().4 m —*■|
SOLUTION
Use member ABC as a free body, and note that member BD is a two-force member.
+)
= 0 : (P cos40°)(l .2) + (P sin 40°)(0.6)
- (F BD cos 30°)(0.6)
- (F BD sin 30°X0.4) = 0
1.30493P - 0.7L962FBD = 0
FBD = 1.81335P = (1.81335)(16 x 103 ) = 29.014 x 103 N
Fy = lOOx 103 N
1 . j.3 .
_--- Fv --- 100 xlO 3
F BD
2 9 . 0 1 4 x l 03
F.S. = 3.45 ◄
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PROBLEM 1.68
A force P is applied as shown to a steel reinforcing bar that has
been embedded in a block of concrete. Determine the smallest
length L for which the full allow able normal stress in the bar can be
developed. Express the result in terms of the diameter d of the bar,
the allowable normal stress cra n in the steel, and the average
allowable bond stress r aU between the concrete and the cylindrical
surface of the bar. (Neglect the normal stresses between the
concrete and the end of the bar.)
SOLUTION
For shear,
A = ndL
P = T M A = T ndL
For tension,
? -
Equating,
all
-
all
- cra n —d
Solving for £,
Anin “
Cr
all
/4r
all
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PROBLEM 1.69
2.4 kip?
The two portions of member AB are glued together along a plane
forming an angle 0 with the horizontal. Knowing that the ultimate stress
for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine
(a) the value of 0 for which the factor of safety of the member is
maximum, (b) the corresponding value of the factor of safety. (Hint:
Equate the expressions obtained for the factors of safety with respect to the
normal and shearing stresses.)
B
SOLUTION
= (2.0)(1.25) = 2.50 in 2
At the optimum angle,
(F.S.) = (F.S.) r
p
7
cr = — cos 0
4,
Normal stress:
Pr „
'
P
Shearing stress: r =-- — sin# cos#
0
&u- Q
----=—
cos 2 #
r
P
U.T =
(F.S.) r = Pu
L' 0
sin# cos#
=
P
Fsin# cos#
r
Equating,
Solving,
(b)
Pu
2
Pcos 0
uA)
Fsin# cos#
sin#
= tan# =
cos#
_ 1.3
=
= 0.520
" 2.5
_ (12.5)(2.50)
cos 2 #
cos 2 27.5°
(«)
#opt = 27.5° ◄
= 7.94 kips
F.S. =
P
_ 7.94
2.4
F.S. = 3.31 ◄
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P R O B L E M 1.70
2.4 kips
The two portions of member AB are glued together along a plane
forming an angle 0 with the horizontal. Knowing that the ultimate stress
for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine
the range of values of #for which the factor of safety of the members is
at least 3.0.
[B
2.0 in.
SOLUTION
Ao = (2.0)(1.25) = 2.50 in.2
P = 2.4 kips
Pu = (F.S.)P = 7.2 kips
Based on tensile stress,
cr L; = —cos 2 #
4
cos 2 ff =
= (2 -5 )(2 -5 0 ) = 0.86806
7.2
Pu
cos# = 0.93169
Based on shearing stress,
# = 21.3°
#>21.3°
R.
Pr
= — s i n # cos# = ——sin 2#
4)
4)
O 2 1 ' 2 5 0 1 " 31 =0.90278
7-2
20 = 64.52°
0 = 32.3°
0 < 32.3°
21.3° < 0 < 32.3° ◄
Hence,
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PROBLEM 1.C1
Element n
A solid steel rod consisting of n cylindrical elements welded together is subjected to the
loading shown. The diameter of element i is denoted by d ! and the load applied to its
lower end by P with the magnitude F, of this load being assumed positive if P, is
directed downward as shown and negative otherwise, (a) Write a computer program
that can be used with either SI or U.S. customary units to determine the average stress
in each element of the rod. (b) Use this program to solve Problems 1.1 and 1.3.
Element 1
SOLUTION
Force in clement f:
It is the sum of the forces applied to that element and all lower ones:
1=1
Average stress in element /:
Area = Ai =
F.
