© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–108. Determine the moment reactions at the supports
A and B, then draw the shear and moment diagrams. Solve
by expressing the internal moment in the beam in terms of
Ay and MA. EI is constant.
w
B
A
L
wx2
2
Elastic Curve and Slope:
M(x) = Ayx - MA -
EI
d2v
wx2
= M(x) = Ayx - MA 2
2
dx
EI
Ayx
dv
wx3
=
- MAx + C1
dx
2
6
2
Ayx3
EIv =
-
6
(1)
MAx2
wx4
+ C1x + C2
2
24
(2)
Boundary Conditions:
dv
= 0
dx
at
x = 0
From Eq. (1)
C1 = 0
x = 0
at
v = 0
From Eq. (2)
C2 = 0
dv
= 0
dx
at
x = L
From Eq. (1)
0 =
AyL2
2
at
v = 0
wL3
6
(3)
MAL2
wL4
2
24
(4)
- MAL x = L
From Eq. (2)
0 =
AyL3
6
-
Solving Eqs. (3) and (4) yields:
wL
2
wL2
MA =
12
Ay =
Ans.
Due to symmetry:
MB =
wL2
12
Ans.
1261
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–109. The beam has a constant E1I1 and is supported by
the fixed wall at B and the rod AC. If the rod has a crosssectional area A2 and the material has a modulus of
elasticity E2, determine the force in the rod.
C
w
L2
B
A
L1
+ c ©Fy = 0
TAC + By - wL1 = 0
c + ©MB = 0
TAC(L1) + MB MB =
wL1 2
= 0
2
(1)
wL1 2
- TACL1
2
(2)
Bending Moment M(x):
wx2
2
M(x) = TACx -
Elastic Curve and Slope:
EI
d2v
wx2
= M(x) = TACx 2
2
dx
EI
TACx2
dv
wx3
=
+ C1
dx
2
6
EIv =
(3)
TACx3
wx4
+ C1x + C2
6
24
(4)
Boundary Conditions:
v =
TACL2
A 2E2
x = 0
From Eq. (4)
-E1I1 a
TACL2
b = 0 - 0 + 0 + C2
A2E2
C2 = a
v = 0
- E1I1L2
b TAC
A 2E2
at
x = L1
From Eq. (4)
0 =
TACL1 3
wL1 4
E1I1L2
+ C1L1 T
6
24
A 2E2 AC
dv
= 0
dx
at
(5)
x = L1
From Eq. (3)
0 =
TACL1 2
wL1 3
+ C1
2
6
(6)
Solving Eqs. (5) and (6) yields:
TAC =
3A 2E2wL1 4
8 A A 2E2L1 3 + 3E1I1L2 B
Ans.
Ans:
TAC =
1262
3A2E2wL41
8(A2E2L31 + 3E1I1L2)
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–110. The beam is supported by a pin at A, a roller at B,
and a post having a diameter of 50 mm at C. Determine the
support reactions at A, B, and C. The post and the beam are
made of the same material having a modulus of elasticity
E = 200 GPa, and the beam has a constant moment of
inertia I = 255(106) mm4.
15 kN/m
A
1m
6m
B
C
6m
Equations of Equilibrium. Referring to the free-body diagram of the entire beam, Fig. a,
+
: ©Fx = 0;
Ax = 0
+ c ©Fy = 0;
A y + By + FC - 15(12) = 0
a + ©MB = 0;
15(12)(6) - FC(6) - Ay(12) = 0
Ans.
(1)
2A y + FC = 180
(2)
Moment Functions. Referring to the free-body diagram of the beam’s segment, Fig. b,
x
M(x) + 15xa b - Ayx = 0
2
a + ©MO = 0;
M(x) = A yx - 7.5x2
Equations of Slope and Elastic Curves.
EI
d2v
= M(x)
dx2
EI
d2v
= A yx - 7.5x2
dx2
EI
Ay
dv
=
x 2 - 2.5x3 + C1
dx
2
EIv =
Ay
6
(3)
x3 - 0.625x4 + C1x + C2
(4)
Boundary Conditions. At x = 0, v = 0. Then Eq. (4) gives
C2 = 0
0 = 0 - 0 + 0 + C2
At x = 6 m, v = - ¢ C = -
E C 255 A 10 - 6 B D a -
FC(1)
1600FC
FCLC
= = . Then Eq. (4) gives
p
A CE
pE
A 0.052 B E
4
Ay
1600FC
b =
A 63 B - 0.625 A 64 B + C1(6)
pE
6
C1 = 135 - 6A y - 0.02165FC
Due to symmetry,
0 =
Ay
2
dv
= 0 at x = 6 m. Then Eq. (3) gives
dx
A 62 B - 2.5 A 63 B + 135 - 6A y - 0.02165FC
12A y - 0.02165FC = 405
(5)
Solving Eqs. (2) and (5),
FC = 112.096 kN = 112 kN
A y = 33.95 kN = 34.0 kN
Ans.
Substituting these results into Eq. (1),
By = 33.95 kN = 34.0 kN
Ans.
1263
Ans:
Ax = 0, FC = 112 kN, Ay = 34.0 kN,
By = 34.0 kN
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–111. Determine the moment
supports A and B. EI is constant.
reactions
at
the
w
B
A
L
–
2
uB>A = 0 =
-MA
1 - wL2 L
1 AyL
a
b (L) + a
b (L) + a
ba b
2 EI
EI
3 8EI
2
0 =
tB>A = 0 =
L
–
2
AyL
2
- MA -
wL2
48
(2)
-MA
1 AyL
L
L
1 -wL2 L L
a
b (L) a b + a
b (L)a b + a
ba ba b
2 EI
3
EI
2
3 8EI
2
8
0 =
AyL
6
-
MA
wL2
2
384
(3)
Solving Eqs. (2) and (3) yields:
3wL
32
5wL2
MA =
192
Ay =
Ans.
2
c + © MB = 0;
MB +
3wL
5wL
wL L
(L) a b = 0
32
192
2 4
MB =
11wL2
192
Ans.
Ans:
MA =
1264
5wL2
11wL2
, MB =
192
192
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–112. Determine the moment reactions at the supports,
and then draw the shear and moment diagrams. EI is
constant.
P
C
B
A
L
–
2
uC>A = 0 =
- MA
1 AyL
1 - PL L
a
b (L) + a
b (L) + a
ba b
2 EI
EI
2 2EI
2
0 =
tC>A = 0 =
PL
1
A L - MA 2 y
8
(1)
-MA
1 AyL
L
L
1 -PL L L
a
b (L) a b + a
b (L)a b + a
ba ba b
2 EI
3
EI
2
2 2EI
2
6
0 =
AyL
6
-
P
MA
PL
2
48
(2)
Solving Eqs. (1) and (2) yields:
P
Ay =
2
PL
MA =
8
Ans.
Ans.
Due to symmetry:
P
By =
2
PL
MB =
8
Ans.
Ans.
Cy = P
Ans.
1265
L
–
2
L
–
2
L
–
2
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–113. Determine the reactions at the bearing support A
and fixed support B, then draw the shear and moment
diagrams for the beam. EI is constant.
3 kip
2 kip
A
B
3 ft
3 ft
3 ft
Equations of Equilibrium. Referring to the free-body diagram of the shaft, Fig. a,
: © Fx = 0;
+
Bx = 0
+ c ©Fy = 0;
Ay + By - 2 - 3 = 0
a+ ©MB = 0;
3(3) + 2(6) - Ay(9) - MB = 0
Ans.
(1)
MB = 21 - 9Ay
(2)
M
Diagram
EI
M
As shown in Fig. b, the
diagrams for 2 kip and 3 kip and Ay on the catilever beam
EI
are drawn separately.
Elastic Curve and
Moment Area Theorems. From the elastic curve, notice that tA>B = 0. Thus,
tA>B = 0 = c 3 +
2
2
1
6
1
15
1
6
2
1 9Ay
b (9) d
(3) d c a - b (3) d + c 6 + (3) d c a - b (3) d + c6 + (3) d c - (3) d + c (9) d c a
3
2
EI
3
2
EI
2
EI
3
2 EI
Ay = 1.4815 kip = 1.48 kip
Ans.
Substituting this result into Eqs. (1) and (2),
By = 3.5185 kip = 3.52 kip
Ans.
MB = 7.6667 kip # ft = 7.67 kip # ft
Ans.
Ans:
Ay = 1.48 kip, Bx = 0, By = 3.52 kip,
MB = 7.67 kip # ft
1266
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–114. Determine the reactions at the supports A and B,
then draw the shear and moment diagrams. EI is constant.
P
P
A
B
L
–
3
(tB>A)1 =
2L
1 - 2PL 2L L
4L
2PL3
1 - PL L 2L
a
ba ba
+
b + a
ba
ba +
b = 2 3EI
3
3
9
2 3EI
3
3
9
9EI
(tB>A)2 =
ByL3
2L
1 ByL
a
b (L)a
b =
2 EI
3
3EI
L
–
3
L
–
3
tB>A = 0 = (tB>A)1 + (tB>A)2
3
0 =-
ByL
2PL3
+
9EI
3EI
By =
2
P
3
Ans.
From the free-body diagram,
MA =
PL
3
Ay =
4
P
3
Ans.
Ans.
Ax = 0
Ans.
Ans:
By =
1267
2
PL
4
P, MA =
, Ay = P, Ax = 0
3
3
3
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–115. Determine the vertical reactions at the bearings
supports, then draw the shear and moment diagrams. EI is
constant.
P
A
B
L
2
L
Support Reactions: FBD(a).
+
: ©Fx = 0;
Bx = 0
C
L
2
Ans.
+ c ©Fy = 0;
- A y + By + Cy - P = 0
[1]
a+ ©MA = 0;
By (L) + Cy (2L) - P a
[2]
3L
b = 0
2
Elastic Curve: As shown.
M/EI Diagrams: M/EI diagrams for P and By acting on a simply supported beam are
drawn separately.
Moment-Area Theorems:
(tA>C)1 =
=
1 3PL 3L 2 3L
1 3PL L 3L
L
a
ba
ba ba
b + a
ba ba
+ b
2 8EI
2
3
2
2 8EI
2
2
6
7PL3
16EI
(tA>C)2 =
By L
By L3
1
ab (2L)(L) = 2
2EI
2EI
(tB>C)1 =
1 PL
L 2 L
PL
L L
a
ba ba ba b + a
ba ba b
2 8EI
2
3
2
4EI
2
4
+
=
(tB>C)2 =
1 3PL L L
L
a
ba ba + b
2 8EI
2
2
6
5PL3
48EI
By L
By L3
1
L
ab (L)a b = 2
2EI
3
12EI
3
tA>C = (tA>C)1 + (tA>C)2 =
By L
7PL3
16EI
2EI
tB>C = (tB>C)1 + (tB>C)2 =
By L3
5PL3
48EI
12EI
From the elastic curve,
tA>C = 2tB>C
By L3
By L3
7PL3
5PL3
= 2a
b
16EI
2EI
48EI
12EI
By =
11P
16
Ans.
Substituting By into Eqs. [1] and [2] yields,
Cy =
13P
32
Ay =
Ans:
3P
32
Ans.
1268
By =
11P
13P
3P
, Cy =
, Ay =
16
32
32
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–116. Determine the reactions at the journal bearing
support A and fixed support B, then draw the shear and
moment diagrams for the shaft. EI is constant.
3 kip⭈ft
A
3 ft
Equations of Equilibrium. Referring to the free-body diagram of the shaft, Fig. a,
+ ©Fx = 0;
:
+ c ©Fy = 0 ;
Ay - By = 0
+ ©MB = 0;
3 + MB - Ay(6) = 0
Bx = 0
Ans.
(1)
MB = 6Ay - 3
(2)
M
M
Diagram. As shown in Fig. b, the
diagrams for the 3 kip # ft
EI
EI
couple moment and Ay are drawn separately.
Elastic Curve and
Moment Area Theorems. From the elastic curve, notice that tA>B = 0. Thus,
1
3
2
1 6Ay
b (6) d + c (6) d c a
b(6) d
tA>B = 0 = c (6) d c a 2
EI
3
2 EI
Ay = 0.75 kip
Ans.
Substituting the result of Ay into Eqs. (1) and (2),
By = 0.75 kip
MB = 1.5 kip # ft
Ans.
The shear and moment diagrams are shown in Figs. c and d, respectively.
1269
B
6 ft
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–117. Determine the reactions at the bearing supports
A, B, and C of the shaft, then draw the shear and moment
diagrams. EI is constant. Each bearing exerts only vertical
reactions on the shaft.
A
C
B
1m
1m
400 N
1m
1m
400 N
Support Reactions: FBD(a).
+ c ©Fy = 0;
A y + By + Cy - 800 = 0
[1]
a + ©MA = 0;
By (2) + Cy (4) - 400(1) - 400(3) = 0
[2]
Method of Superposition: Using the table in Appendix C, the required
displacements are
yB œ =
Pbx
A L2 - b2 - x2 B
6EIL
=
400(1)(2) 2
A 4 - 12 - 2 2 B
6EI(4)
=
366.67 N # m3
EI
T
By A 4 B
1.3333By m
PL3
=
=
48EI
48EI
EI
3
yB fl =
3
c
The compatibility condition requires
(+ T)
0 = 2yB ¿ + yB –
0 = 2a
1.3333By
366.67
b + ab
EI
EI
By = 550 N
Ans.
Substituting By into Eqs. [1] and [2] yields,
A y = 125 N
Cy = 125 N
Ans.
Ans:
By = 550 N, Ay = 125 N, Cy = 125 N
1270
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–118. Determine the reactions at the supports A and B.
EI is constant.
P
A
B
L
Referring to the FBD of the beam, Fig. a
+
: ©Fx = 0;
Ax = 0
L
2
Ans.
By - P - A y = 0
+ c ©Fy = 0;
A y = By - P
(1)
3
a + ©MA = 0; -MA + By L - P a Lb = 0
2
MA = By L -
3
PL
2
(2)
Referring to Fig. b and the table in Appendix C, the necessary deflections at B are
computed as follow:
yP =
Px 2
(3LAC - x)
6EI
=
P(L2)
3
c3a Lb - L d
6EI
2
=
7PL3
12EI
T
3
yBy =
By L
PL3AB
c
=
3EI
3EI
The compatibility condition at support B requires that
(+ T )
0 = vP + vBy
3
-By L
7PL3
+ a
b
0 =
12EI
3EI
By =
7P
4
Ans.
Substitute this result into Eq (1) and (2)
Ay =
3P
4
MA =
PL
4
Ans.
Ans:
Ax = 0, By =
1271
7P
3P
PL
, Ay =
, MA =
4
4
4
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–119. Determine the reactions at the supports A, B, and
C, then draw the shear and moment diagrams. EI is constant.
12 kip
A
3 kip/ ft
C
B
6 ft
6 ft
12 ft
Support Reaction: FBD(b).
+ ©F = 0;
:
x
Cx = 0
Ans.
+ c ©Fy = 0;
A y + By + Cy - 12 - 36.0 = 0
[1]
a + ©MA = 0;
By (12) + Cy (24) - 12(6) - 36.0(18) = 0
[2]
Method of Superposition: Using the table in Appendix C, the required
displacements are
5(3) A 24 B
6480 kip # ft3
5wL4
yB ¿ =
=
=
768EI
768EI
EI
4
yB – =
=
Pbx
A L2 - b2 - x2 B
6EIL
2376 kip # ft3
12(6)(12)
A 24 2 - 62 - 12 2 B =
6EI(24)
EI
By A 24 B
288By ft
PL3
=
=
48EI
48EI
EI
3
yB –¿ =
T
T
3
c
The compatibility condition requires
(+ T)
0 = yB ¿ + yB – + yB –¿
0 =
288By
2376
6480
+
+ ab
EI
EI
EI
By = 30.75 kip
Ans.
Substituting By into Eqs. [1] and [2] yields,
Ay = 2.625 kip
Cy = 14.6 kip
Ans.
Ans:
Cx = 0, By = 30.75 kip, Ay = 2.625 kip,
Cy = 14.6 kip
1272
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–120. Determine the moment
supports A and B. EI is constant.
reactions
at
the
w
B
A
L
–
2
- MA
1 Ay L
1 - wL2 L
a
b (L) + a
b (L) + a
ba b
2
EI
EI
3 8EI
2
uB>A = 0 =
Ay L
0 =
tB>A = 0 =
0 =
2
- MA -
wL2
48
(1)
-MA
1 Ay L
L
L
1 -wL2 L L
a
b (L)a b + a
b (L) a b + a
ba ba b
2
EI
3
EI
2
3 8EI
2
8
Ay L
6
-
MA
wL2
2
384
(2)
Solving Eqs. (1) and (2) yields:
Ay =
3wL
32
MA =
5wL2
192
c+ ©MB = 0;
Ans.
MB +
3wL
5wL2
wL L
(L) a b = 0
32
192
2
4
MB =
11wL2
192
Ans.
1273
L
–
2
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–121. Determine the deflection at the end B of the
clamped A-36 steel strip. The spring has a stiffness of
k = 2 N> mm. The strip is 5 mm wide and 10 mm high. Also,
draw the shear and moment diagrams for the strip.
50 N
200 mm
B
A
I =
10 mm
k= 2 N/mm
1
(0.005)(0.01)3 = 0.4166 (10 - 9) m4
12
(¢ B)1 =
50(0.23)
PL3
= 0.0016 m
=
3EI
3(200)(109)(0.4166)(10 - 9)
(¢ B)2 =
2000¢ B(0.23)
PL3
= 0.064 ¢ B
=
3EI
3(200)(109)(0.4166)(10 - 9)
Compatibility Condition:
+T
¢ B = (¢ B)1 - (¢ B)2
¢ B = 0.0016 - 0.064¢ B
¢ B = 0.001504 m = 1.50 mm
Ans.
By = k¢ B = 2(1.5) = 3.01 N
Ans:
¢ B = 1.50 mm T
1274
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–122. Determine the reactions at the supports A and B.
EI is constant.