Ave. stress = —
Program outputs:
Problem 1.1
Element
Stress (MPa)
1
2
Problem 1.3
Element
Stress (ksi)
84.883
1
22.635
-96.766
2
17.927
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PROBLEM 1.C2
A 20-kN load is applied as shown to the horizontal member ABC.
Member ABC has a !0x50-mm uniform rectangular cross section
and is supported by four vertical links, each of 8x36-mm uniform
rectangular cross section. Each of the four pins at A, B, C, and D has the
same diameter d and is in double shear, (a) Write a computer program
to calculate for values of d from 10 to 30 mm, using 1-mm increments,
(i) the maximum value of the average normal stress in the links
connecting pins B and D, (ii) the average normal stress in the links
connecting pins C and E, (iii) the average shearing stress in pin B,
(iv) the average shearing stress in pin C, (v) the average bearing stress
at B in member ABC, and (vi) the average bearing stress at C in member
ABC. (b) Check your program by comparing the values obtained for
d = 16 mm with the answers given for Probs. 1.7 and 1.27. (c) Use this
program to find the permissible values of the diameter d of the pins,
knowing that the allowable values of the normal, shearing, and bearing
stresses for the steel used are, respectively, 150 MPa, 90 MPa, and
230 MPa. (d) Solve Part c, assuming that the thickness of member ABC
has been reduced from 10 to 8 mm.
0.4 m
0.25 m
€
E
20 kN
SOLUTION
P = 20 kN
Forces in links.
F.B. diagram of ABC'.
+ £ M C = 0: 2FBD (5C) - P( AC) = 0
FBD = P(AC)!2(BC) (tension)
= 0: 2FCE ( B Q - P(AB) = 0
(i)
A
a
(iii)
FCE = P(AB)!2(BC) (comp.)
(ii)
Link CE.
Thickness = tL
Link BD.
Thickness = t L
BL> = t d
BD =
+F
w
CE =
L ~ d '>
BD
&CE
BD
(iv)
Pin B.
T
(v)
~ CE
CE
C - FcE
d
)
Shearing stress in ABC under Pin B.
-
r
AC t Ac( W AC ')
.=0:
Sig Bear B = FBD /(dt AC )
(vi)
LW L
Pin C.
T
B = F BD /( 2 /4)
Bearing stress at B.
Thickness of member AC = t AC
t
=
Bearing stress at C.
rT
Sig Bear C = F CE /(dt AC )
2F B =2F BD
_
AC ~ -------------T
AC™AC
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PROBLEM 1.C2 (Continued)
Program Outputs
Input data for Parts (a), (h), (c):
d
10 .CO
11.00
12.00
13.00
14 . 00
15 . 00
16 . 00
1'7 70 0
18 . 00
19 . 00
20 . 00
21 . 00
22 . 00
23 . 00
24 . 00
25 . 00
26 . 00
27 . 00
28 . 00
29.00
30 . 00
P = 20kN,
5 = 0.25 m, BC= 0.40 m, AC = 0.65 m,
7Z = 8mm,
FPL = 36 mm, E4C=10mm, IE4C=50mm
Sigma BD Sigma CE
78 . 13
81.25
84 . 6 4
88 . 32
92.33
96.73
101.56
TOK. 9 1
112.85
119.49
126 . 95
135.42
1 4 5 . 09
6
3
- 2 1 . 70
- 2 1 . 70
- 2 1 . 70
-21.70
-21.70
- 2 1 . 70
- 2 1 . 70
- 2 1 . 7U
- 2 1 . 70
- 2 1 . 70
-21.70
- 2 1 . 70
-21.70
-21.70
- 2 1 . 70
- 2 1 . 70
-21.70
-21.70
-21.70
-21.70
- 2 1 . 70
Tau B
Tau C
79.58
65 . 77
55.26
47.09
40 . 60
35.37
31 . 08
27 . 54
24.56
22 . 04
19.89
18 . 04
16.44
15 . 04
13.82
12.73
1 1 . 77
10 . 92
10.15
9.46
8.84
80.82
71 • 5 9
63 . 86
57.31
5 1 . 73
46.92
42 . 75
39 . 1 1
35.92
33 . 10
30 . 61
28.3’8
26.39
24 . 60
22 . 99
SigBear B SigBear C
0>TO
216 . 67
203.12
191 . 18
180.56
171.05
1 6 2 . 50
154 . 76
147.73
141.30
135.42
1 3 0 . 00
1 2 5 . 00
1 2 0 . 37
1 1 6 . 07
1 1 2 . 07
108.33
125.00
1 1 3 . 64
1 0 4 . 17
96.15
89.29
83.33
78 . 13
73.53
69.44
65 . 79
62.50
59 . 52
56.82
54 . 3 5
52 . 08
50 . 00
48 . 08
46 . 3 0
44 . 64
43 . 10
41.67
(c) Answer: 16 mm < d < 22 mm ◄ (c)
Check: For d = 22 mm, Tau J C = 65 MPa < 90 MPa O.K.