M0
A
B
L
Referring to the FBD of the beam, Fig. a,
+
: ©Fx = 0;
Ax = 0
Ans.
+ c ©Fy = 0;
By - A y = 0
(1)
a + ©MA = 0;
By(L) - M0 - MA = 0
MA = ByL - M0
(2)
Referring to Fig. b and the table in the appendix, the necessary deflections are:
vM0 =
M0L2
2EI
vBy =
ByL3
PL3
=
3EI
3EI
T
c
Compatibility condition at roller support B requires
0 = vM + (vB)y
(+ T )
0
0 =
ByL3
M0L2
+ ab
2EI
3EI
By =
3M0
2L
Ans.
Substitute this result into Eq. (1) and (2)
Ay =
3M0
2L
MA =
M0
2
Ans.
Ans:
Ax = 0, By =
1275
3M0
3M0
M0
, Ay =
, MA =
2L
2L
2
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–123. Determine the reactions at support C. EI is the
same for both beams.
P
D
B
A
C
L
2
L
2
Support Reactions: FBD (a).
+ ©F = 0;
:
x
Cx = 0
a + ©MA = 0;
Cy(L) - By a
Ans.
L
b = 0
2
[1]
Method of Superposition: Using the table in Appendix C, the required
displacements are
yB =
By L3
PL3
=
48EI
48EI
T
PA 2 B
PL3BD
PL3
=
=
3EI
3EI
24EI
L 3
yB ¿ =
yB – =
By L3
PL3BD
=
3EI
24EI
T
c
The compatibility condition requires
yB = yB ¿ + yB –
(+ T)
By L3
48EI
=
By =
By L3
PL3
+ ab
24EI
24EI
2P
3
Substituting By into Eq. [1] yields,
Cy =
P
3
Ans.
Ans:
Cx = 0 , Cy =
1276
P
3
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–124. Before the uniform distributed load is applied on
the beam, there is a small gap of 0.2 mm between the beam
and the post at B. Determine the support reactions at A, B,
and C. The post at B has a diameter of 40 mm, and the
moment of inertia of the beam is I = 875(106) mm4. The
post and the beam are made of material having a modulus
of elasticity of E = 200 GPa.
30 kN/m
A
6m
Equations of Equilibrium. Referring to the free-body diagram of the beam, Fig. a,
+ ©F = 0;
:
x
Ax = 0
+ c ©Fy = 0;
A y + FB + Cy - 30(12) = 0
(1)
a + ©MA = 0;
FB(6) + Cy(12) - 30(12)(6) = 0
(2)
Ans.
Method of Superposition: Referring to Fig. b and the table in the Appendix, the
necessary deflections are
5(30) A 12 B
5wL4
8100 kN # m3
=
=
T
(vB)1 =
384EI
384EI
EI
4
FB A 12 3 B
36FB
PL3
=
=
(vB)2 =
48EI
48EI
EI
c
The deflection of point B is
vB = 0.2 A 10 - 3 B +
FB(1)
FBLB
= 0.2 A 10 - 3 B +
AE
AE
T
The compatibility condition at support B requires
A+TB
vB = (vB)1 + (vB)2
0.2 A 10 - 3 B +
FB (1)
36FB
8100
=
+ ab
AE
EI
EI
0.2 A 10 - 3 B E +
FB
p
A 0.04 2 B
4
+
FB
36FB
8100
=
A
I
I
36FB
875 A 10 - 6 B
8100
=
875 A 10 - 6 B
-
C
1m
0.2 A 10 - 3 B C 200 A 109 B D
1000
FB = 219.78 kN = 220 kN
Ans.
Substituting the result of FB into Eqs. (1) and (2),
A y = Cy = 70.11 kN = 70.1 kN
Ans.
1277
B
0.2 mm
6m
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–125. The fixed supported beam AB is strengthened
using the simply supported beam CD and the roller at F
which is set in place just before application of the load P.
Determine the reactions at the supports if EI is constant.
P
A
B
C
L
—
4
D
F
L
—
4
L
—
4
L
—
4
dF = Deflection of top beam at F
d¿ F = Deflection of bottom beam at F
dF = d¿ F
2M A L2 B
Q A L2 B
(P - Q)(L3)
L 2
2
cL - a b d =
(+ T)
48 EI
6 EIL
2
48EI
3
(P - Q)L
QL
1
3
- M =
48
6
4
48(8)
(1)
8PL - 48M = 9QL
uA = u¿ A + u– A = 0
c+
-
(P - Q)L2
ML
ML
+
= 0
6EI
3EI
16EI
8M = (P - Q)L
(2)
Solving Eqs. (1) and (2):
M = QL>16
Q = 2P>3
S = P>3
R = P>6
M = PL>24
Thus,
MA = MB =
1
PL
24
Ans.
Ay = By =
1
P
6
Ans.
Cy = Dy =
1
P
3
Ans.
Ans:
MA = MB =
Cy = Dy =
1278
1
1
PL, Ay = By = P,
24
6
1
P, Dx = 0
3
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
w
12–126. Determine the force in the spring. EI is constant.
A
B
k
L
¢ B¿ =
wL4
;
8EI
dB =
FspL3
3EI
By Superposition:
¢ B = ¢¿B - dB
+T
3
Fsp
FspL
wL4
=
k
8EI
3EI
24 EIFsp
k
24 EIFsp
k
Fsp c
= 3wL4 - 8FspL3
+ 8FspL3 = 3wL4
24EI + 8kL3
d = 3wL4
k
Fsp =
3kwL4
24EI + 8kL3
Ans.
Ans:
Fsp =
1279
3kwL4
24EI + 8kL3
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–127. The beam is supported by the bolted supports at
its ends. When loaded these supports initially do not provide
an actual fixed connection, but instead allow a slight
rotation a before becoming fixed after the load is fully
applied. Determine the moment at the connections and the
maximum deflection of the beam.
P
L
—
2
L
—
2
u - u¿ = a
ML
ML
PL2
= a
16EI
3EI
6EI
ML = a
M =
PL2
- ab (2EI)
16EI
2EI
PL
a
8
L
¢ max = ¢ - ¢¿ =
Ans.
M(L2 )
PL3
- 2c
C L2 - (L>2)2 D d
48EI
6EIL
¢ max =
PL3
L2 PL
2EIa
a
b
48EI
8EI
8
L
¢ max =
aL
PL3
+
192EI
4
Ans.
Ans:
M =
1280
PL
2EI
PL3
aL
a , ¢ max =
+
8
L
192EI
4
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–128. Each of the two members is made from 6061-T6
aluminum and has a square cross section 1 in. * 1 in. They
are pin connected at their ends and a jack is placed between
them and opened until the force it exerts on each member is
50 lb. Determine the greatest force P that can be applied to
the center of the top member without causing either of the
two members to yield. For the analysis neglect the axial
force in each member. Assume the jack is rigid.
P
B
A
E
6 ft
The jack force will cause a spread, ¢ , between the bars. After P is applied, this
spread is the difference between dE and dF.
¢ = dF - dE
Let R be the final reaction force of the jack on the bar above and the bar below.
From Appendix C,
2a
(P - R)L3
RL3
50L3
b =
48EI
48EI
48EI
R =
P
+ 50
2
The bottom member will yield first, since it will be subject to greater deformation
after P is applied. The moment due to the support reactions, R> 2 at each end, is
greatest in the middle:
Mmax =
P
R L
a b = a + 25 b (6)(12) = 18P + 1800
2 2
4
smax =
Mc
I
37(103) =
(18P + 1800) A 12 B
1
3
12 (1)(1 )
P = 243 lb
Ans.
1281
D
F
C
6 ft
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–129. The beam is made from a soft linear elastic
material having a constant EI. If it is originally a distance
¢ from the surface of its end support, determine the length
a that rests on this support when it is subjected to the
uniform load W0, which is great enough to cause this to
happen.
w0
⌬
a
L
The curvature of the beam in region BC is zero, therefore there is no bending
moment in the region BC. The reaction R is at B where it touches the support. The
slope is zero at this point and the deflection is ¢ where
¢ =
w0(L - a)4
R(L - a)3
8EI
3EI
ux = 0 =
w0(L - a)3
R(L - a)2
6EI
2EI
Thus,
R =
w0(L - a)
3
¢ =
w0(L - a)4
(72EI)
1
L - a = a
72¢EI 4
b
w0
a = L - a
72¢EI 4
b
w0
1
Ans.
Ans:
a = L - a
1282
1
72¢EI 4
b
wo
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–130. The A992 steel beam and rod are used to support
the load of 8 kip. If it is required that the allowable normal
stress for the steel is sallow = 18 ksi, and the maximum
deflection not exceed 0.05 in., determine the smallest
diameter rod that should be used. The beam is rectangular,
having a height of 5 in. and a thickness of 3 in.
C
5 ft
A
dr = db
B
4 ft
(8 - F)(48)3
F(5)(12)
=
AE
3E A 1 B (3)(5)3
12
8 kip
Assume rod reaches its maximum stress.
s =
F
= 18(103)
A
18(5)(12)
1179.648(8 - F)
=
E
E
F = 7.084 kip
Maximum stress in beam,
s =
(8 - 7.084)(48)(2.5)
Mc
=
= 3.52 ksi < 18 ksi
1
3
I
12 (3)(5)
OK
Maximum deflection
d =
(8 - 7.084)(48)3
PL3
=
= 0.0372 in. < 0.05 in.
1
3EI
3(29)(103) A 12
B (3)(5)3
A =
7.084
1
= 0.39356 in2 = pd2
18
4
OK
Thus,
d = 0.708 in.
Ans.
Ans:
d = 0.708 in.
1283
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–131. The 1-in.-diameter A-36 steel shaft is supported by
bearings at A and C. The bearing at B rests on a simply
supported A-36 steel wide-flange beam having a moment of
inertia of I = 500 in4. If the belt loads on the pulley are 400 lb
each, determine the vertical reactions at A, B, and C.
3 ft
5 ft
A
2 ft
5 ft
B
Es = Eb = E
400
lb
For the Shaft:
(¢ b)1 =
(¢ b)2 =
800(3)(5)
13200
T
A - 52 - 32 + 102 B =
6EIs(10)
EIs
By A 103 B
48EIs
400
lb
C
5 ft
20.833By
c
=
EIs
For the Beam:
¢b =
By A 103 B
48EIb
20.833By
T
=
EIb
Compatibility Condition:
+ T ¢ b = (¢ b)1 - (¢ b)2
20.833By
EIb
Is =
=
20.833By
13200
EIs
EIs
p
(0.5)4 = 0.04909 in4
4
20.833By (0.04909)
500
= 13200 - 20.833By
By = 634 lb c
Ans.
Form the free-body diagram,
A y = 243 lb c
Ans.
Cy = 76.8 lb T
Ans.
Ans:
By = 634 lb, Ay = 243 lb, Cy = 76.8 lb
1284
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–132. The assembly consists of three simply supported
beams for which the bottom of the top beam rests on the
top of the bottom two. If a uniform load of 3 kN> m is
applied to the beam, determine the vertical reactions at
each of the supports. EI is constant.
3 kN/m
B
D
F
A
3m
3m
da = da¿
da = d1 + d2 + d3
d1 =
w(L>3)
((L>3)3 - 2L(L>3)2 + L3)
24EI
Set L = 8 m, w = 3 kN>m
d1 =
d2 =
Set L = 8 m ,
P A 13L B A 13L B
6EI(L)
2
2
1
1
a L3 - a L b - a L b b
3
3
P = R
d1 =
d1 =
Set L = 8 m,
139.062
T
EI
7.374R
c
EI
P A 23L B A 13L B
6EIL
2
2
2
1
a L3 - a L b - a Lb b
3
3
P = R
d1 =
8.428 R
c
EI
Thus,
R(6)3
7.374 R
8.428 R
139.062
=
EI
EI
EI
48 EI
R = 6.849 ksi
Thus,
Ay = By = 9 - 6.849 = 2.15 kN
Ans.
1285
H
C
G
E
2m
2m
2m
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–133. The shaft supports the two pulley loads shown.
Using discontinuity functions, determine the equation of
the elastic curve. The bearings at A and B exert only vertical
reactions on the shaft. EI is constant.
x
A
12 in.
B
12 in.
36 in.
70 lb
180 lb
M = - 1808x - 09 - ( - 277.5)8x - 129 - 708x - 249
M = - 180x + 277.58x - 129 - 708x - 249
Elastic Curve and Slope:
EI
d2v
= M = - 180x + 277.58x - 129 - 708x - 249
dx2
EI
dv
= -90x2 + 138.758x - 1292 - 358x - 2492 + C1
dx
EIv = - 30x3 + 46.258x - 1293 - 11.678x - 2493 + C1x + C2
(1)
Boundary Conditions:
v= 0
at
x = 12 in.
From Eq. (1)
0 = - 51,840 + 12C1 + C2
12C1 + C2 = 51 840
v = 0
at
(2)
x = 60 in.
From Eq.(1)
0 = - 6 480 000 + 5 114 880 - 544 320 + 60C1 + C2
60C1 + C2 = 1909440
(3)
Solving Eqs. (2) and (3) yields:
C1 = 38 700
v =
C2 = - 412 560
1
[ - 30x3 + 46.258x - 1293 - 11.78x - 2493
EI
+ 38 700x - 412 560] lb # in3
Ans.
Ans:
1
( -30x3 + 46.25 8x - 1239
EI
- 11.7 8x - 2493 + 38,700x - 412,560) lb # in3
v =
1286
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–134. The shaft is supported by a journal bearing at A,
which exerts only vertical reactions on the shaft, and by a
thrust bearing at B, which exerts both horizontal and
vertical reactions on the shaft. Draw the bending-moment
diagram for the shaft and then, from this diagram, sketch
the deflection or elastic curve for the shaft’s centerline.
Determine the equations of the elastic curve using the
coordinates x1 and x2. EI is constant. Use the method of
integration.
80 lb
A
B
4 in.
x1
80 lb
12 in.
4 in.
x2
12 in.
For M1 (x) = 26.67 x1
EI
d2v1
dx21
= 26.67x1
dv1
= 13.33x21 + C1
dx1
(1)
EIv1 = 4.44x31 + C1x1 + C2
(2)
EI
For M2 (x) = - 26.67x2
EI
d2v2
dx22
= - 26.67x2
dv2
= - 13.33x22 + C3
dx2
(3)
EIv2 = - 4.44x32 + C3x2 + C4
(4)
EI
Boundary Conditions:
v1 = 0
at
x1 = 0
at
x2 = 0
From Eq. (2)
C2 = 0
v2 = 0
C4 = 0
Continuity Conditions:
dv1
dv2
= dx1
dx2
at
x1 = x2 = 12
From Eqs. (1) and (3)
1920 + C1 = - (- 1920 + C3)
C1 = - C3
v1 = v2
(5)
x1 = x2 = 12
at
7680 + 12C1 = - 7680 + 12C3
C3 - C1 = 1280
(6)
Solving Eqs. (5) and (6) yields:
C3 = 640
C1 = - 640
v1 =
1
A 4.44x31 - 640x1 B lb # in3
EI
Ans
v2 =
1
A - 4.44x32 + 640x2 B lb # in3
EI
Ans.
Ans:
1
(4.44x31 - 640x1) lb # in3,
EI
1
v2 =
( -4.44x32 + 640x2) lb # in3
EI
v1 =
1287
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–135. Determine the moment reactions at the supports
A and B. Use the method of integration. EI is constant.
w0
A
B
L
Support Reactions: FBD(a).
w0L
= 0
2
+ c ©Fy = 0;
Ay + By -
a+ ©MA = 0;
ByL + MA - MB -
[1]
w0L L
a b = 0
2
3
[2]
Moment Function: FBD(b).
a+ ©MNA = 0;
- M(x) -
x
1 w0
a x b x a b - MB + Byx = 0
2 L
3
M(x) = Byx -
w0 3
x - MB
6L
Slope and Elastic Curve:
EI
EI
EI
EI y =
d2y
= M(x)
dx2
w0 3
d2y
= By x x - MB
6L
dx2
By
w0 4
dy
=
x2 x - MBx + C1
dx
2
24L
By
6
[3]
w0 5
MB 2
x x + C1x + C2
120L
2
x3 -
[4]
Boundary Conditions:
At x = 0,
dy
= 0
dx
From Eq. [3], C1 = 0
At x = 0, y = 0.
At x = L,
0 =
From Eq. [4],
dy
= 0.
dx
By L2
2
-
C2 = 0
From Eq. [3].
w0L3
- MBL
24
0 = 12By L - w0 L2 - 24MB
At x = L, y = 0.
0 =
By L3
6
-
[5]
From Eq. [4],
w0 L4
MB L2
120
2
0 = 20By L - w0 L2 - 60MB
[6]
1288
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–135.
Continued
Solving Eqs. [5] and [6] yields,
MB =
w0 L2
30
By =
3w0L
20
Ans.
Substituting By and MB into Eqs. [1] and [2] yields,
MA =
w0L2
20
Ay =
7w0 L
20
Ans.
(a)
(b)
Ans:
MB =
1289
w0L2
w0L2
, MA =
30
20
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
w
*12–136. Determine the equations of the elastic curve for
the beam using the x1 and x2 coordinates. Specify the slope
at A and the maximum deflection. EI is constant. Use the
method of integration.
B
C
A
x1
x2
L
Elastic Curve and Slope:
EI
d2v
= M(x)
dx2
- wx11
2
For M1(x) =
EI
EI
d2v1
- wx21
2
=
dx21
dv1
- wx31
+ C1
=
dx1
6
(1)
- wx41
+ C1x1 + C2
24
- wLx2
For M2(x) =
2
EIv1 =
EI
EI
d2v2
dx22
(2)
-wLx2
2
=
dv2
- wLx32
+ C3
=
dx2
4
EIv2 =
(3)
- wLx32
+ C3x2 + C4
12
(4)
Boundary Conditions:
v2 = 0
at
x2 = 0
at
x2 = L
From Eq. (4):
C4 = 0
v2 = 0
From Eq. (4):
0 =
- wL4
+ C3L
12
C3 =
wL3
12
v1 = 0
at
x1 = L
From Eq. (2):
0 = -
wL4
+ C1L + C2
24
(5)
1290
L
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–136.