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PROBLEM 1.C2 (Continued)
Input data for Part (d). P -= 20 kN,
AB = 0.25 m, BC = 0.40 m,
AC = 0.65 m, TL = 8 mm, WL = 36 mm,
TAC = 8 mm, WAC = 50 mm
d
10.00
11 ,00
12.00
13.00
14.00
15.00
16.00
17.00
18.00
19.00
20.00
21.00
22.00
23 .00
24.00
25.00
26.00
27.00
28.00
29 .00
30.00
Sigma BD Sigma CE
78.13
81.25
84.64
88.32
92.33
96.73
101.56
106.91
112.85
119.49
126.95
135.42
145.09
-21.70
-21,70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
-21.70
Tau B
Tau C
79.58
65.77
55.26
47.09
40 .60
35.37
31.08
27.54
24.56
22.04
19.89
18.04
16.44
15.04
13.82
12.73
11.77
10.92
10.15
9.46
8.84
71.59
63.86
57.31
51.73
46.92
42.75
39.11
35.92
33.10
30.61
28.38
26.39
24.60
22.99
SigBear B SigBear C
8
725 .69
213 .82
203.12
193.45
184.66
176.63
169.27
162.50
156.25
150.46
145.09
140.09
135.42
156.25
142.05
130.21
120.19
111.61
104.17
97.66
91.91
86.81
82.24
78.13
74.40
71.02
67.93
65.10
62.50
60.10
57.87
55.80
53.88
52.08
(d) Answer: 18 mm < d < 22 mm
◄
(d)
Check: For d = 22 mm, Tau AC = 81.25 MPa < 90 MPa O.K.
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PROBLEM 1.C3
0.5 in.
5 kips
5 kips
. 0.5 in.
60° \
i
45°J
Two horizontal 5-kip forces are applied to Pin B of the assembly
shown. Each of the three pins at J, B, and C has the same diameter d
and is double shear, (a) Write a computer program to calculate for
values of d from 0.50 to 1.50 in., using 0.05-in. increments, (i) the
maximum value of the average normal stress in member AB, (ii) the
average normal stress in member BC, (iii) the average shearing stress
in pin A, (iv) the average shearing stress in pin C, (v) the average
bearing stress at A in member AB, (vi) the average bearing stress at C
in member BC, and (vii) the average bearing stress at B in member BC.
(b) Check your program by comparing the values obtained for
d = 0.8 in. with the answers given for Problems 1.60 and 1.61. (c) Use
this program to find the permissible values of the diameter d of the
pins, knowing that the allowable values of the normal, shearing, and
bearing stresses for the steel used are, respectively, 22 ksi, 13 ksi, and
36 ksi. (d) Solve Part c, assuming that a new design is being
investigated in which the thickness and width of the two members are
changed, respectively, from 0.5 to 0.3 in. and from 1.8 to 2.4 in.
SOLUTION
Forces in members AB and BC.
Free body: Pin B.