Continued
Continuity Conditions:
dv1
dv2
=
dx1
- dx2
x1 = x2 = L
at
From Eqs. (1) and (3)
-
wL3
wL3
wL3
+ C1 = - a+
b
6
4
12
C1 =
wL3
3
Substitute C1 into Eq. (5)
C2 = -
7wL4
24
dv1
w
=
(2L3 - x31)
dx1
6EI
dv2
w
=
(L3 - 3Lx22)
dx2
12EI
(6)
uA =
dv1
dv2
wL3
`
= `
=
dx1 x1 = L
dx2 x2 = L
6EI
Ans.
v1 =
w
(- x41 + BL3x1 - 7L4)
24EI
Ans.
(v1)max =
-7wL4
24EI
(x1 = 0)
The negative sign indicates downward displacement.
v2 =
wL
(L2x2 - x32)
12EI
(v2)max occurs when
(7)
Ans.
dv2
= 0
dx2
From Eq. (6)
L3 - 3Lx22 = 0
x2 =
L
23
Substitute x2 into Eq. (7),
(v2)max =
wL4
(8)
18 23EI
vmax = (v1)max =
7wL4
24EI
(9)
Ans.
1291
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–137. Determine the maximum deflection between the
supports A and B. EI is constant. Use the method of
integration.
w
B
C
A
x1
x2
L
Elastic Curve and Slope:
EI
d2v
= M(x)
dx2
For M1(x) =
EI
EI
- wx21
2
d2v1
=
dx21
dv1
- wx31
+ C1
=
dx1
6
EIv1 =
For M2(x) =
EI
EI
-wx21
2
(1)
-wx41
+ C1x1 + C2
24
(2)
- wLx2
2
d2v2
dx22
=
-wLx2
2
dv2
- wLx22
=
+ C3
dx2
4
EIv2 =
(3)
-wLx32
+ C3x2 + C4
12
(4)
Boundary Conditions:
v2 = 0
at
x2 = 0
at
x2 = L
From Eq. (4):
C4 = 0
v2 = 0
From Eq. (4):
0 =
-wL4
+ C3L
12
C3 =
wL3
12
v1 = 0
at
x1 = L
From Eq. (2):
0 = -
wL4
+ C1L + C2
24
(5)
1292
L
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–137.
Continued
Continuity Conditions:
dv1
dv2
=
dx1
- dx2
x1 = x2 = L
at
From Eqs. (1) and (3)
-
wL3
wL3
wL3
+ C1 = - ab
+
6
4
12
C1 =
wL3
3
Substitute C1 into Eq. (5)
C2 = -
7wL4
24
dv1
w
=
(2L3 - x31)
dx1
6EI
dv2
w
=
(L3 - 3Lx22)
dx2
12EI
uA =
dv1
dv2
wL3
`
= `
=
dx1 x1 = L
dx2 x2 = L
6EI
v1 =
w
( -x41 + 8L3x1 - 7L4)
24EI
(v1)max =
-7wL4
24EI
(6)
(x1 = 0)
The negative sign indicates downward displacement.
v2 =
wL
(L2x2 - x32)
12EI
(v2)max occurs when
(7)
dv2
= 0
dx2
From Eq. (6)
L3 - 3Lx22 = 0
x2 =
L
23
Substitute x2 into Eq. (7),
(v2)max =
wL4
Ans.
18 23EI
Ans:
(v2)max =
1293
wL4
1823EI
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–138. If the bearings at A and B exert only vertical
reactions on the shaft, determine the slope at B and the
deflection at C. Use the moment-area theorems.
P
A
B
a
a
C
a
Support Reaction and Elastic Curve: As shown.
M/ EI Diagram: As shown.
Moment-Area Theorems:
uB>D =
Pa2
1 Pa
a
b (a) =
2 2EI
4EI
Due to symmetry, the slope at point D is zero. Hence, the slope at B is
uB = |uB>D| =
Pa2
4EI
Ans.
The displacement at C is
¢ C = uB LBC =
Pa2
Pa3
c
(a) =
4EI
4EI
Ans.
Ans:
uB =
1294
Pa2
Pa3
c
, ¢C =
4EI
4EI
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–139. The bearing supports A, B, and C exert only
vertical reactions on the shaft. Determine these reactions,
then draw the shear and moment diagrams. EI is constant.
Use the moment-area theorems.
B
A
1m
1m
C
2m
200 N
(tB>A)1 =
1 3PL L L
L
1 PL
L L
5PL3
PL L L
a
a ba b =
ba ba + b + a
ba ba b +
2 8EI
2
2
6
2 8EI
2
3
4EI 2
4
48EI
(tC>A)1 =
1 3PL L 3L
L
1 3PL 3L
7PL3
a
ba ba
+ b + a
ba
b(L) =
2 8EI
2
2
6
2 8EI
2
16EI
(tB>A)2 =
- ByL3
L
1 - ByL
a
b (L) a b =
2 2EI
3
12EI
(tC>A)2 =
-ByL3
1 -ByL
a
b (2L)(L) =
2 2EI
2EI
2tB>A = tC>A
2[(tB>A)1 + (tB>A)2] = (tC>A)1 + (tC>A)2
2c
- ByL3
- ByL3
7PL3
5PL3
+ a
bd =
+ a
b
48EI
12EI
16EI
2EI
By =
11
P
16
Thus,
By =
11
(200) = 138 N c
16
Ans.
As shown on the free-body diagram
Ay = 81.3 N c
Ans.
Cy = 18.8 N T
Ans.
Ans:
By = 138 N, Ay = 81.3 N, Cy = 18.8 N
1295
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
w
*12–140. Using the method of superposition, determine
the magnitude of M0 in terms of the distributed load w and
dimension a so that the deflection at the center of the beam
is zero. EI is constant.
M0
M0
a
(¢ C)1 =
5wa4
T
384EI
(¢ C)2 = (¢ C)3 =
M0a2
c
16EI
¢ C = 0 = (¢ C)1 + (¢ C)2 + (¢ C)3
+c
0 =
M0a2
-5wa4
+
384EI
8EI
M0 =
5wa2
48
Ans.
1296
a
a
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–141. Using the method of superposition, determine the
deflection at C of beam AB. The beams are made of wood
having a modulus of elasticity of E = 1.5(103) ksi.
100 lb/ ft
a
A
B
C
D
E
a
4 ft
a
4 ft
6 ft
a
6 ft
3 in.
6 in.
Section a – a
Support Reactions: The reaction at B is shown on the free-body diagram of
beam AB, Fig. a.
Method of superposition. Referring to Fig. b and the table in the appendix, the
deflection of point B is
600 A 8 B
PLDE 3
6400 lb # ft3
=
=
T
48EI
48EI
EI
3
¢B =
Subsequently, referring to Fig. c,
(¢ C)1 = ¢ B a
6
6400 6
3200 lb # ft3
b =
a b =
T
12
EI 12
EI
5(100) A 12 B
27000 lb # ft3
5wL4
=
=
T
384EI
384EI
EI
4
(¢ C)2 =
Thus, the deflection of point C is
A+TB
¢ C = (¢ C)1 + (¢ C)2
=
3200
27000
+
EI
EI
=
30200 lb # ft3
=
EI
30200 A 12 3 B
1.5 A 106 B c
1
(3) A 63 B d
12
= 0.644 in T
Ans.
Ans:
¢ C = 0.644 in T
1297
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–142. The rim on the flywheel has a thickness t, width b,
and specific weight g. If the flywheel is rotating at a constant
rate of v, determine the maximum moment developed in the
rim. Assume that the spokes do not deform. Hint: Due to
symmetry of the loading, the slope of the rim at each spoke is
zero. Consider the radius to be sufficiently large so that the
segment AB can be considered as a straight beam fixed at
both ends and loaded with a uniform centrifugal force per
unit length. Show that this force is w = btgv2r>g.
A
t
B
v
r
Centrifugal Force: The centrifugal force acting on a unit length of the rim rotating at
a constant rate of v is
g
btgv2r
w = mv2 r = bt a b v2r =
g
g
(Q.E.D.)
Elastic Curve: Member AB of the rim is modeled as a straight beam with both of
its ends fixed and subjected to a uniform centrifugal force w.
Method of Superposition: Using the table in Appendix C, the required displacements are
uB ¿ =
wL3
6EI
yB ¿ =
wL4
c
8EI
uB – =
yB – =
MBL
EI
uB ¿– =
MBL2
c
2EI
yB – ¿ =
By L2
2EI
ByL3
3EI
T
Compatibility requires,
0 = uB ¿ + uB – + uB ¿–
0 =
By L2
M BL
wL3
+
+ ab
6EI
EI
2EI
0 = wL2 + 6MB - 3By L
(+ c)
(1)
0 = yB ¿ + yB – + yB – ¿
By L3
MB L2
wL4
0 =
+
+ ab
8EI
2EI
3EI
0 = 3wL2 + 12MB - 8By L
(2)
Solving Eqs. (1) and (2) yields,
By =
wL
2
MB =
wL2
12
Due to symmetry, Ay =
wL
2
MA =
wL2
12
Maximum Moment: From the moment diagram, the maximum moment occurs at
btgv2r
pr
the two fixed end supports. With w =
and L = ru =
.
g
3
wL2
=
Mmax =
12
A B
btgv2r gr 2
g
3
12
=
p2btgv2r3
108g
Ans.
Ans:
Mmax =
1298
p2brgv2r3
108g
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–1. Determine the critical buckling load for the column.
The material can be assumed rigid.
P
L
2
k
Equilibrium: The disturbing force F can be determined by summing moments
about point A.
a + ©MA = 0;
P(Lu) - F a
L
b = 0
2
L
2
A
F = 2Pu
Spring Formula: The restoring spring force F1 can be determine using spring
formula Fs = kx.
Fs = k a
L
kLu
ub =
2
2
Critical Buckling Load: For the mechanism to be on the verge of buckling, the
disturbing force F must be equal to the restoring spring force Fs .
2Pcr u =
kLu
2
Pcr =
kL
4
Ans.
Ans:
Pcr =
1299
kL
4
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–2. The column consists of a rigid membr that is pinned
at its bottom and attached to a spring at its top. If the spring
is unstretched when the column is in the vertical position,
determine the critical load that can be placed on the column.
P
k
B
L
A
a + ©MA = 0;
PL sin u - (kL sin u)(L cos u) = 0
P = kL cos u
Since u is small
cos u = 1
Pcr = kL
Ans.
Ans:
Pcr = kL
1300
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4 kip
13–3. The aircraft link is made from an A992 steel rod.
Determine the smallest diameter of the rod, to the nearest
1
16 in., that will support the load of 4 kip without buckling.
The ends are pin connected.
I =
4 kip
18 in.
p d 4 p d4
a b =
4 2
64
K = 1.0
Pcr =
p2 E I
(KL)2
p2(29)(103) A pd
64 B
4
4 =
((1.0)(18))2
d = 0.551 in.
Use d =
9
in.
16
Ans.
Check:
scr =
Pcr
4
= 16.2 ksi 6 sY
= p
2
A
(0.562
)
4
Therefore, Euler’s formula is valid.
Ans:
Use d =
1301
9
in.
16
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
*13–4. Rigid bars AB and BC are pin connected at B. If
the spring at D has a stiffness k, determine the critical load
Pcr for the system.
A
a
Equilibrium. The disturbing force F can be related P by considering the equilibrium
of joint A and then the equilibrium of member BC,
Joint A (Fig. b)
+ c ©Fy = 0;
B
a
FAB cos f - P = 0
FAB =
k
P
cos f
D
a
Member BC (Fig. c)
C
P
P
cos f (2a sin u) sin f(2a cos u) = 0
©MC = 0; F(a cos u) cos f
cos f
F = 2P(tan u + tan f)
Since u and f are small, tan u ⬵ u and tan f ⬵ f. Thus,
F = 2P(u + f)
(1)
Also, from the geometry shown in Fig. a,
2au = af
f = 2u
Thus Eq. (1) becomes
F = 2P(u + 2u) = 6Pu
Spring Force. The restoring spring force Fsp can be determined using the spring
formula, Fsp = kx, where x = au, Fig. a. Thus,
Fsp = kx = kau
Critical Buckling Load. When the mechanism is on the verge of buckling the
disturbing force F must be equal to the restoring spring force Fsp.
F = Fsp
6Pcru = kau
Pcr =
ka
6
Ans.
1302
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
s (ksi)
13–5. A rod made from polyurethane has a stress-strain
diagram in compression as shown. If the rod is pinned at its
ends and is 37 in. long, determine its smallest diameter so it
does not fail from elastic buckling.
8
0.003
E =
P (in./in.)
8(103)
s
=
= 2.667(106) psi
P
0.003
Pcr =
p2 EI
(KL)2
8(103)p(d> 2)2 =
p2(2.667)(106) A p4 BA d2 B 4
(1.0(37))2
d = 2.58 in.
Ans.
Ans:
d = 2.58 in.
1303
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
s (ksi)
13–6. A rod made from polyurethane has a stress-strain
diagram in compression as shown. If the rod is pinned at its
top and fixed at its base, and is 37 in. long, determine its
smallest diameter so it does not fail from elastic buckling.
8
0.003
E=
P (in./in.)
8(103)
= 2.667(106) psi
0.003
Pcr =
p2EI
(KL)2
8(103)p(d> 2)2 =
p2(2.667)(106) A p4 BA d2 B 4
[(0.7)(37)]2
d = 1.81 in.
Ans.
Ans:
d = 1.81 in.
1304
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–7. A 2014-T6 aluminum alloy hollow circular tube
has an outer diameter of 150 mm and inner diameter of
100 mm. If it is pinned at both ends, determine the largest
axial load that can be applied to the tube without causing
it to buckle. The tube is 6 m long.
Section Properties. The cross-sectional area and moment of inertia of the tube are
A = p(0.0752 - 0.052) = 3.125(10 - 3)p m2
I =
p
(0.0754 - 0.054) = 19.9418(10 - 6) m4
4
Critical Buckling Load. Applying Euler’s formula,
Pcr =
p2[73.1(109)][19.9418(10 - 6)]
p2EI
=
(KL)2
[1(6)]2
= 399.65 kN = 400 kN
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
Pcr
399.65
= 40.71 MPa 6 sY = 414 MPa
=
A
3.125(10 - 3)p
(O.K.)
Ans:
Pcr = 400 kN
1305
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–8. A 2014-T6 aluminum alloy hollow circular tube has
an outer diameter of 150 mm and inner diameter of 100 mm.
If it is pinned at one end and fixed at the other end,
determine the largest axial load that can be applied to the
tube without causing it to buckle. The tube is 6 m long.
Section Properties. The cross-sectional area and moment of inertia of the tube are
A = p(0.0752 - 0.052) = 3.125(10 - 3)p m2
I =
p
(0.0754 - 0.054) = 19.9418(10 - 6) m4
4
Critical Buckling Load. Applying Euler’s formula,
Pcr =
p2[73.1(109)][19.9418(10 - 6)]
p2EI
=
(KL)2
[0.7(6)]2
= 815.61 kN = 816 kN
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY .
scr =
815.61(103)
Pcr
= 83.08 MPa 6 sY = 414 MPa
=
A
3.125(10 - 3)p
1306
(O.K.)
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
0.932 in. 2.068 in.
d
0.932 in.
x
C
C
2.068 in.
13–9. A column is constructed using four A992 steel angles
that are laced together as shown. The length of the column is
to be 25 ft and the ends are assumed to be pin connected.
Each angle shown below has an area of A = 2.75 in2 and
moments of inertia of Ix = Iy = 2.22 in4. Determine the
distance d between the centroids C of the angles so that the
column can support an axial load of P = 350 kip without
buckling. Neglect the effect of the lacing.
x
d
C
C
y
d 2
Ix = Iy = 4 c 2.22 + 2.75 a b d = 8.88 + 2.75 d2
2
scr =
Pcr
350
=
= 31.8 ksi 6 sY
A
4(2.75)
OK
Therefore, Euler’s formula is valid.
Pcr =
350 =
p2E I
(K L)2
p2 (29)(103)(8.88 + 2.75 d2)
[1.0 (300)]2
d = 6.07 in.
Ans.
Check dimension:
d 7 2(2.068) = 4.136 in.
OK
Ans:
d = 6.07 in.
1307
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
0.932 in. 2.068 in.
d
0.932 in.
x
C
C
2.068 in.
13–10. A column is constructed using four A992 steel
angles that are laced together as shown.The length of the column
is to be 40 ft and the ends are assumed to be fixed connected.
Each angle shown below has an area of A = 2.75 in2
and moments of inertia of Ix = Iy = 2.22 in4. Determine the
distance d between the centroids C of the angles so that the
column can support an axial load of P = 350 kip without
buckling. Neglect the effect of the lacing.
x
d
C
C
y
d 2
Ix = Iy = 4 c 2.22 + 2.75a b d = 8.88 + 2.75 d2
2
scr =
Pcr
350
=
= 31.8 ksi 6 sY
A
4(2.75)
OK
Therefore, Euler’s formula is valid.
Pcr =
p2 E I
;
(K L)2
350 =
p2 (29)(103)(8.88 + 2.75 d2)
[0.5 (12)(40)]2
d = 4.73 in.
Ans.
Check dimension:
d 7 2 (2.068) = 4.136 in.
OK
Ans:
d = 4.73 in.
1308
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
13–11. The A992 steel angle has a cross-sectional area of
A = 2.48 in2 and a radius of gyration about the x axis of
rx = 1.26 in. and about the y axis of ry = 0.879 in. The
smallest radius of gyration occurs about the a–a axis and is
ra = 0.644 in. If the angle is to be used as a pin-connected
10-ft-long column, determine the largest axial load that can
be applied through its centroid C without causing it to buckle.
a
C
x
y
x
a
The Least Radius of Gyration:
r2 = 0.644 in.
scr =
p2E
2
A KL
r B
;
controls.