W
A
From force triangle:
AB
sin 45°
F
_
BC
sin 60°
_
sin 75°
=2P(sin457sin75°)
Fs c = 2P(sm607sin75°)
(i)
ABC = wt
Thickness = t
&BC ~
AB = ( w
AB =
—
AB
BC
BC
d)t
AB
(iv)
Pin A.
r A = (FAB /2)/(xd 2 /4)
(v)
Ave, stress in BC.
Width = w
a
(iii)
(ii)
Max, ave. stress in AB.
Pin C.
T C = (F BC l2)l(7rd 2l4)
(vi)
Bearing stress at J.
Bearing stress at C.
Sig Bear C = F BC /dt
Sig Bear A = F AB /dt
(vii) Bearing stress at B in member BC.
Sig Bear B = F BC /2dt
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PROBLEM 1.C3 (Continued)
Program Outputs
Input data for Parts (a), (b), (c):
P = 5kips, w = 1 . 8 i n . ,
= 0.5 in.
SIGBRGA SIGBRGC SIGBRGB
D
SIGAB
SIGBC
TAUA
TAUC
in.
ksi
ksi
ksi
ksi
ksi
ksi
ksi
0.500
0 . 550
C. 600
0.650
0.700
0 . 750
0 .800
0 . 850
0.900
0 . 950
1.000
1 . 050
1 . 100
1 . 150
1.200
1.250
1,300
1.350
1.400
1.450
1 . 500
11.262
11.713
12 . 2 0 1
12.731
13.310
13.944
-9 . 962
-9.962
-9.962
-9.962
-9.962
-9.962
-9.962
-9.962
-9 . 962
-9.962
-9.962
-9 . 962
-9.962
-9.962
-9.962
-9.962
-9.962
-9.962
-9 . 962
-9.962
-9.962
13< 1
11.54
10.147
8.918
7.900
7.047
6.324
5.708
5.177
4.717
4.316
3.964
3.653
3.377
3.132
2.912
2 . 715
2.537
29.282
26 . 6 2 0
24.402
22 . 525
20.916
19.521
18 . 3 0 1
17.225
16.268
15.412
14.641
13.944
13.310
12.731
12.201
11.713
11.262
10.845
10.458
10.097
9.761
35.863
32.603
29.886
27.587
25.616
23 . 909
22.414
21 . 0 9 6
19.924
18.875
1 7 . 932
17.078
16.301
15.593
14 . 943
14.345
13 . 7 9 3
13.283
12.808
12.367
1 1 . 954
17 . 932
16.301
14 . 943
13 . 793
12.808
11.954
11.207
10 . 5 4 8
9.962
9.438
8.966
8.539
8.151
7.796
7.471
7.173
6.897
6.641
6.404
6 .183
5.977
14.641
15.412
16 . 2 6 8
17.225
18.301
19.521
20.916
0
8.285
7.282
6.450
5.754
5.164
4.660
4.227
3.852
3.524
3.236
2.983
2.758
2.557
2.378
2.217
2.071
(c) Answer: 0.70 in. < d< 1.10 in.
(*>)
◄ (c)
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PROBLEM 1.C3 (Continued)
Input data for Part (d),
P = 5 k i p s , w = 2.4in., t = 0.3 in.
D
SIGAB
SIGBC
TAUA
TAUC
in.