K = 1.0
p2 (29)(103)
(120) 2
C 1.00.644
D
= 8.243 ksi 6 sg
O.K.
Pcr = scr A = 8.243 (2.48) = 20.4 kip
Ans.
=
Ans:
Pcr = 20.4 kip
1309
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–12. The control linkage for a machine consists of
two L2 steel rods BE and FG, each with a diameter of 1 in.
If a device at G causes the end G to freeze up and become
pin connected, determine the maximum horizontal force
P that could be applied to the handle without causing
either of the two rods to buckle. The members are pin
connected at A, B, D, E, and F.
C
P
12 in.
E
4 in.
A
a + ©MA = 0; FBE (4) - P(16) = 0
FBE = 4P
a + ©MD = 0; FFG (6) - 4P(4) = 0
FFG = 2.6667 P
For rod BE,
4P =
p2 E I
;
(K L)2
K = 1.0
p2 (29)(103) A p4 B(0.54)
[1.0 (15)]2
P = 15.6 kip
Check Stress:
scr =
4(15.6)
Pcr
= p 2 = 79.5 ksi 6 sY = 102 ksi
A
4 (1 )
OK
For rod FG:
Pcr =
p2 E I
;
(K L)2
2.6667 P =
K = 1.0
p2 [(29)(103)] p4 (0.54)
P = 13.2 kip
[1.0 (20)]2
(controls)
Ans.
Check Stress:
scr =
2.6667 (13.2)
Pcr
= 44.7 ksi 6 sY = 102 ksi
=
p 2
A
4 (1 )
OK
Hence, Euler’s equation is still valid.
1310
2 in.
4 in.
D
15 in.
Pcr =
G
F
B
20 in.
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10 mm
13–13. An A992 steel column has a length of 5 m and is
fixed at both ends. If the cross-sectional area has the
dimensions shown, determine the critical load.
10 mm
50 mm
100 mm
I =
1
1
(0.1)(0.053) (0.08)(0.033) = 0.86167 (10 - 6) m4
12
12
Pcr =
p2(200)(109)(0.86167)(10 - 6)
p2EI
=
2
(KL)
[(0.5)(5)]2
= 272 138 N
= 272 kN
scr =
=
Pcr
;
A
Ans.
A = (0.1)(0.05) - (0.08)(0.03) = 2.6(10 - 3) m2
272 138
= 105 MPa 6 sg
2.6 (10 - 3)
Therefore, Euler’s formula is valid.
Ans:
Pcr = 272 kN
1311
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–14. The two steel channels are to be laced together
to form a 30-ft-long bridge column assumed to be pin
connected at its ends. Each channel has a cross-sectional
area of A = 3.10 in2 and moments of inertia Ix = 55.4 in4,
Iy = 0.382 in4 . The centroid C of its area is located in the
figure. Determine the proper distance d between the
centroids of the channels so that buckling occurs about
the x–x and y¿– y¿ axes due to the same load. What is the
value of this critical load? Neglect the effect of the lacing.
Est = 29(103) ksi, sy = 50 ksi.
y
y¿
0.269 in.
1.231 in.
x
C
C
x
d
y
y¿
Ix = 2(55.4) = 110.8 in.4
d 2
Iy = 2(0.382) + 2 (3.10)a b = 0.764 + 1.55 d2
2
In order for the column to buckle about x – x and y – y at the same time, Iy must be
equal to Ix.
Iy = Ix
0.764 + 1.55 d2 = 110.8
d = 8.43 in.
Ans.
Check:
d 7 2(1.231) = 2.462 in.
2
Pcr =
O.K.
3
p (29)(10 )(110.8)
p2 EI
=
2
(KL)
[1.0(360)]2
= 245 kip
Ans.
Check Stress:
scr =
Pcr
245
=
= 39.5 ksi 6 sg
A
2(3.10)
Therefore, Euler’s formula is valid.
Ans:
d = 8.43 in., Pcr = 245 kip
1312
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–15. An A992 steel W200 * 46 column of length 9 m
is pinned at both of its ends. Determine the allowable
axial load the column can support if F.S. = 2 is to be used
against buckling.
Section Properties. From the table listed in the appendix, the cross-sectional area
and moment of inertia about the y axis for a W200 * 46 are
A = 5890 mm2 = 5.89(10 - 3) m2
Iy = 15.3(106) mm4 = 15.3(10 - 6) m4
Critical Buckling Load. The column will buckle about the weak (y axis). Applying
Euler’s formula,
Pcr =
p2EIy
p2[200(109)][15.3(10 - 6)]
=
(KL)2
[1(9)]2
= 372.85 kN
Thus, the allowable centric load is
Pallow =
Pcr
372.85
=
= 186.43 kN = 186 kN
F.S.
2
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY .
scr =
372.85(103)
Pcr
= 63.30 MPa 6 sY = 345 MPa
=
A
5.89(10 - 3)
(O.K.)
Ans:
Pallow = 186 kN
1313
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–16. An A992 steel W200 * 46 column of length 9 m
is fixed at one end and free at its other end. Determine the
allowable axial load the column can support if F.S. = 2 is to
be used against buckling.
Section Properties. From the table listed in the appendix, the cross-sectional area
and moment of inertia about the y axis for a W200 * 46 are
A = 5890 mm2 = 5.89(10 - 3) m2
Iy = 15.3(106) mm4 = 15.3(10 - 6) m4
Critical Buckling Load. The column will buckle about the weak (y axis). Applying
Euler’s formula,
Pcr =
p2 E Iy
p2[200(109)][15.3(10 - 6)]
=
(KL)2
[2(9)]2
= 93.21 kN
Thus, the allowable centric load is
Pallow =
Pcr
93.21
=
= 46.61 kN = 46.6 kN
F.S.
2
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY .
scr =
93.21(103)
Pcr
= 15.83 MPa 6 sY = 345 MPa
=
A
5.89(10 - 3)
(O.K.)
1314
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–17. The 10-ft wooden rectangular column has the
dimensions shown. Determine the critical load if the ends are
assumed to be pin connected. Ew = 1.6(103) ksi, sY = 5 ksi.
10 ft
4 in.
2 in.
Section Properties:
A = 4(2) = 8.00 in2
Ix =
1
(2) A 43 B = 10.667 in4
12
Iy =
1
(4) A 23 B = 2.6667 in4 (Controls !)
12
Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s
formula,.
Pcr =
p2EI
(KL)2
p2(1.6)(103)(2.6667)
=
[1(10)(12)]2
Ans.
= 2.924 kip = 2.92 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
2.924
=
= 0.3655 ksi 6 sg = 5 ksi
A
8.00
O.K.
Ans:
Pcr = 2.92 kip
1315
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–18. The 10-ft wooden column has the dimensions
shown. Determine the critical load if the bottom is fixed
and the top is pinned. Ew = 1.6(103) ksi, sY = 5 ksi.
10 ft
4 in.
2 in.
Section Properties:
A = 4(2) = 8.00 in2
Ix =
1
(2) A 43 B = 10.667 in4
12
Iy =
1
(4) A 23 B = 2.6667 in4 (Controls!)
12
Critical Buckling Load: K = 0.7 for column with one end fixed and the other end
pinned. Applying Euler’s formula.
Pcr =
p2EI
(KL)2
p2 (1.6)(103)(2.6667)
=
[0.7(10)(12)]2
= 5.968 kip = 5.97 kip
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
5.968
=
= 0.7460 ksi 6 sg = 5 ksi
A
8.00
O.K.
Ans:
Pcr = 5.97 kip
1316
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–19. Determine the maximum force P that can be
applied to the handle so that the A992 steel control rod AB
does not buckle. The rod has a diameter of 1.25 in. It is pin
connected at its ends.
3 ft
2 ft
A
P
3 ft
B
a + ©MC = 0;
FAB (2) - P(3) = 0
P =
2
F
3 AB
(1)
Bucking Load for Rod AB:
I =
p
(0.6254) = 0.1198 in4
4
A = p(0.6252) = 1.2272 in2
Pcr =
p2EI
(KL)2
FAB = Pcr =
p2(29)(103)(0.1198)
[1.0(3)(12)]2
= 26.47 kip
From Eq. (1)
P =
2
(26.47) = 17.6 kip
3
Ans.
Check:
scr =
Pcr
26.47
=
= 21.6 ksi 6 sY
A
1.2272
OK
Therefore, Euler’s formula is valid.
Ans:
P = 17.6 kip
1317
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–20. The A992 steel tube has the cross-sectional area
shown. If it has a length of 15 ft and is pinned at both ends,
determine the maximum axial load that the tube can
support without causing it to buckle.
0.5 in.
3 in.
0.5 in.
3 in.
0.5 in.
Section Properties. The cross-sectional area and moment of inertia of the tube are
A = 4(4) - 3(3) = 7 in2
I =
1
1
(4)(43) (3)(33) = 14.5833 in4
12
12
Critical Buckling Load. Applying Euler’s formula,
Pcr =
p2[29(103)](14.5833)
p2EI
=
= 128.83 kip = 129 kip
(KL)2
[1(15)(12)]2
Critical Stress. Euler’s formula is valid only if scr 6 sY
scr =
Pcr
128.83
=
= 18.40 MPa 6 sY = 50 ksi
A
7
(O.K.)
1318
Ans.
0.5 in.
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–21. The A992 steel tube has the cross-sectional area
shown. If it has a length of 15 ft and is fixed at one end and
free at the other end, determine the maximum axial load
that the tube can support without causing it to buckle.
0.5 in.
3 in.
0.5 in.
3 in.
0.5 in.
0.5 in.
Section Properties. The cross-sectional area and moment of inertia of the tube are
A = 4(4) - 3(3) = 7 in2
I =
1
1
(4)(43) (3)(33) = 14.5833 in4
12
12
Critical Buckling Load. Applying Euler’s formula,
Pcr =
p2[29(103)](14.5833)
p2EI
=
= 32.21 kip = 32.2 kip
(KL)2
[2(15)(12)]2
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY .
scr =
Pcr
32.21
=
= 4.60 MPa 6 sY = 50 ksi
A
7
(O.K.)
Ans:
Pcr = 32.2 kip
1319
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
13–22. The linkage is made using two A992 steel rods, each
having a circular cross section. Determine the diameter of each
rod to the nearest 43 in. that will support a load of P = 6 kip.
Assume that the rods are pin connected at their ends. Use a
factor of safety with respect to buckling of 1.8.
B
p d 4
pd4
a b =
4 2
64
I =
12 ft
45⬚
A
30⬚
C
Joint B:
+
: ©Fx = 0;
FAB cos 45° - FBC sin 30° = 0
FAB = 0.7071 FBC
(1)
FAB sin 45° + FBC cos 30° - 6 = 0
+ c ©Fy = 0;
(2)
Solving Eqs. (1) and (2) yields:
FAB = 3.106 kip
FBC = 4.392 kip
For rod AB:
Pcr = 3.106 (1.8) = 5.591 kip
K = 1.0
LAB =
Pcr =
5. 591 =
12(12)
= 203.64 in.
cos 45°
p2EI
(KL)2
dAB4
p2(29)(103) a
b
64
[(1.0)(203.64)]2
dAB = 2.015 in.
1
Use dAB = 2 in.
8
Ans.
Check:
scr =
Pcr
5.591
= 1.58 ksi 6 sY
= p
2
A
4 (2.125 )
OK
For rod BC:
Pcr = 4.392 (1.8) = 7.9056 kip
K = 1.0
Pcr =
LBC =
12(12)
= 166.28 in.
cos 30°
p2EI
(KL)2
7.9056 =
p dBC4
p2(29)(103) a
b
64
[(1.0)(166.28)]2
dBC = 1.986 in.
Use dBC = 2 in.
Ans.
Check:
scr =
Pcr
7.9056
= p 2 = 2.52 ksi 6 sY
A
4 (2 )
Ans:
1
Use dAB = 2 in., dBC = 2 in.
8
OK
1320
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–23. The linkage is made using two A992 steel rods,
each having a circular cross section. If each rod has a
diameter of 34 in., determine the largest load it can support
without causing any rod to buckle. Assume that the rods are
pin connected at their ends.
P
B
12 ft
A
+
: ©Fx = 0;
FAB sin 45° - FBC sin 30° = 0
+ c ©Fy = 0;
FAB cos 45° + FBC cos 30° - P = 0
45⬚
30⬚
C
FAB = 0.5176 P
FBC = 0.73205 P
LAB =
12
= 16.971 ft
cos 45°
LBC =
12
= 13.856 ft
cos 30°
Assume Rod AB Buckles:
Pcr =
0.5176 P =
p2EI
(KL)2
p 3 4
p2(29)(106) a b a b
4
8
(1.0 (16.971)(12))2
P = 207 lb
scr =
(controls)
Pcr
207
=
= 469 psi 6 sY
A
p A3B2
Ans.
OK
8
Assume Rod BC Buckles:
0.73205 P =
p 3 4
p2(29)(106) a b a b
4
8
(1.0 (13.856)(12))2
P = 220 lb
Ans:
P = 207 lb
1321
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–24. An L-2 tool steel link in a forging machine is pin
connected to the forks at its ends as shown. Determine the
maximum load P it can carry without buckling. Use a factor
of safety with respect to buckling of F.S. = 1.75. Note from
the figure on the left that the ends are pinned for buckling,
whereas from the figure on the right the ends are fixed.
P
P
1.5 in.
0.5 in.
24 in.
Section Properties:
A = 1.5(0.5) = 0.750 in2
Ix =
1
(0.5) A 1.53 B = 0.140625 in4
12
Iy =
1
(1.5) A 0.53 B = 0.015625 in4
12
P
Critical Buckling Load: With respect to the x – x axis, K = 1 (column with both
ends pinned). Applying Euler’s formula,
Pcr =
p2EI
(KL)2
p2(29.0)(103)(0.140625)
=
[1(24)]2
= 69.88 kip
With respect to the y – y axis, K = 0.5 (column with both ends fixed).
Pcr =
p2EI
(KL)2
p2(29.0)(103)(0.015625)
=
[0.5(24)]2
= 31.06 kip
(Controls!)
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
31.06
=
= 41.41 ksi 6 sg = 102 ksi
A
0.75
O.K.
Factor of Safety:
F.S =
Pcr
P
1.75 =
31.06
P
P = 17.7 kip
Ans.
1322
P
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–25. The W14 * 30 is used as a structural A992 steel
column that can be assumed pinned at both of its ends.
Determine the largest axial force P that can be applied
without causing it to buckle.
P
25 ft
From the table in appendix, the cross-sectional area and the moment of inertia
about weak axis (y-axis) for W14 * 30 are
A = 8.85 in2
Iy = 19.6 in4
Critical Buckling Load: Since the column is pinned at its base and top, K = 1. For
A992 steel, E = 29.0(103) ksi and sg = 50 ksi . Here, the buckling occurs about the
weak axis (y-axis).
P = Pcr =
p2EIy
=
(KL)2
p2 C 29.0(103) D (19.6)
C 1(25)(12) D 2
Ans.
= 62.33 kip = 62.3 kip
Euler’s formula is valid only if scr 6 sg.
scr =
Pcr
62.33
=
= 7.04 ksi 6 sg = 50 ksi
A
8.85
O.K.
Ans:
P = 62.3 kip
1323
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–26. The A992 steel bar AB has a square cross section.
If it is pin connected at its ends, determine the maximum
allowable load P that can be applied to the frame. Use a
factor of safety with respect to buckling of 2.
C
A
1.5 in.
30⬚
B
1.5 in
1.5 in.
10 ft
P
a + ©MA = 0;
FBC sin 30°(10) - P(10) = 0
FBC = 2 P
+
: ©Fx = 0;
FA - 2P cos 30° = 0
FA = 1.732 P
Buckling Load:
Pcr = FA(F.S.) = 1.732 P(2) = 3.464 P
L = 10(12) = 120 in.
I =
1
(1.5)(1.5)3 = 0.421875 in4
12
Pcr =
p2 EI
(KL)2
3.464 P =
p2 (29)(103)(0.421875)
[(1.0)(120)]2
P = 2.42 kip
Ans.
Pcr = FA(F.S.) = 1.732(2.42)(2) = 8.38 kip
Check:
scr =
Pcr
8.38
=
= 3.72 ksi 6 sg
A
1.5 (1.5)
O.K.
Ans:
P = 2.42 kip
1324
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–27. The strongback BC is made of an A992 steel hollow
circular section with do = 60 mm and di = 40 mm. Determine
the allowable maximum lifting force P without causing the
strong back to buckle. F.S. = 2 against buckling is desired.
P
A
do
45⬚
a
45⬚
B
C
a
di
Section a – a
4m
D
E
Equilibrium. The compressive force developed in the strongback can be determined
by analyzing the equilibrium of joint A followed by joint B.
Joint A (Fig. a)
+
: ©Fx = 0;
FAC cos 45° - FAB cos 45° = 0
FAC = FAB = F
+ c ©Fy = 0;
P - 2F sin 45° = 0
FAB = FAC = F = 0.7071 P
0.7071P cos 45° - FBC = 0
FBC = 0.5P
Joint B (Fig. b)
+
: ©Fx = 0;
Section Properties. The cross-sectional area and moment of inertia are
A = p(0.032 - 0.022) = 0.5(10 - 3)p m2
I =
p
(0.034 - 0.024) = 0.1625(10 - 6)p m4
4
Critical Buckling Load. Both ends can be considered as pin connections. Thus,
K = 1. The critical buckling load is
Pcr = FBC (F.S.) = 0.5P(2) = P
Applying Euler’s formula,
Pcr =
P =
p2EI
(KL)2
p2[200(109)][0.1625(10 - 6)p]
[1(4)]2
P = 62.98 kN = 63.0 kN
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
62.98(103)
Pcr
= 40.10 MPa 6 sY = 345 MPa
=
A
0.5(10 - 3)p
(O.K.)
Ans:
P = 63.0 kN
1325
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–28. The strongback is made of an A992 steel hollow
circular section with the outer diameter of do = 60 mm. If it is
designed to withstand the lifting force of P = 60 kN, determine
the minimum required wall thickness of the strong back so
that it will not buckle. Use F.S. = 2 against buckling.
P
A
do
45⬚
a
45⬚
B
C
a
di
Section a – a
4m
D
Equilibrium. The compressive force developed in the strongback can be determined
by analyzing the equilibrium of joint A followed by joint B.
Joint A (Fig. a)
+
: ©Fx = 0;
FAC cos 45° - FAB cos 45° = 0 FAC = FAB = F
+ c ©Fy = 0;
60 - 2F sin 45° = 0
FAB = FAC = F = 42.43 kN (T)
42.43 cos 45° - FBC = 0
FBC = 30 kN (C)
Joint B (Fig. b)
+
: ©Fx = 0;
Section Properties. The cross-sectional area and moment of inertia are
A =
p
(0.062 - di 2)
4
I =
di 4
p
p
c 0.034 - a b d =
(0.064 - di 4 )
4
2
64
Critical Buckling Load. Both ends can be considered as pin connections. Thus,
K = 1. The critical buckling load is
Pcr = FBC (F.S.) = 30(2) = 60 kN
Applying Euler’s formula,
Pcr =
p2EI
(KL)2
60(103) =
p2[200(109)] c
p
(0.064 - di 4d
64
[1(4)]2
di = 0.04180 m = 41.80 mm
Thus, t =
da - di
60 - 41.80
=
= 9.10 mm
2
2
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY .
scr =
60(103)
Pcr
=
= 41.23 MPa 6 sY = 345 MPa
p
A
2
2
(0.06 - 0.04180 )
4
1326
(O.K.)
E
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–29. The beam supports the load of P = 6 kip. As a result,
the A992 steel member BC is subjected to a compressive load.
Due to the forked ends on the member, consider the supports
at B and C to act as pins for x–x axis buckling and as fixed
supports for y–y axis buckling. Determine the factor of safety
with respect to buckling about each of these axes.
P
4 ft
4 ft
A
B
3 ft
C x
y
a + ©MA = 0;
y
3 in.
1 in.
x
3
FBC a b (4) - 6000(8) = 0
5
FBC = 20 kip
x –x axis Buckling:
Pcr =
1
)(1)(3)3
p2(29)(103)(12
p2EI
=
= 178.9 kip
(KL)2
(1.0(5)(12))2
F.S. =
178.9
= 8.94
20
Ans.
y – y axis Buckling:
Pcr =
1
)(3)(1)3
p2 (29)(103)(12
p2EI
=
= 79.51
(KL)2
(0.5(5)(12))2
F.S. =
79.51
= 3.98
20
Ans.
Ans:
x- x axis buckling: F.S = 8.94
y -y axis buckling: F.S = 3.98
1327
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–30. Determine the greatest load P the beam will support
without causing the A992 steel member BC to buckle. Due to
the forked ends on the member, consider the supports at
B and C to act as pins for x–x axis buckling and as fixed
supports for y–y axis buckling.
P
4 ft
4 ft
A
3 ft
B
y
3 in.
C x
y
1 in.
x
3
FBC a b (4) - P(8) = 0
5
a + ©MA = 0;
FBC = 3.33 P
x –x axis Buckling:
2
Pcr =
3
1
3
3
1
3
p (29)(10 )(12)(1)(3)
p2EI
=
= 178.9 kip
(KL)2
(1.0(5)(12))2
y –y axis Buckling:
2
Pcr =
p (29)(10 )(12)(3)(1)
p2EI
=
= 79.51 kip
(KL)2
(0.5(5)(12))2
Thus,
3.33 P = 79.51
P = 23.9 kip
Ans.
Ans:
P = 23.9 kip
1328
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
w
13–31. The steel bar AB has a rectangular cross section. If
it is pin connected at its ends, determine the maximum
allowable intensity w of the distributed load that can be
applied to BC without causing bar AB to buckle. Use a
factor of safety with respect to buckling of 1.5. Est = 200 GPa,
sY = 360 MPa.
B
C
5m
y
3m
30 mm
x
20 mm
x
A
y
20 mm
Buckling Load:
Pcr = FAB (F.S.) = 2.5 w(1.5) = 3.75 w
I =
1
(0.03)(0.02)3 = 20 (10 - 9) m4
12
K = 1.0
Pcr =
p2 E I
(K L)2
3.75 w =
p2 (200)(109)(20)(10 - 9)
[(1.0)(3)]2
Ans.
w = 1170 N>m = 1.17 kN>m
Pcr = 4.39 kN
Check:
scr =
4.39 (103)
Pcr
=
= 7.31 MPa 6 sY
A
0.02 (0.03)
OK
Ans:
w = 1.17 kN>m
1329
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
*13–32. The frame supports the load of P = 4 kN. As a
result, the A992 steel member BC is subjected to a
compressive load. Due to the forked ends on this member,
consider the supports at B and C to act as pins for x–x axis
buckling and as fixed supports for y–y axis buckling.
Determine the factor of safety with respect to buckling
about each of these axes.
B
25 mm
1m
P
x
y
2m
35 mm
x
A
C
4m
a + ©MA = 0;
4
4(2) - FBC a b (3) = 0
5
FBC = 3.333 kN
x – x axis Buckling:
Pcr =
1
p2(200)(109) A 12
B (0.025)(0.035)3
p2 E I
=
= 7.053 kN
(K L)2
(1.0(5))2
F.S. =
7.053
= 2.12
3.333
Ans.
y – y axis Buckling:
Pcr =
1
p2(200)(109) A 12
B(0.035)(0.025)3
p2 E I
=
= 14.39 kN
(K L)2
(0.5(5))2
F.S. =
14.39
= 4.32
3.333
Ans.
1330
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
13–33. Determine the greatest load P the frame will
support without causing the A992 steel member BC to
buckle. Due to the forked ends on the member, consider the
supports at B and C to act as pins for x–x axis buckling and
as fixed supports for y–y axis buckling.
B
25 mm
1m
P
x
y
2m
35 mm
x
A
C
4m
4
P(2) - 3 a b FBC = 0
5
a + ©MA = 0;
FBC = 0.8333 P
x – x axis Buckling:
p (200)(10 ) A 12 B (0.025)(0.035)
p2EI
Pcr =
=
= 7.053 kN
(KL)2
(1.0(5))2
2
9
1
3
y – y axis Buckling:
1
p2(200)(109) A 12
B (0.035)(0.025)3
p2EI
Pcr =
=
= 14.39 kN
2
(KL)
(0.5(5))2
Thus,
0.8333 P = 7.053
P = 8.46 kN
Ans.
Ans:
P = 8.46 kN
1331
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–34. A 6061-T6 aluminum alloy solid circular rod of
length 4 m is pinned at both of its ends. If it is subjected to
an axial load of 15 kN and F.S. = 2 is required against
buckling, determine the minimum required diameter of the
rod to the nearest mm.
Section Properties. The cross-sectional area and moment of inertia of the solid rod are
A =
p 2
d
4
I =
p d 4
p 4
d
a b =
4 2
64
Critical Buckling Load. The critical buckling load is
Pcr = Pallow (F.S.) = 15(2) = 30 kN
Applying Euler’s formula,
Pcr =
p2EIy
(KL)2
30(103) =
p2 c 68.9(109) d c
p 4
d d
64
[1(4)]2
d = 0.06158 m = 61.58 mm
Use d = 62 mm
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY .
scr =
30(103)
Pcr
=
= 9.94 MPa 6 sY = 255 MPa
p
A
(0.0622)
4
(O.K.)
Ans:
Use d = 62 mm
1332
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–35. A 6061-T6 aluminum alloy solid circular rod of
length 4 m is pinned at one end while fixed at the other end.
If it is subjected to an axial load of 15 kN and F.S. = 2
is required against buckling, determine the minimum
required diameter of the rod to the nearest mm.
Section Properties. The cross-sectional area and moment of inertia of the solid rod are
A =
p 2
d
4
I =
p d 4
p 4
a b =
d
4 2
64
Critical Buckling Load. The critical buckling load is
Pcr = Pallow (F.S.) = 15(2) = 30 kN
Applying Euler’s formula,
Pcr =
p2EIy
(KL)2
30(103) =
p2 c 68.9(109) d c
p 4
d d
64
[0.7(4)]2
d = 0.05152 m = 51.52 mm
Use d = 52 mm
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY .
scr =
30(103)
Pcr
=
= 14.13 MPa 6 sY = 255 MPa
p
A
(0.0522)
4
(O.K.)
Ans:
Use d = 52 mm
1333
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–36. The members of the truss are assumed to be pin
connected. If member BD is an A992 steel rod of radius
2 in., determine the maximum load P that can be supported
by the truss without causing the member to buckle.
B
D
C
F
12 ft
A
G
16 ft
P
a + ©MC = 0;
FBD (12) - P(16) = 0
FBD =
4
P
3
Buckling Load:
A = p(22) = 4p in2
I =
p 4
(2 ) = 4p in4
4
L = 16(12) = 192 in.
K = 1.0
Pcr =
p2 EI
(KL)2
FBD =
p2(29)(103)(4p)
4
P =
3
[(1.0)(192)]2
P = 73.2 kip
Ans.
Pcr = FBD = 97.56 kip
Check:
scr =
Pcr
97.56
=
= 7.76 ksi 6 sY
A
4p
OK
1334
16 ft
16 ft
P
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–37. Solve Prob. 13–36 in the case of member AB, which
has a radius of 2 in.
B
D
C
F
12 ft
A
G
16 ft
P
P -
+ c ©Fy = 0;
16 ft
16 ft
P
3
F = 0
5 AB
FAB = 1.667 P
Buckling Load:
A = p(2)2 = 4p in2
I =
p 4
(2) = 4p in4
4
L = 20(12) = 240 in.
K = 1.0
Pcr =
p2(29)(103)(4p)
p2EI
= 62.443 kip
=
(KL)2
(1.0(240))2
Pcr = FAB = 1.667 P = 62.443
P = 37.5 kip
Ans.
Check:
scr =
62.443
P
=
= 4.97 ksi 6 sY
A
4p
OK
Ans:
P = 37.5 kip
1335
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–38. The truss is made from A992 steel bars, each of
which has a circular cross section with a diameter of
1.5 in. Determine the maximum force P that can be applied
without causing any of the members to buckle. The members
are pin connected at their ends.
C
4 ft
D
3 ft
B
A
4 ft
I =
4 ft
P
p
(0.754) = 0.2485 in4
4
A = p(0.752) = 1.7671 in2
Members AB and BC are in compression:
Joint A:
+ c ©Fy = 0;
3
F - P = 0
5 AC
FAC =
+
; ©Fx = 0;
FAB -
5P
3
4 5P
a
b = 0
5 3
FAB =
4P
3
Joint B:
+
: ©Fx = 0;
4
8P
4P
= 0
FBC +
5
3
3
FBC =
5P
3
Failure of rod AB:
K = 1.0
L = 8(12) = 96 in.
Pcr =
p2EI
(KL)2
FAB =
p2(29)(103)(0.2485)
4P
=
3
((1.0)(96))2
P = 5.79 kip (controls)
Ans.
Check:
Pcr = FAB = 7.72 kip
scr =
Pcr
7.72
=
= 4.37 ksi 6 sY
A
1.7671
OK
Failure of rod BC:
K = 1.0
FBC =
L = 5(12) = 60 in.
p2(29)(103)(0.2485)
5P
=
3
[(1.0)(60)]2
Ans:
P = 5.79 kip
P = 11.9 kip
1336
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–39. The truss is made from A992 steel bars, each of
which has a circular cross section. If the applied load
P = 10 kip, determine the diameter of member AB to
the nearest 18 in. that will prevent this member from buckling.
The members are pin connected at their ends.
C
4 ft
D
3 ft
B
A
4 ft
4 ft
P
Joint A:
+ c ©Fy = 0;
3
- 10 + FAC a b = 0;
5
FAC = 16.667 kip
+
: ©Fx = 0;
4
-FAB + 16.667a b = 0;
5
FAB = 13.33 kip
Pcr =
13.33 =
p2EI
(KL)2
p
p2(29)(103) a b (r)4
4
(1.0(8)(12))2
r = 0.8599 in.
d = 2r = 1.72 in.
Use:
3
d = 1 in.
4
Ans.
Check:
scr =
Pcr
13.33
=
= 5.54 ksi 6 sY
p
A
(1.75)2
4
OK
Ans:
3
Use d = 1 in.
4
1337
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–40. The steel bar AB of the frame is assumed to be
pin connected at its ends for y–y axis buckling. If P = 18 kN,
determine the factor of safety with respect to buckling
about the y–y axis due to the applied loading. Est = 200 GPa,
sY = 300 MPa.
3m
A
P
50 mm
C
1
(0.10)(0.053) = 1.04167 (10 - 6) m4
12
B
Joint A:
+
; ©Fx = 0;
3
F - 18 = 0
5 AC
FAC = 30 kN
+ c ©Fy = 0;
FAB -
4
(30) = 0
5
FAB = 24 kN
Pcr =
p2(200)(109)(1.04167)(10 - 6)
p2E I
=
= 57116 N = 57.12 kN
2
(K L)
[(1.0)(6)]2
F.S. =
Pcr
57.12
=
= 2.38
FAB
24
Ans.
57.12 (103)
Pcr
=
= 11.4 MPa 6 sY
A
0.1 (0.05)
OK
Check:
scr =
50 mm
x
50 mm
x
4m
6m
Iy =
y
1338
y
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–41. The ideal column has a weight w (force> length) and
rests in the horizontal position when it is subjected to the axial
load P. Determine the maximum moment in the column at
midspan. EI is constant. Hint: Establish the differential
equation for deflection, Eq. 13–1, with the origin at the mid
span. The general solution is v = C1 sin kx + C2 cos kx +
(w>(2P))x2 - (wL>(2P))x - (wEI>P2) where k2 = P>EI.
w
P
L
Moment Functions: FBD(b).
a + ©Mo = 0;
M(x) =
x
wL
wx a b - M(x) - a
b x - Pv = 0
2
2
w 2
A x - Lx B - Pv
2
[1]
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
EI
d2y
w 2
=
A x - Lx B - Py
2
dx2
d2y
P
w
+
y =
A x2 - Lx B
2
EI
2EI
dx
The solution of the above differential equation is of the form
P
w 2
wL
wEI
P
xb +
x x x b + C2 cos ¢
A EI
2P
2P
A EI
P2
[2]
wL
w
dv
P
P
P
P
xx ≤ - C2
sin ¢
x≤ +
cos ¢
= C1
A EI
A EI
P
2P
A EI
A EI
dx
[3]
v = C1 sin a
and
The integration constants can be determined from the boundary conditions.
Boundary Condition:
At x = 0, y = 0. From Eq. [2],
0 = C2 -
wEI
P2
C2 =
wEI
P2
At x =
L dy
= 0. From Eq.[3],
,
2 dx
0 = C1
P
P L
wEI
P
P L
w L
wL
cos ¢
sin ¢
≤ ≤ + a b A EI 2
A EI 2
P 2
2P
A EI
P2 A EI
C1 =
wEI
P L
tan ¢
≤
A EI 2
P2
1339
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–41.
Continued
Elastic Curve:
y =
w EI
L
EI
EI
x2
P L
P
P
tan ¢
B
x≤ +
cos ¢
x≤ +
- x ≤ sin ¢
R
P P
A EI 2
A EI
P
A EI
2
2
P
However, y = ymax at x =
ymax =
=
L
. Then,
2
EI
w EI
EI
L2
P L
P L
P L
tan ¢
cos ¢
B
≤ sin ¢
≤ +
≤ R
P P
A EI 2
A EI 2
P
A EI 2
8
P
wEI
PL2
P L
sec ¢
- 1R
≤ 2 B
A EI 2
8EI
P
Maximum Moment: The maximum moment occurs at x =
Mmax =
L
. From, Eq.[1],
2
w L2
L
- L a b R - Pymax
B
2 4
2
= -
wL2
PL2
wEI
P L
- P b 2 B sec ¢
- 1R r
≤ 8
A EI 2
8EI
P
= -
P L
wEI
B sec ¢
≤ - 1R
P
A EI 2
Ans.
Ans:
Mmax = -
1340
wEI
L
P
csec a
b - 1d
P
2 A EI
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–42. The ideal column is subjected to the force F at its
midpoint and the axial load P. Determine the maximum
moment in the column at midspan. EI is constant. Hint:
Establish the differential equation for deflection, Eq. 13–1.
The general solution is v = C1 sin kx + C2 cos kx - c2x>k2,
where c2 = F>2EI, k2 = P>EI.
F
P
L
2
Moment Functions: FBD(b).
c + ©Mo = 0;
F
x + P(v) = 0
2
M(x) +
M(x) = -
F
x - Pv
2
[1]
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
EI
F
d2y
= - x - Py
2
dx2
d2y
P
F
+
y = x
EI
2EI
dx2
The solution of the above differential equation is of the form,
v = C1 sin a
P
P
F
x b + C2 cos ¢
xb x
A EI
2P
A EI
[2]
and
dv
P
P
P
P
F
= C1
cos ¢
x ≤ - C2
sin ¢
x≤ dx
A EI
A EI
A EI
A EI
2P
The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x = 0, y = 0. From Eq.[2], C2 = 0
At x =
L dy
= 0. From Eq.[3],
,
2 dx
0 = C1
F
P
P L
cos ¢
≤ A EI
A EI 2
2P
C1 =
F
EI
P L
sec ¢
≤
2P A P
A EI 2
Elastic Curve:
y =
F
F
EI
P L
P
sec ¢
x≤ x
≤ sin ¢
2P A P
A EI 2
A EI
2P
=
F
EI
P L
P
sec ¢
x≤ - xR
B
≤ sin ¢
2P A P
A EI 2
A EI
1341
[3]
L
2
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–42.