ksi
ksi
ksi
ksi
0.500
550
C. 600
0 . 650
0 , 700
0.750
0 . 800
0 . 850
0.900
0.950
1 . 000
1 . 050
1.100
1 . 150
1.200
1.250
1.300
1.350
1 .400
1.450
1.500
12 . 8 4 3
13 . 190
13.556
13 . 944
14 . 354
14 . 789
15 . 2 5 1
15.743
16.268
16.829
17.430
18 . 075
18 . 7 7 1
19 . 5 2 1
20.335
21 . 2 1 9
-12.452
-12.452
-12.452
-12.452
-12.452
-12 .452
-12 . 4 5 2
-12.452
-12.452
- 1 2 . 452
- 1 2 . 452
-12.452
-12.452
-12.452
-12 .452
-12.452
-12.452
-12.452
-12.452
-12 .452
-12.452
11.030
9.511
8.285
7 . 282
6 .450
5.754
5.164
4 . 660
4 . 227
3 . 852
3.524
3.236
2 . 983
2 .758
2.557
2.378
2.217
2 . 071
10 . 1 4 7
8 . 918
7.900
7 . 047
6.324
5.708
5.177
4 . 717
4.316
3.964
3 . 653
3.377
3.132
2 . 912
2 . 715
2.537
SIGBRGA SIGBRGC SIGBRGB
ksi
28.708
27 . 1 1 3
25.686
24.402
23.240
22.183
21.219
20.335
19 . 5 2 1
18.771
18.075
17.430
16.829
16.268
ksi
ksi
35.160
33 . 206
31.459
29.886
28.463
27 . 1 6 9
25 . 988
24 . 905
23 . 909
22 . 989
22.138
21.347
20 . 6 1 1
19.924
29.886
27 . 1 6 9
24 . 905
22 . 989
21.347
1 9 . 924
18 . 679
17.580
16 . 603
1 5 . 729
14 . 943
14.231
13.584
12 . 994
12.452
11.954
11.495
11.069
10 . 674
10 . 305
9.962
(d) Answer: 0.85 in. < d < 1.25 in.
◄
(d)
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PROBLEM 1.C4
A 4-kip force P forming an angle a with the vertical is applied as
shown to member ABC, which is supported by a pin and bracket at C
and by a cable BD forming an angle fl with the horizontal (a) Knowing
that the ultimate load of the cable is 25 kips, write a computer program
to construct a table of the values of the factor of safety of the cable for
values of a and fl from 0 to 45°, using increments in a and fl
corresponding to 0.1 increments in tan a and tan fl . (b) Check that
for any given value of a, the maximum value of the factor of safety is
obtained for fl- 38.66° and explain why. (<?) Determine the smallest
possible value of the factor of safety for fl - 38.66°, as well as the
corresponding value of of, and explain the result obtained.
A
r
15 in.
I
SOLUTION
(a)
Draw F.B. diagram of ABC'.
+p.M c =0:
(Fsina)(1.5 in.) + (F cos «)(30 in.)
Al
T
- (F cos /?)(1 5 in.) - (F sin /?)(! 2 in.) = 0
p- _ p 15 sin cr + 30cosa
15 cos /?+ 12 sin/?
P>
ISin,
1
F.S. = Fult /F
/Pin.
Output for P = 4 kips and Fuk =20 kips:
0
ALPHA
0.000
5.711
11.310
16.699
,21.801
,26.565
30.964
34.992
38 . 660
41 . 987
45.000
3.125
2.991
2.897
2 . 837
2 .805
2.795
2 ,803
2 .826
2 . 859
2 . 899
2.946
5.71
11.31
3.358 3.555
3.214 3.402
3.113 3.295
3 . 049 3 . 2 2 7
3,014 3.190
3 . 004 3 . 1 7 9
3 .013 3.189
3 .036 3 .214
3 . 0 7 2 3 . 252
3 .116 3.298
3 .166 3.351
VALUES OF FS
BETA
16.70 21.80 26.56
3 .712
3.552
3.441
3.370
3 . 331
3 . 320
3 .’33 0
3.356
3.395
3.444
3.499
3.830
3.666
3.551
3.477
3.438
3.426
3'436'
3 .463
3.503
3.554
3 . 611
30.96
34.99
3 . 913 3 . 966
3.745 3.796
3.628 3.677
3.553 3.600
3.512 3.560
3.500 3.547
3 . 5 1 0 "3 7 5 5 8
3 . 538 3 . 586
3.579 3.628
3.631 3.680
3 . 6 8 9 3 . 739
3 .994
3.823
3.703
3.626
3.585
3.572
37'583
3 .611
3.653
3.706
3.765
4 . 002
3 . 830
3.710
3.633
3.592
3 . 579
5 . 590
3.619
3 . 661
3.713
3.773
41.99
45.00
3 . 995
3 . 824
3.704
3.627
3 .586
3 . 573
3 . 584
3.612
3 . 655
3.707
3.767
3.977
3 . 807
3.687
3.611
3 . 570,
3 . 558|
T.TSK
3 .596
3 .638
3.690
3 .750
kb)
(b)
When fl = 38.66°, tan fl - 0.8 and cable BD is perpendicular to the lever arm BC.