Continued
However, y = ymax at x =
L
. Then,
2
F
L
EI
P L
P L
sec ¢
B
≤ sin ¢
≤ - R
2P A P
A EI 2
A EI 2
2
ymax =
=
F
L
EI
P L
tan ¢
B
≤ - R
2P A P
A EI 2
2
Maximum Moment: The maximum moment occurs at x =
F L
a b - Pymax
2 2
Mmax = -
= -
L
. From Eq.[1],
2
F
FL
L
EI
P L
- Pb
tan ¢
B
≤ - Rr
4
2P A P
A EI 2
2
= -
F EI
P L
tan ¢
≤
2 AP
A EI 2
Ans.
Ans:
Mmax = -
1342
F EI
L
P
tan a
b
2 AP
2 A EI
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–43. The column with constant EI has the end constraints
shown. Determine the critical load for the column.
P
L
Moment Function. Referring to the free-body diagram of the upper part of the
deflected column, Fig. a,
a + ©MO = 0;
M + Pv = 0
M = - Pv
Differential Equation of the Elastic Curve.
EI
d2v
= M
dx2
EI
d2v
= - Pv
dx2
d2v
P
v = 0
+
EI
dx2
The solution is in the form of
v = C1 sin a
P
P
x b + C2 cos ¢
xb
A EI
A EI
(1)
dv
P
P
P
P
= C1
cos ¢
x ≤ - C2
sin ¢
x≤
dx
A EI
A EI
A EI
A EI
(2)
Boundary Conditions. At x = 0, v = 0. Then Eq. (1) gives
0 = 0 + C2
At x = L,
C2 = 0
dv
= 0. Then Eq. (2) gives
dx
0 = C1
P
P
cos ¢
L≤
A EI
A EI
C1 = 0 is the trivial solution, where v = 0. This means that the column will remain
straight and buckling will not occur regardless of the load P. Another possible
solution is
cos ¢
P
L≤ = 0
A EI
np
P
L =
A EI
2
n = 1, 3, 5
The smallest critical load occurs when n = 1, then
p
Pcr
L =
A EI
2
Pcr =
p2EI
4L2
Ans.
Ans:
Pcr =
1343
p2EI
4L2
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–44. Consider an ideal column as in Fig. 13–10c,
having both ends fixed. Show that the critical load on the
column is given by Pcr = 4p2EI>L2. Hint: Due to the
vertical deflection of the top of the column, a constant
moment M¿ will be developed at the supports. Show that
d2v>dx2 + (P>EI)v = M¿>EI. The solution is of the form
v = C1 sin ( 2P>EIx) + C2 cos ( 2P>EIx) + M¿>P.
Moment Functions:
M(x) = M¿ - Py
Differential Equation of The Elastic Curve:
EI
EI
d2y
= M(x)
dx2
d2y
= M¿ - Py
dx2
P
d2y
M¿
+
y =
EI
EI
dx2
(Q.E.D.)
The solution of the above differential equation is of the form
v = C1 sin a
P
P
M¿
x b + C2 cos ¢
xb +
A EI
A EI
P
[1]
and
P
P
dv
P
P
= C1
cos ¢
x ≤ - C2
sin ¢
x≤
A EI
A EI
A EI
A EI
dx
[2]
The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x = 0, y = 0. From Eq.[1], C2 = At x = 0,
M¿
P
dy
= 0. From Eq.[2], C1 = 0
dx
Elastic Curve:
y =
M¿
P
x≤ R
B 1 - cos ¢
P
A EI
and
dy
M¿
P
P
=
sin ¢
x≤
dx
P A EI
A EI
However, due to symmetry
sin B
L
dy
= 0 at x = . Then,
dx
2
P L
a bR = 0
A EI 2
or
P L
a b = np
A EI 2
where n = 1, 2, 3,...
The smallest critical load occurs when n = 1.
Pcr =
4p2EI
L2
(Q.E.D.)
1344
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–45. Consider an ideal column as in Fig. 13–10d, having
one end fixed and the other pinned. Show that the critical
load on the column is given by Pcr = 20.19EI>L2. Hint:
Due to the vertical deflection at the top of the column, a
constant moment M¿ will be developed at the fixed support
and horizontal reactive forces R¿ will be developed at both
supports.Show that d2v>dx2 + 1P>EI2v = 1R¿>EI21L - x2.
The solution is of the form v = C1 sin 11P>EIx2 +
C2 cos 1 1P>EIx2 + 1R¿>P21L - x2. After application of the
boundary conditions show that tan 11P>EIL2 = 1P>EI L.
Solve by trial and error for the smallest nonzero root.
Equilibrium. FBD(a).
Moment Functions: FBD(b).
M(x) = R¿(L - x) - Py
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
EI
d2y
= R¿(L - x) - Py
dx2
d2y
P
R¿
+
y =
(L - x)
EI
EI
dx2
(Q.E.D.)
The solution of the above differential equation is of the form
v = C1 sin a
R¿
P
P
(L - x)
x b + C2 cos ¢
xb +
A EI
P
A EI
[1]
and
dv
R¿
P
P
P
P
= C1
cos ¢
x ≤ - C2
sin ¢
x≤ dx
A EI
A EI
A EI
A EI
P
The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x = 0, y = 0. From Eq.[1], C2 = -
At x = 0,
R¿L
P
dy
R¿ EI
= 0. From Eq.[2], C1 =
dx
P AP
Elastic Curve:
y =
R¿ EI
R¿L
R¿
P
P
sin ¢
x≤ cos ¢
x≤ +
(L - x)
P AP
A EI
P
A EI
P
=
EI
P
P
R¿
sin ¢
x ≤ - L cos ¢
x ≤ + (L - x) R
B
P AP
A EI
A EI
1345
[2]
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–45.
Continued
However, y = 0 at x = L. Then,
0 =
P
P
EI
sin ¢
L ≤ - L cos ¢
L≤
A EI
A EI
AP
tan ¢
P
P
L≤ =
L
A EI
A EI
(Q.E.D.)
By trial and error and choosing the smallest root, we have
P
L = 4.49341
A EI
Then,
Pcr =
20.19EI
L2
(Q.E.D.)
1346
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
13–46. The wood column has a square cross section with
dimensions 100 mm by 100 mm. It is fixed at its base and free
at its top. Determine the load P that can be applied to the
edge of the column without causing the column to fail either
by buckling or by yielding. Ew = 12 GPa, sY = 55 MPa.
100 mm
120 mm
100 mm
2m
Section properties:
A = 0.1(0.1) = 0.01 m2
r =
I =
1
(0.1)(0.1)3 = 8.333(10 - 6) m4
12
I
8.333(10 - 6)
=
= 0.02887 m
A
0.01
AA
Buckling:
Pcr =
p2(12)(109)(8.333)(10 - 6)
p2EI
=
= 61.7 kN
2
(KL)
[2.0(2)]2
Check: scr =
61.7(103)
Pcr
=
= 6.17 MPa 6 sY
A
0.01
OK
Yielding:
smax =
ec
P
KL
P
c 1 + 2 sec a
bd
A
2r A EA
r
0.12(0.05)
ec
=
= 7.20
r2
(0.02887)2
2.0(2)
P
KL P
= 0.0063242P
=
2r A EA
2(0.02887) A 12(109)(0.01)
55(106)(0.01) = P[1 + 7.20 sec (0.006324 2P)]
By trial and error:
P = 31400 N = 31.4 kN
Ans.
controls
Ans:
P = 31.4 kN
1347
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–47. The hollow red brass C83400 copper alloy shaft is
fixed at one end but free at the other end. Determine the
maximum eccentric force P the shaft can support without
causing it to buckle or yield. Also, find the corresponding
maximum deflection of the shaft.
2m
a
a
P
150 mm
30 mm
Section Properties.
A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2
I =
20 mm
p
A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4
4
Section a – a
0.1625 A 10 B p
I
=
= 0.01803 m
C 0.5 A 10 - 3 B p
AA
-6
r =
e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus,
KL = 2(2) = 4 m
Yielding. In this case, yielding will occur before buckling. Applying the secant
formula,
smax =
P
ec
KL P
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
70.0 A 106 B =
70.0 A 106 B =
P
0.5 A 10 - 3 B p
P
0.5 A 10
-3
Bp
D1 +
0.15(0.03)
0.018032
secC
4
P
ST
2(0.01803)A 101 A 109 B C 0.5 A 10 - 3 B p D
a 1 + 13.846 sec 8.8078 A 10 - 3 B 2Pb
Solving by trial and error,
P = 5.8697 kN = 5.87 kN
Ans.
Maximum Deflection.
vmax = e B sec ¢
P KL
≤ - 1R
A EI 2
= 0.15 D sec C
5.8697 A 103 B
4
a b S - 1T
C 101 A 109 B C 0.1625 A 10 - 6 B p D 2
= 0.04210 m = 42.1 mm
Ans.
Ans:
P = 5.87 kN, vmax = 42.1 mm
1348
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–48. The hollow red brass C83400 copper alloy shaft is
fixed at one end but free at the other end. If the eccentric
force P = 5 kN is applied to the shaft as shown, determine
the maximum normal stress and the maximum deflection.
2m
a
a
P
150 mm
30 mm
Section Properties.
A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2
I =
20 mm
p
A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4
4
Section a – a
0.1625 A 10 B p
I
= 0.01803 m
=
C 0.5 A 10 - 3 B p
AA
-6
r =
e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus,
KL = 2(2) = 4 m
Yielding. Applying the secant formula,
smax =
=
P
ec
KL P
B 1 + 2 sec ¢
≤R
A
2r A EA
r
5 A 103 B
0.5 A 10 - 3 B p
D1 +
5 A 103 B
4
ST
secC
2(0.01803) C 101 A 109 B C 0.5 A 10 - 3 B p D
0.018032
0.15(0.03)
Ans.
= 57.44 MPa = 57.4 MPa
Since smax 6 sY = 70 MPa, the shaft does not yield.
Maximum Deflection.
P KL
vmax = e B sec ¢ A EI 2 ≤ - 1 R
= 0.15D sec C
5 A 103 B
4
a b S - 1T
C 101 A 109 B C 0.1625 A 10 - 6 B p D 2
= 0.03467 m = 34.7 mm
Ans.
1349
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–49. The tube is made of copper and has an outer diameter
of 35 mm and a wall thickness of 7 mm. Determine the
eccentric load P that it can support without failure.The tube is
pin supported at its ends. Ecu = 120 GPa, sY = 750 MPa.
2m
P
P
14 mm
Section Properties:
A =
p
(0.0352 - 0.0212) = 0.61575(10 - 3) m2
4
I =
p
(0.01754 - 0.01054) = 64.1152(10 - 9) m4
4
r =
I
64.1152(10 - 9)
=
= 0.010204 m
AA
A 0.61575(10 - 3)
For a column pinned at both ends, K = 1. Then KL = 1(2) = 2 m.
Buckling: Applying Euler’s formula,
Pmax = Pcr =
p2 (120)(109) C 64.1152(10 - 9) D
p2EI
=
= 18983.7 N = 18.98 kN
(KL)2
22
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
18983.7
= 30.83 MPa 6 sg = 750 MPa
=
A
0.61575(10 - 3)
O.K.
Yielding: Applying the secant formula,
smax =
(KL) Pmax
Pmax
ec
B 1 + 2 sec ¢
≤R
A
2r A EA
r
750 A 106 B =
0.61575(10 - 3)
750 A 106 B =
0.61575(10 - 3)
Pmax
Pmax
B1 +
0.014(0.0175)
0.0102042
sec ¢
Pmax
2
≤R
2(0.010204)A 120(109)[0.61575(10 - 3)]
A 1 + 2.35294 sec 0.01140062Pmax B
Solving by trial and error,
Ans.
Pmax = 16 884 N = 16.9 kN (Controls!)
Ans:
Pmax = 16.9 kN
1350
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–50. Solve Prob. 13–49 if instead the left end is free and
the right end is fixed-supported.
2m
P
P
14 mm
Section Properties:
A =
p
A 0.0352 - 0.0212 B = 0.61575 A 10 - 3 B m2
4
I =
p
A 0.01754 - 0.01054 B = 64.1152 A 10 - 9 B m4
4
r =
I
64.1152(10 - 9)
=
= 0.010204 ms
AA
A 0.61575(10 - 3)
For a column fixed at one end and free at the other, K = 2 . Then KL = 2(2) = 4 m.
Buckling: Applying Euler’s formula,
Pmax = Pcr =
p2(120)(109) C 64.1152(10 - 9) D
p2EI
=
= 4746 N = 4.75 kN
(KL)2
42
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
4746
= 7.71 MPa 6 sg = 750 MPa
=
A
0.61575(10 - 3)
O. K.
Yielding: Applying the secant formula,
smax =
(KL) Pmax
Pmax
ec
B 1 + 2 sec ¢
≤R
A
2r A EA
r
750 A 106 B =
0.61575(10 )
750 A 106 B =
0.61575(10 - 3)
Pmax
-3
Pmax
B1 +
0.014(0.0175)
0.0102042
sec ¢
4
Pmax
≤R
2(0.010204)A 120(109)[0.61575(10 - 3)]
A 1 + 2.35294 sec 22.801 A 10 - 3 B 2P B
Solving by trial and error,
Pmax = 4604 N = 4.60 kN (Controls!)
Ans.
Ans:
Pmax = 4.60 kN
1351
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–51. Assume that the wood column is pin connected at
its base and top. Determine the maximum eccentric load P
that can be applied without causing the column to buckle or
yield. Ew = 1.8(103) ksi, sY = 8 ksi.
P
y
4 in.
x
x
P
y 10 in.
10 ft
Section Properties:
A = 10(4) = 40 in2
ry =
Iy =
1
(4)(103) = 333.33 in4
12
Ix =
1
(10)(43) = 53.33 in4
12
Ix
333.33
=
= 2.8868 in.
AA
A 40
Buckling about y -y axis:
P = Pcr =
p2(1.8)(103)(333.33)
p2EI
=
= 102.8 kip
(KL)2
[(2)(10)(12)]2
Buckling about x -x axis:
P = Pcr =
p2(1.8)(103)(53.33)
p2EI
=
= 65.8 kip (controls)
(KL)2
[(1)(10)(12)]2
Check: scr =
Pcr
65.8
=
= 1.64 ksi 6 sY
A
40
Ans.
O.K.
Yielding about y- y axis:
smax =
P
ec
KL
P
a 1 + 2 seca
b
b
A
2r
A
EA
r
5(5)
ec
=
= 3.0
2
r
2.88682
a
(1)(10)(12)
P
P
KL
b
=
= 0.077460 2P
2r A EA
2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0774602P)]
By trial and error:
P = 67.6 kip
Ans:
smax = 65.8 kip
1352
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–52. Assume that the wood column is pinned top and
bottom for movement about the x - x axis, and fixed at the
bottom and free at the top for movement about the y -y axis.
Determine the maximum eccentric load P that can be
applied without causing the column to buckle or yield.
Ew = 1.8(103) ksi, sY = 8 ksi .
P
y
4 in.
x
y 10 in.
10 ft
Section Properties:
A = 10(4) = 40 in2
ry =
Iy =
1
(4)(103) = 333.33 in4
12
Ix =
1
(10)(43) = 53.33 in4
12
Iy
333.33
=
= 2.8868 in.
AA
A 40
Buckling about x -x axis:
P = Pcr =
p2(1.8)(103)(53.33)
p2EI
=
= 65.8 kip
(KL)2
[(1)(10)(12)]2
Check: scr =
Pcr
65.8
=
= 1.64 ksi 6 sY
A
40
O.K.
Yielding about y - y axis:
smax =
ec
P
KL P
a 1 + 2 seca
bb
A
2r A EA
r
5(5)
ec
=
= 3.0
2
r
2.88682
a
(2)(10)(12)
P
P
KL
b
=
= 0.15492 2P
2r A EA
2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.154922P)]
By trial and error:
P = 45.7 kip
x
P
Ans.
(controls)
1353
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–53. A W12 * 26 structural A992 steel column is pin
connected at its ends and has a length L = 11.5 ft. Determine
the maximum eccentric load P that can be applied so the
column does not buckle or yield. Compare this value with
an axial critical load P¿ applied through the centroid of
the column.
P
6 in.
L
Section properties for W12 * 26 :
A = 7.65 in2
Ix = 204 in4
rx = 5.17 in.
d = 12.22 in.
Iy = 17.3 in4
P
Buckling about y -y axis:
Pcr =
p2EI
(KL)2
Pcr = Pcr =
p2(29)(103)(17.3)
[1(11.5)(12)]2
= 260 kip
Pcr
260
Check: scr = A = 7.65 = 34.0 ksi 6 sY
OK
Yielding about x-x axis:
smax =
ec
P
KL P
bd
c 1 + 2 sec a
A
2r A EA
r
6A 2 B
ec
=
= 1.37155
2
r
5.172
12.22
1(11.5)(12)
KL P
P
=
= 0.028335 2P
3
2r A EA
2(5.17) A 29(10 )(7.65)
50(7.65) = P[1 + 1.37155 sec (0.028335 2P)]
By trial and error:
P = 155.78 = 156 kip
controls
Ans.
Ans:
P = 156 kip
1354
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–54. A W14 * 30 structural A-36 steel column is pin
connected at its ends and has a length L = 10 ft.
Determine the maximum eccentric load P that can be
applied so the column does not buckle or yield. Compare
this value with an axial critical load P¿ applied through the
centroid of the column.
P
6 in.
L
Section properties for W 14 : 30
A = 8.85 in2
Ix = 291 in4
d = 13.84 in.
rx = 5.73 in.
Iy = 19.6 in4
P
Buckling about y - y axis:
Pcr =
P¿ =
p2EI
(KL)2
K = 1
p2(29)(103)(19.6)
[1(10)(12)]2
Ans.
= 390 kip
Yielding about x - x axis:
smax =
P
ec
KL P
c 1 + 2 sec a
bd
A
2 r A EA
r
6A 2 B
1(10)(12)
P
P
c1 +
sec a
bd
2
8.85
2(5.73) A 29(103)(8.85)
5.73
13.84
36 =
Solving by trial and error:
P = 139 kip
controls
Ans.
Ans:
P¿ = 390 kip, P = 139 kip
1355
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–55. The wood column is pinned at its base and top. If
the eccentric force P = 10 kN is applied to the column,
investigate whether the column is adequate to support this
loading without buckling or yielding. Take E = 10 GPa
and sY = 15 MPa.
x
P
150 mm
25 mm
yx
25 mm
75 mm
75 mm
3.5 m
Section Properties.
A = 0.05(0.15) = 7.5 A 10 - 3 B m2
Ix =
rx =
1
(0.05) A 0.153 B = 14.0625 A 10 - 6 B m4
12
14.0625 A 10 - 6 B
Ix
=
= 0.04330 m
AA
C 7.5 A 10 - 3 B
1
(0.15) A 0.053 B = 1.5625 A 10 - 6 B m4
12
e = 0.15 m
c = 0.075 m
Iy =
For a column that is pinned at both ends, K = 1. Then,
(KL)x = (KL)y = 1(3.5) = 3.5 m
Buckling About the Weak Axis. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
=
2
p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D
3.52
= 12.59 kN
Euler’s formula is valid if scr 6 sY .
12.59 A 10 B
Pcr
=
= 1.68 MPa 6 sY = 15 MPa
A
7.5 A 10 - 3 B
3
scr =
O.K.
Since Pcr 7 P = 10 kN, the column will not buckle.
Yielding About Strong Axis. Applying the secant formula.
smax =
=
(KL)x P
P
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
10 A 103 B
7.5 A 10 - 3 B
D1 +
0.15(0.075)
0.043302
secC
10 A 103 B
3.5
ST
2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D
= 10.29 MPa
Since smax 6 sY = 15 MPa , the column will not yield.
Ans.
Ans:
Yes.
1356
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–56. The wood column is pinned at its base and top.
Determine the maximum eccentric force P the column can
support without causing it to either buckle or yield. Take
E = 10 GPa and sY = 15 MPa .
x
P
150 mm
25 mm
yx
25 mm
75 mm
3.5 m
Section Properties.
A = 0.05(0.15) = 7.5 A 10 - 3 B m2
Ix =
1
(0.05) A 0.153 B = 14.0625 A 10 - 6 B m4
12
14.0625 A 10
Ix
=
C
AA
7.5 A 10 - 3 B
-6
rx =
B
= 0.04330 m
1
(0.15) A 0.053 B = 1.5625 A 10 - 6 B m4
12
e = 0.15 m
c = 0.075 m
Iy =
For a column that is pinned at both ends, K = 1. Then,
(KL)x = (KL)y = 1(3.5) = 3.5 m
Buckling About the Weak Axis. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
=
2
p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D
3.52
= 12.59 kN = 12.6 kN
Ans.
Euler’s formula is valid if scr 6 sY .
12.59 A 10 B
Pcr
= 1.68 MPa 6 sY = 15 MPa
=
A
7.5 A 10 - 3 B
3
scr =
O.K.
Yielding About Strong Axis. Applying the secant formula with P = Pcr = 12.59 kN,
smax =
=
(KL)x P
P
ec
C B 1 + 2 sec B
RS
A
2rx A EA
rx
12.59 A 103 B
7.5 A 10 - 3 B
D1 +
0.15(0.075)
0.043302
12.59 A 103 B B
3.5
ST
secC
2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D
= 13.31 MPa 6 sY = 15 MPa
O.K.
1357
75 mm
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–57. The 6061-T6 aluminum alloy solid shaft is fixed
at one end but free at the other end. If the shaft has a
diameter of 100 mm, determine its maximum allowable
length L if it is subjected to the eccentric force P = 80 kN.
L
P
100 mm
Section Properties.
A = p(0.052) = 2.5(10 - 3)p m2
p
(0.054) = 1.5625(10 - 6)p m4
4
I
1.5625(10 - 6)p
r =
=
= 0.025 m
AA
A 2.5(10 - 3)p
I =
e = 0.1 m
c = 0.05 m
For a column that is fixed at one end and free at the other, K = 2. Thus,
KL = 2L
Buckling. The critical buckling load is Pcr = 80 kN. Applying Euler’s equation,
Pcr =
p2EI
(KL)2
80(103) =
p2[68.9(109)][1.5625(10 - 6)p]
(2L)2
L = 3.230 m
Euler’s equation is valid only if scr 6 sY.
scr =
80(103)
Pcr
= 10.19 MPa 6 sY = 255 MPa
=
A
2.5(10 - 3)p
(O.K.)
Yielding. Applying the secant formula,
ec
P
KL P
c 1 + 2 sec c
dd
A
2r A EA
r
0.1(0.05)
80(103)
2L
80(103)
255(106) =
c1 +
sec c
dd
-3
2
2(0.025)A
2.5(10 )p
0.025
68.9(109)[2.5(10 - 3)p]
smax =
sec 0.4864L = 3.0043
L = 2.532 m = 2.53 m (controls)
Ans.
Ans:
L = 2.53 m
1358
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–58. The 6061-T6 aluminum alloy solid shaft is fixed at
one end but free at the other end. If the length is L = 3 m,
determine its minimum required diameter if it is subjected
to the eccentric force P = 60 kN.
L
P
100 mm
Section Properties.
A =
p 2
d
4
I =
I
r =
=
AA
p 4
p d 4
a b =
d
4 2
64
p 4
d
64
d
=
p 2
4
d
a4
e = 0.1 m
c =
d
2
For a column that is fixed at one end and free at the other, K = 2. Thus,
KL = 2(3) = 6 m
Buckling. The critical buckling load is Pcr = 60 kN. Applying Euler’s equation,
Pcr =
p2EI
(KL)2
60(103) =
p2 c 68.9(109) d a
p 4
d b
64
62
d = 0.08969 m = 89.7 mm
Yielding. Applying the secant formula,
smax =
P
ec
KL P
c 1 + 2 sec c
dd
A
2r A EA
r
d
0.1 a b
3
)
60(10
6
60(103)
2
¥¥
255(106) =
≥1 +
sec
≥
d
p
p 2
d 2
2a b
8.9(109) a d2b
d
6
a b
4 Q
4
4
4
255(106) =
240(103)
pd2
c1 +
0.8
0.012636
sec a
bd
d
d2
Solving by trial and error,
d = 0.09831 m = 98.3 mm (controls)
Ans.
Ans:
d = 98.3 mm
1359
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–59. The wood column is pinned at its base and top. If
L = 7 ft, determine the maximum eccentric load P that
can be applied without causing the column to buckle or
yield. Ew = 1.8 A 103 B ksi, sY = 8 ksi.
P
y
4 in.
x
x
P
y 10 in.
L
Section Properties:
A = 10(4) = 40 in2
ry =
Iy =
1
(4)(103) = 333.33 in4
12
Ix =
1
(10)(43) = 53.33 in4
12
333.33
Iy
=
= 2.8868 in.
AA
A 40
Buckling about x- x axis:
P = Pcr =
p2(1.8)(103)(53.33)
p2EI
=
= 134 kip
2
(KL)
[1(7)(12)]2
Check: scr =
Pcr
134
=
= 3.36 ksi 6 sY
A
40
OK
Yielding about y–y axis:
smax =
P
ec
KL
P
c 1 + 2 sec a
b
d
A
A
2r
EA
r
5(5)
ec
=
= 3.0
2
r
2.88682
a
1(7)(12)
P
P
KL
b
=
= 0.054221 2P
2r A EA
2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0542212P)]
By trial and error:
P = 73.5 kip
controls
Ans.
Ans:
P = 73.5 kip
1360
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–60. The wood column is pinned at its base and top.
If L = 5 ft, determine the maximum eccentric load P that
can be applied without causing the column to buckle or
yield. Ew = 1.8(103) ksi, sY = 8 ksi.
P
y
4 in.
x
x
P
y 10 in.
L
Section Properties:
A = 10(4) = 40 in2
ry =
Iy =
1
(4)(103) = 333.33 in4
12
Ix =
1
(10)(43) = 53.33 in4
2
Iy
333.33
=
= 2.8868 in.
AA
A 40
Buckling about x–x axis:
P = Pcr =
p2(1.8)(103)(53.33)
p2EI
=
= 263 kip
(KL)2
[1(5)(12)]2
Check: scr =
Pcr
263
=
= 6.58 ksi 6 sY
A
40
OK
Yielding about y–y axis:
smax =
P
ec
KL P
c 1 + 2 sec a
bd
A
2r A EA
r
5(5)
ec
=
= 3.0
2
r
2.88682
a
1(5)(12)
KL
P
P
b
= 0.0387291 P
=
2r A EA
2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0387291P)]
By trial and error:
P = 76.6 kip
(controls)
Ans.
1361
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–61. The A992 steel rectangular hollow section
column is pinned at both ends. If it has a length of
L = 14 ft, determine the maximum allowable eccentric
force P it can support without causing it to either buckle
or yield.
0.5 in.
2 in.
0.5 in.
5 in.
0.5 in.
Section a – a
P
a
6 in.
a
6 in.
L
Section Properties.
A = 3(6) - 2(5) = 8 in2
Ix =
1
1
(3)(63) (2)(53) = 33.167 in4
12
12
rx =
Ix
33.167
=
= 2.0361 in.
AA
A 8
Iy =
1
1
(6)(33) (5)(23) = 10.167 in4
12
12
For a column that is pinned at both ends, K = 1. Then,
(KL)x = (KL)y = 1(14)(12) = 168 in.
Buckling About the Weak Axis. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
=
2
p2[29(103)](10.167)
1682
= 103.10 kip
Euler’s formula is valid if scr 6 sY .
scr =
Pcr
103.10
=
= 12.89 ksi 6 sY = 50 ksi
A
8
(O.K.)
Yielding About Strong Axis. Applying the secant formula,
smax =
50 =
(KL)x Pmax
Pmax
ec
c 1 + 2 sec c
dd
A
2rx A EA
rx
6(3)
Pmax
Pmax
168
c1 +
sec c
dd
2
A
8
2(2.0361) 29(103)(8)
2.0361
Pmax = 61.174 = 61.2 kip
Ans.
Ans:
Pmax = 61.2 kip
1362
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–62. The A992 steel rectangular hollow section column
is pinned at both ends. If it is subjected to the eccentric
force P = 45 kip, determine its maximum allowable length L
without causing it to either buckle or yield.
0.5 in.
2 in.
0.5 in.
5 in.
0.5 in.
Section a – a
P
a
6 in.
a
6 in.
L
Section Properties.
A = 3(6) - 2(5) = 8 in2
Ix =
1
1
(3)(63) - (2)(53) = 33.167 in4
12
2
rx =
33.167
Ix
=
= 2.0361 in.
AA
A 8
Iy =
1
1
(6)(33) (5)(23) = 10.167 in4
12
12
For a column that is pinned at both ends, K = 1. Then,
(KL)x = (KL)y = 1L
Buckling About the Weak Axis. The critical load is
Pcr = 45 kip
Applying Euler’s formula,
Pcr =
45 =
p2EIy
(KL)y2
p2[29(103)](10.167)
L2
L = 254.3 in. = 21.2 ft
Ans.
(controls)
Euler’s formula is valid if scr 6 sY.
scr =
Pcr
45
=
= 5.625 ksi 6 sY = 50 ksi
A
8
(O. K.)
Yielding About Strong Axis. Applying the secant formula,
smax =
50 =
(KL)x P
P
ec
dd
c 1 + 2 sec c
A
2rx A EA
rx
6(3)
45
L
45
dd
c1 +
sec c
8
2(2.0361) A 29(103)(8)
2.03612
sec [3.420(10 - 3)L] = 1.817
L = 288.89 in. = 24.07 ft
Ans:
L = 21.2 ft
1363
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–63. The W10 * 30 structural A992 steel column is
pinned at its top and bottom. Determine the maximum load
P it can support.
P
8 in.
y
x
x
y
15 ft
680 kip⭈in.
Section properties for W 10 : 30:
2
P
4
A = 8.84 in
Ix = 170 in
d = 10.47 in.
Iy = 16.7 in4
rx = 4.38 in.
Yielding about x–x axis:
smax =
P
ec
KL P
c 1 + 2 sec a
bd
A
2r A EA
r
8A 2 B
ec
=
= 2.1830
r2
4.382
10.47
1.0(15)(12)
KL P
P
=
= 0.040583 2P
2r A EA
2(4.38) A 29(103)(8.84)
50(8.84) = P[1 + 2.1830 sec (0.040583 2P)]
By trial and error:
P = 129 kip
controls
Ans.
Buckling about y–y axis:
Pcr =
p2(29)(103)(16.7)
p2EI
=
= 147.5 kip
2
(KL)
[(1.0)(15)(12)]2
Check: scr =
Pcr
147.5
=
= 16.7 ksi 6 sY
A
8.84
OK
Ans:
P = 129 kip
1364
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–64. The W10 * 30 structural A992 steel column is
fixed at its bottom and free at its top. If it is subjected to the
eccentric load of P = 85 kip, determine if the column fails by
yielding. The column is braced so that it does not buckle
about the y–y axis.
P
8 in.
y
x
x
y
15 ft
680 kip⭈in.
P
Section properties for W10 : 30:
A = 8.84 in2
Ix = 170 in4
d = 10.47 in.
Iy = 16.7 in4
rx = 4.38 in.
Yielding about x–x axis:
smax =
ec
P
KL
P
c 1 + 2 sec a
bd
A
2r A EA
r
8A 2 B
ec
=
= 2.1830
2
r
4.382
10.47
(2)(15)(12)
P
KL
P
=
= 0.0811662P
2(4.38) A 29(103)(8.84)
2r A EA
50(8.84) = P[1 + 2.1830 sec (0.8166)2P]
By trial and error:
P = 104 kip
Since 104 kip 7 85 kip, the column does not fail.
No
Ans.
1365
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–65. Determine the maximum eccentric load P the
2014-T6-aluminum-alloy strut can support without
causing it either to buckle or yield. The ends of the strut
are pin connected.
P
150 mm
100 mm
a
P
150 mm
a
3m
50 mm
100 mm
Section a – a
Section Properties. The necessary section properties are
A = 0.05(0.1) = 5 A 10 - 3 B m2
Iy =
1
(0.1) A 0.053 B = 1.04167 A 10 - 6 B m4
12
4.1667 A 10
Ix
=
C
AA
5 A 10 - 3 B
-6
rx =
B
= 0.02887 m
For a column that is pinned at both of its ends K = 1. Thus,
(KL)x = (KL)y = 1(3) = 3 m
Buckling About the Weak Axis. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
=
2
p2 C 73.1 A 109 B D C 1.04167 A 10 - 6 B D
32
= 83.50 kN = 83.5 kN
Ans.
Critical Stress: Euler’s formula is valid only if scr 6 sY.
83.50 A 10 B
Pcr
=
= 16.70 MPa 6 sY = 414 MPa
A
5 A 10 - 3 B
3
scr =
O.K.
Yielding About Strong Axis. Applying the secant formula,
smax =
=
(KL)x P
P
ec
1 + 2 sec c
d
AJ
2rx A EA K
rx
83.50 A 103 B ≥
5 A 10 - 3 B
1 +
0.15(0.05)
0.02887
sec
2
= 229.27 MPa 6 sY = 414 MPa
83.50 A 103 B
3
¥
J 2(0.02887) C 73.1 A 109 B C 5 A 10 - 3 B D K
O.K.
Ans:
Pcr = 83.5 kN
1366
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–66. The W10 * 45 structural A992 steel column is
assumed to be pinned at its top and bottom. If the 12-kip
load is applied at an eccentric distance of 8 in., determine
the maximum stress in the column. Take L = 12.6 ft.
12 kip
8 in.
L
Section Properties for W10 : 45:
A = 13.3 in2
Ix = 248 in4
rx = 4.32 in.
d = 10.10 in.
Iy = 53.4 in4
Secant Formula:
smax =
P
ec
KL P
c 1 + 2 sec a
bd
A
2r A EA
r
P
12
=
= 0.90226 ksi
A
13.3
8A 2 B
ec
=
= 2.16478
2
r
4.322
10.10
a
1(12.6)(12)
KL
P
12
b
=
= 0.097612
2r A EA
2(4.32) A 29(103)(13.3)
smax = 0.90226 [1 + 2.16478 sec (0.097612)] = 2.86 ksi
Ans.
Ans:
smax = 2.86 ksi
1367
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–67. The W10 * 45 structural A992 steel column is
assumed to be pinned at its top and bottom. If the 12-kip
load is applied at an eccentric distance of 8 in., determine
the maximum stress in the column. Take L = 9 ft.
12 kip
8 in.
L
Section properties for W10 : 45:
A = 13.3 in2
Ix = 248 in4
rx = 4.32 in.
d = 10.10 in.
Iy = 53.4 in4
Secant Formula:
smax =
P
ec
KL P
bd
c 1 + 2 sec a
A
2r AEA
r
P
12
=
= 0.90226 ksi
A
13.3
8A 2 B
ec
=
= 2.16478
2
r
4.322
10.10
a
1(9)(12)
KL
P
12
b
=
= 0.069723
2r A EA
2(4.32) A 29(103)(13.3)
smax = 0.90226[1 + 2.16478 sec (0.069723)] = 2.86 ksi
Ans.
Ans:
smax = 2.86 ksi
1368
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–68. The W14 * 53 structural A992 steel column is
fixed at its base and free at its top. If P = 75 kip, determine
the sidesway deflection at its top and the maximum stress in
the column.
10 in.
P
18 ft
Section properties for a W14 : 53:
A = 15.6 in2
Ix = 541 in4
rx = 5.89 in.
d = 13.92 in.
Iy = 57.7 in4
Maximum Deflection:
vmax = e c sec a
P KL
b - 1d
A EI 2
2.0(18)(12)
P KL
75
=
a
b = 0.472267
3
A EI 2
A 29(10 )541
2
Ans.
vmax = 10 [sec (0.472267) - 1] = 1.23 in.