(c)
F.S. = 3.579 for a = 26.6°; P is perpendicular to the lever arm AC.
Note: The value F.S.= 3.579 is the smallest of the values of F.S. corresponding to /? = 38.66° and the
largest of those corresponding to a = 26.6°. The point a - 26.6°, fl - 38.66° is a “saddle point,” or
“minimax” of the function F.S. (a, fl).
PROPRIETARY MATERIAL. Copyright £ 2 0 1 5 McGraw-Hill Education. This is p r o p r i e t a r y material solely tor a u t h o r i z e d i n s t r u c t o r use.
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PROBLEM 1.C5
A load P is supported as shown by two wooden members of uniform rectangular cross
section that are joined by a simple glued scarf splice, (a) Denoting by
and T v ,
respectively, the ultimate strength of the joint in tension and in shear, write a
computer program which, for given values of a, b, P, cr L; and
, expressed in either
SI or U.S. customary units, and for values of a from 5 to 85° at 5° intervals, can be
used to calculate (i) the normal stress in the joint, (ii) the shearing stress in the joint,
(iii) the factor of safety relative to failure in tension, (iv) the factor of safety relative
to failure in shear, and (v) the overall factor of safety for the glued joint, (b) Apply
this program, using the dimensions and loading of the members of Probs. 1.29 and
1.31, knowing that tfy =150 psi and
= 214 psi for the glue used in Prob. 1.29, and
that tty =1.26 MPa and
= 1 .50 MPa for the glue used in Prob. 1.31. (c) Verify in
each of these two cases that the shearing stress is maximum for a = 45°.
SOLUTION
(i) and (ii) Draw the F.B. diagram of lower member:
0: - V + P cos a
/lF=0:
f-Psina
P coscr
=0
F - Ps\na
<X
Area = o6/sin a
Normal stress:
Shearing stress:
(iii)
F
Area
= (P/ab) sin 2 a
------= (P/ab) sin a cos a
F.S. for tension (normal stresses):
FSN = a u /<
(iv)
F.S. for shear:
FSS=T U /T
(v)
Overall F.S. :
F.S. = The smaller of FSN and FSS.
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PROBLEM 1.C5 (Continued)
Program Outputs
ALPHA
SIG (MPa)
TAG (MPa)
II
II
II
Ln NJ
o os K □ °
2 2 § 1 |
ns PD
no
»
PD
o
II
II
<5 <5
ft
Problem 1.29
FSN
FSS
FS
5
0.007
0.085
169.644
17.669
17.669
10
0.029
0.167
42.736
8.971
8.971
15
0.065
0.244
19.237
6.136
6.136
20
0.114
0.314
11.016
4.773
4.773
25
0.175
0.375
7.215
4.005
4.005
30
0.244
0.423
5.155
3.543
3.543
35
0.322
0.459
3.917
3.265
3.265
40
0.404
0.481
3.119
3.116
3.116
45
0.489
0.489
2.577
3.068
2.577
50
0.574
0.481
2.196
3.116
2.196
55
0.656
0.459
1.920
3.265
1.920
60
0.733
0.423
1.718
3.543
1.718
65
0.803
0.375
1.569
4.005
1.569
70
0.863
0.314
1.459
4.773
1.459
75
0.912
0.244
1.381
6.136
1.381
80
0.948
0.167
1.329
8.971
1.329
85
0.970
0.085
1.298
17.669
1.298
◄ (b), (c)
PROPRIETARY MATERIAL. Copyright £ 2 0 1 5 McGraw-Hill Education. This is p r o p r i e t a r y m a t e r i a ] solely for a u t h o r i z e d i n s t r u c t o r use.
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PROBLEM 1.C5 (Continued)
Problem 13 1
a = 5 in.
b = 3 in.
P = 1400 lb
<J L! =150 psi
Ty =214 psi
ALPHA
SIG (psi)
TAU (psi)
FSN
FSS
FS
5
0.709
8.104
211.574
26.408
26.408
10
2.814
15.961
53.298
13.408
13.408
15
6.252
23.333
23.992
9.171
9.171
20
10.918
29.997
13.739
7.134
7.134
25
16.670
35.749
8.998
5.986
5.986
30
23.333
40.415
6.429
5.295
5.295
35
30.706
43.852
4.885
4.880
4.880
40
38.563
45.958
3.890
4.656
3.890
45
46.667
46.667
3.214
4.586
3.214
50
54.770
45.958
2.739
4.656
2.739
55
62.628
43.852
2.395
4.880
2.395
60
70.000
40.415
2.143
5.295
2.143
65
76.663
35.749
1.957
5.986
1.957
70
82.415
29.997
1.820
7.134
1.820
75
87.081
23.333
1.723
9.171
1.723
80
90.519
15.961
1.657
13.408
1.657
85
92.624
8.104
1.619
26.408
1.619
◄ (C)
◄ (b)
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PROBLEM 1.C6
12 mm
D
(TT
B
pB
20 mm
■8 mm
8 mm
12 m m —>- —
Side view
F ront view
Member ABC is supported by a pin and
bracket at A and by two links, which are pinconnected to the member at B and to a fixed
support at D. (a) Write a computer program to
calculate the allowable load
for any given
values of (i) the diameter
of the pin at A,
(ii) the common diameter (fa of the pins at B
and D, (iii) the ultimate normal stress
in
each of the two links, (iv) the ultimate shearing
stress ru in each of the three pins, and (v) the
desired overall factor of safety F.S. (b) Your
program should also indicate which of the
following three stresses is critical: the normal
stress in the links, the shearing stress in the pin
at A, or the shearing stress in the pins at B and
D. (c) Check your program by using the data
of Probs. 1.55 and 1.56, respectively, and
comparing the answers obtained for P a n with
those given in the text, (d) Use your program to
determine the allowable load P a n, as well as
which of the stresses is critical, when d x =
d 2 = 15 mm,
= 1 10 MPa for aluminum links,
T LI = 100 MPa for steel pins, and F.S. = 3.2.
SOLUTION
(a)
F.B. diagram
ABC.
ZM A =0:
P = — F BD
BD
380
XM B =0:
P = ----- F A
180
ZOO
F*
200
of Pin J:
n
(i)
Forgiven
= 2{T U IFS)(
I4),
(ii)
For given d 2 of Pins B and D :
FBD =2(T v /FS
/4),
(iii)
For ultimate stress in links /?£>: F BD = 2(<7L/ /FS)(0.02)(0.008),
(iv)
For ultimate shearing stress in pins: P4 is the smaller of Px andP2 -
(v)
For desired overall F.S. :
200
----- FA
180
2
200
" 380
BD
200
Fr.3 = ----- FBD
380
P5 is the smaller of P3 and P4 .
If P3 < P4 , stress is critical in links.
If P4 < P3 and P x < P2 , stress is critical in Pin A.
If P4 < P3 and P2 < Px, stress is critical in Pins B and D.
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PROBLEM 1.C6 (Continued)
Program Outputs
(b)
Problem 1 .55.
Data:
=8 mm, d 2 =12mm,
07,,= 250 MPa,
r L! =100 MPa,
F,S. = 3.0
◄
Pall = 3.72 kN. Stress in Pin A is critical.
(c)
Problem 1.56.
Data:
=10 mm, d 2 = 12 mm, <7 =250 MPa,
r L. =100 MPa,
◄
= 3.97 kN. Stress in Pins B and D is critical.
(d)
Data:
d x = d 2 - 15mm,
=110 MPa,
ru = 100 MPa,
F.S. = 3.0
F.S. = 3.2
Pal i = 5.79 kN. Stress in links is critical.
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on a website, in whole or part.
90
Full file at https://TestbankDirect.eu/
2/21/2025, 2:25 PM