Maximum Stress:
smax =
P
ec
KL P
c 1 + 2 sec a
bd
A
2r A EA
r
P
75
=
= 4.808 ksi
A
15.6
10 A 2 B
ec
=
= 2.0062
2
r
5.892
13.92
2.0(18)(12)
75
KL P
=
= 0.47218
A
A
2r EA
2(5.89)
29(103)(15.6)
smax = 4.808[1 + 2.0062 sec (0.47218)] = 15.6 ksi 6 sY
Ans.
1369
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–69. The W14 * 53 column is fixed at its base and
free at its top. Determine the maximum eccentric load P
that it can support without causing it to buckle or yield.
Est = 29(103) ksi, sY = 50 ksi.
10 in.
P
18 ft
Section Properties for a W14 : 53:
A = 15.6 in2
Ix = 541 in4
rx = 5.89 in.
d = 13.92 in.
Iy = 57.7 in4
Buckling about y–y axis:
P = Pcr =
p2(29)(103)(57.7)
p2EI
=
= 88.5 kip
(KL)2
[(2.0)(18)(12)]2
Check: scr =
Pcr
88.5
=
= 5.67 ksi 6 sY
A
15.6
controls
Ans.
OK
Yielding about x–x axis:
smax =
P
ec
KL
P
c 1 + 2 sec a
bd
A
2r A EA
r
10 A 2 B
ec
=
= 2.0062
2
r
5.892
13.92
a
2.0(18)(12)
KL
P
P
b
=
= 0.054523 2P
A
2r
EA
2(5.89) A 29(103)(15.6)
50(15.6) = P[1 + 2.0062 sec (0.0545232P)]
By trial and error:
P = 204 kip
Ans:
P = 88.5 kip
1370
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–70. A column of intermediate length buckles when the
compressive stress is 40 ksi. If the slenderness ratio is 60,
determine the tangent modulus.
scr =
40 =
p2 Et
A
KL
r
2
B
2
;
a
KL
b = 60
r
p Et
(60)2
Et = 14590 ksi = 14.6 (103) ksi ‚
Ans.
Ans:
Et = 14.6(103) ksi
1371
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–71. The aluminum rod is fixed at its base free at its top.
If the eccentric load P = 200 kN is applied, determine the
greatest allowable length L of the rod so that it does not
buckle or yield. Eal = 72 GPa, sY = 410 MPa.
P
5 mm
200 mm
L
Section Properties:
A = p (0.12) = 0.031416 m2
r =
I =
p
(0.14) = 78.54 (10 - 6) m4
4
I
78.54(10 - 6)
=
= 0.05 m
A 0.031416
AA
Yielding:
smax =
P
ec
KL P
c 1 + 2 sec a
bd
A
2r A EA
r
200(103)
P
=
= 6.3662(104) Pa
A
0.031416
0.005(0.1)
ec
=
= 0.2
2
r
(0.05)2
a
P
2.0(L)
KL
200(103)
b
=
= 0.188063L
2r A EA
2(0.05) A 72(109)(0.031416)
410(104) = 6.3662(106)[1 + 0.2 sec (0.188063 L)]
L = 8.34 m
(controls)
Ans.
Buckling about x–x axis:
P
= 6.36 MPa 6 sY
A
Pcr =
Euler formula is valid.
p2 EI
(KL)2
200(103) =
p2(72)(109)(78.54)(10 - 4)
[(2.0)(L)]2
L = 8.35 m
Ans:
L = 8.34 m
1372
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–72. The aluminum rod is fixed at its base free at its
top. If the length of the rod is L = 2 m, determine the
greatest allowable load P that can be applied so that the
rod does not buckle or yield. Also, determine the largest
sidesway deflection of the rod due to the loading. Eal =
72 GPa, sY = 410 MPa.
P
5 mm
200 mm
L
Section Properties:
A = p(0.12) = 0.031416 m2
r =
I =
p
(0.14) = 78.54(10 - 6) m4
4
I
78.54 (10 - 6)
=
= 0.05 m
AA
A 0.031416
Yielding:
smax =
P
ec
KL P
c1 + 2 sec a
bd
A
2r A EA
r
(0.005)(0.1)
ec
=
= 0.2
r2
0.052
a
2(2)
KL
P
P
b
=
= 0.8410(10 - 3) 1P
2r A EA
2(0.05) A 72 (109)(0.031416)
410(104)(0.031416) = P [(1 + 0.2 sec (0.8410(10 - 3) 2P )]
By trial and Error:
P = 3.20 MN
(controls)
Ans.
Buckling:
P = Pcr =
p2(72)(109)(78.54)(10 - 6)
p2EI
=
= 3488 kN
(KL)2
[(2.0)(2)]2
Check: scr =
3488(103)
Pcr
=
= 111 MPa < sY
A
0.031416
OK
Maximum deflection:
P KL
vmax = e c sec a
b - 1d
A EI 2
2.0(2)
P KL
3.20(106)
=
a
b = 1.5045
A EI 2
A 72(109)(78.54)(10 - 6)
2
vmax = 5[sec (1.5045) - 1] = 70.5 mm
Ans.
1373
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
s(MPa)
13–73. The stress-strain diagram of the material of a
column can be approximated as shown. Plot P> A vs. KL> r
for the column.
350
200
Tangent Moduli. From the stress - strain diagram,
(Et)1 =
(Et)2 =
200 A 106 B
0.001
0 … s 6 200 MPa
= 200 GPa
(350 - 200) A 106 B
0.004 - 0.001
= 50 GPa
0
200 MPa 6 s … 350 MPa
Critical Stress. Applying Engesser’s equation,
scr =
p2Et
P
=
A
a
(1)
KL 2
b
r
If Et = (Et)1 = 200 GPa, Eq. (1) becomes
p C 200 A 10 B D
1.974 A 10 B
P
=
=
MPa
2
A
KL
KL 2
a
b
a
b
r
r
2
When scr =
9
6
P
= sY = 200 MPa , this equation becomes
A
200 A 106 B =
p2 C 200 A 109 B D
a
KL 2
b
r
KL
= 99.346 = 99.3
r
If Et = (Et)2 = 50 GPa, Eq. (1) becomes
P
=
A
p2 c 50 A 109 B d
0.4935 A 106 B
MPa
KL 2
KL 2
¢
≤
¢
≤
r
r
P
= sY = 200 MPa , this
when scr =
A
equation gives
200 A 106 B =
=
p2 C 50 A 109 B D
a
KL 2
b
r
KL
= 49.67 = 49.7
r
Using these results, the graphs of
P
KL
as shown in Fig. a can be plotted.
vs.
r
A
1374
P (in./in.)
0.001
0.004
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–74. Construct the buckling curve, P> A versus L> r, for
a column that has a bilinear stress–strain curve in
compression as shown.
s (ksi)
18
13
0.002
E1 =
13
= 6.5(103) ksi
0.002
E2 =
18 - 13
= 1.6667(103) ksi
0.005 - 0.002
For Et = E1
scr =
p2 (6.5)(103)
p2Er
P
64152
=
=
=
2
2
2
L
L
A
A B
A B
ALB
r
scr = 13 =
r
p2 (6.5)(103)
A B
L 2
r
r
L
= 70.2
r
;
For Et = E2
scr =
p2 (1.6667)(103)
P
16449
=
=
2
L 2
A
A B
ALB
r
r
scr = 13 =
p2 (1.6667)(103)
A B
L 2
r
;
L
= 35.6
r
1375
0.005
P (in./in.)
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
s (MPa)
13–75. The stress–strain diagram of the material can be
approximated by the two line segments shown. If a bar
having a diameter of 80 mm and a length of 1.5 m is made
from this material, determine the critical load provided the
ends are pinned. Assume that the load acts through the axis
of the bar. Use Engesser’s equation.
550
100
0.001
E1 =
100(106)
= 100 GPa
0.001
E2 =
550(106) - 100(106)
= 75 GPa
0.007 - 0.001
0.007
P (mm/mm)
Section Properties:
I =
p 4
c;
4
r =
p 4
I
c
= 4 2 = c = 0.04 = 0.02 m
AA Ap c
2
2
A = p c2
Engesser’s Equation:
KL 1.0 (1.5)
=
= 75
r
0.02
scr =
p2 Er
2
(KL
r )
=
p2 Er
(75)2
= 1.7546 (10 - 3) Et
Assume Et = E1 = 100 GPa
scr = 1.7546 (10 - 3)(100)(109) = 175 MPa 7 100 MPa
Therefore, inelastic buckling occurs:
Assume Et = E2 = 75 GPa
scr = 1.7546 (10 - 3) (75)(109) = 131.6 MPa
100 MPa 6 scr 6 550 MPa
OK
Critical Load:
Pcr = scr A = 131.6(106)(p)(0.042) = 661 kN
Ans.
Ans:
Pcr = 661 kN
1376
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
s (MPa)
*13–76. The stress–strain diagram of the material can be
approximated by the two line segments shown. If a bar
having a diameter of 80 mm and a length of 1.5 m is made
from this material, determine the critical load provided the
ends are fixed. Assume that the load acts through the axis of
the bar. Use Engesser’s equation.
550
100
0.001
E1 =
100 (106)
= 100 GPa
0.001
E2 =
550 (106) - 100 (106)
= 75 GPa
0.007 - 0.001
Section Properties:
I =
p 4
c;
4
r =
p 4
c
0.04
c
I
= 0.02 m
= A4 2 = =
2
2
AA
pc
A = p c2
Engesser’s Equation:
KL 0.5 (1.5)
=
= 37.5
r
0.02
scr =
p2 Er
2
(KL
r )
=
p2 Er
(37.5)2
= 7.018385 (10 - 3) Et
Assume Et = E1 = 100 GPa
scr = 7.018385 (10 - 3)(100)(109) = 701.8 MPa 7 100 MPa
NG
Assume Er = E2 = 75 GPa
scr = 7.018385 (10 - 3)(75)(109) = 526.4 MPa
100 MPa 6 scr 6 550 MPa
OK
Critical Load:
Pcr = scr A = 526.4 (106)(p)(0.042) = 2645.9 kN
Ans.
1377
0.007
P (mm/mm)
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
s (MPa)
13–77. The stress–strain diagram of the material can be
approximated by the two line segments shown. If a bar
having a diameter of 80 mm and a length of 1.5 m is made
from this material, determine the critical load provided one
end is pinned and the other is fixed. Assume that the load
acts through the axis of the bar. Use Engesser’s equation.
550
100
0.001
E1 =
100 (106)
= 100 GPa
0.001
E2 =
550 (106) - 100 (106)
= 75 GPa
0.007 - 0.001
0.007
P (mm/mm)
Section Properties:
I =
p 4
c;
4
r =
p 4
I
c
0.04
4 c
=
= =
= 0.02 m
A A A p c2
2
2
A = p c2
Engesser’s Equation:
0.7 (1.5)
KL
=
= 52.5
r
0.02
scr =
p2 Er
p2 Er
=
= 3.58081 (10 - 3) Et
2
(52.5)2
(KL
r )
Assume Et = E1 = 100 GPa
scr = 3.58081 (10 - 3)(100)(109) = 358.1 MPa 7 100 MPa
NG
Assume Et = E2 = 75 GPa
scr = 3.58081 (10 - 3)(75)(109) = 268.6 MPa
100 MPa 6 scr 6 550 MPa
OK
Critical Load:
Pcr = scr A = 268.6 (106)(p)(0.042) = 1350 kN
Ans.
Ans:
Pcr = 1350 kN
1378
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–78. Determine the largest length of a W10 * 12
structural A992 steel section if it is pin supported and is
subjected to an axial load of 28 kip. Use the AISC equations.
ry = 0.785 in.
For a W10 * 12,
s =
A = 3.54 in2
28
P
=
= 7.91 ksi
A
3.54
Assume a long column:
sallow =
12p2E
23(KL>r)2
a
KL
12p2 (29)(103)
12p2 E
b =
=
= 137.4
A 23 sallow A 23(7.91)
r
a
KL
2p2(29)(103)
2p2 E
b =
=
= 107,
A
r c A sY
50
KL
KL
7 a
b
r
r c
Long column.
KL
= 137.4
r
L = 137.4 a
r
0.785
b = 137.4a
b = 107.86 in.
K
1
Ans.
= 8.99 ft
Ans.
Ans:
L = 8.99 ft
1379
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–79. Using the AISC equations, select from Appendix B
the lightest-weight structural A992 steel column that is 14 ft
long and supports an axial load of 40 kip. The ends are fixed.
A = 2.68 in2
Try W6 * 9
a
ry = 0.905 in.
KL
2p2(29)(103)
2p2 E
b =
=
= 107
A
r c A sY
50
0.5(14)(12)
KL
=
= 92.82
ry
0.905
KL
KL
6 a
b
ry
r c
Intermediate column
KL>r
sallow =
[1 - 12((KL>r)c)2]sy
KL>r
KL>r
[53 + 38((KL>r)c) - 18((KL>r)c)3]
=
2
[1 - 12(92.82
107 ) ]50
1 92.82 3
[53 + 38(92.82
107 ) - 8 ( 107 ) ]
= 16.33 ksi
Pallow = sallow A
= 16.33(2.68)
= 43.8 kip 7 40 kip
OK
Use W6 * 9
Ans.
Ans:
Use W6 * 9
1380
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–80. Using the AISC equations, select from Appendix B
the lightest-weight structural A992 steel column that is 14 ft
long and supports an axial load of 40 kip. The ends are
pinned. Take sY = 50 ksi .
Try W6 * 15
(A = 4.43 in2
ry = 1.46 in.)
a
KL
2p2(29)(103)
2p2E
b =
=
= 107
A
A
r c
sY
50
a
(1.0)(14)(12)
KL
KL
KL
b =
b 7 a
b
= 115.1, a
ry
ry
r c
1.46
Long column
sallow =
12p2(29)(103)
12 p2E
=
= 11.28 ksi
23(KL>r)2
23(115.1)2
Pallow = sallowA
= 11.28(4.43) = 50.0 kip 7 40 kip
OK
Use W6 * 15
Ans.
1381
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–81. Determine the largest length of a W8 * 31
structural A992 steel section if it is pin supported and is
subjected to an axial load of 130 kip. Use the AISC
equations.
A = 9.13 in2
For a W 8 * 31,
s =
ry = 2.02 in.
130
P
=
= 14.239 ksi
A
9.13
Assume a long column:
sallow =
12p2 E
23(KL>r)2
a
KL
12p2(29)(103)
12 p2E
b =
=
= 102.4
A 23 sallow
A 23(14.239)
r
a
2
KL
KL
KL
2p2(29)(103)
b = 2p E =
6 a
b
= 107,
A
r c A sY
r
r e
50
Intermediate column
C 1 - 12 A (KL>r)c B 2 D sY
KL>r
sallow =
C 53 + 38 A (KL>r)c B - 13 A (KL>r)c B 3 D
KL>r
KL>r
C 1 - 12 A 107 B 2 D 50
KL>r
14.239 =
C 53 + 38 A 107 B - 13 A 107 B 3 D
1.5722(10 - 3)a
KL>r
KL>r
KL 2
KL
KL 3
b + 0.049902a
b - 1.4529(10 - 6) a
b = 26.269
r
r
r
By trial and error:
KL
= 120.4
r
L = 120.4 a
2.02
b = 243.24 in. = 20.3 ft
1.0
Ans.
Ans:
L = 15.1 ft
1382
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–82. Using the AISC equations, select from Appendix B
the lightest-weight structural A992 steel column that is 12 ft
long and supports an axial load of 20 kip. The ends are
pinned.
Try W6 * 12
A = 3.55 in2
ry = 0.918 in.
KL
2p2(29)(103)
2p2E
a
b =
=
= 107
A
A
r C
sY
50
a
(1.0)(12)(12)
KL
KL
KL
b 7 a
b
b =
= 156.9, a
ry
r c
ry
0.918
Long column
sallow =
12p2(29)(103)
12p2E
=
= 6.069 ksi
2
23(KL>r)
23(156.9)2
Pallow = sallow A
= 6.069(3.55) = 21.5 kip 7 20 kip
OK
Use W6 * 12
Ans.
Ans:
Use W6 * 12
1383
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–83. Determine the largest length of a W10 * 12
structural A992 steel section if it is fixed supported and is
subjected to an axial load of 28 kip. Use the AISC equations.
ry = 0.785 in.
For a W10 * 12,
s =
A = 3.54 in2
P
28
=
= 7.91 ksi
A
3.54
Assume a long column:
sallow =
12 p2E
23 (KL>r)2
a
KL 2
12p2E
12p2(29)(103)
12p2E
= 137.4
=
b =
=
A 23(7.91)
A 23 sallow
r
23 sallow
a
2(29)( 3)
KL
KL
KL
2p2E
10
b =
7 a
b
= 2p
= 107,
A
r
r c
r c A sY
50
Long column.
KL
= 137.4
r
L = 137.4a
0.785
r
b = 137.4 a
b = 215.72 in.
K
0.5
L = 18.0 ft
Ans.
Ans:
L = 18.0 ft
1384
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–84. Using the AISC equations, select from Appendix B
the lightest-weight structural A992 steel column that is
30 ft long and supports an axial load of 200 kip. The ends
are fixed.
Try W8 * 40
a
A = 11.7 in2
ry = 2.04 in.
KL
2 p2 (29)(103)
2 p2E
b =
=
= 107
A
r c A sY
50
0.5(30)(12)
KL
=
= 88.24
ry
2.04
a
KL
KL
b intermediate column.
b 6 a
ry
r c
sallow =
L
1 - 12 ≥
KL
r
2
A KL
r Bc M
¥
sY
3
KL
KL
r
r
5
3≥
1≥
¥
¥
- 8
L 3 + 8 KL
KL
M
r c
r c
A
B
C
A
D
B
e 1 - 1 86.54 2 f50
2 107
=
C
3 86.54
5
e 3 + 8 107
3f
D - 18 C 86.54
107 D
= 17.315 ksi
Pallow = sallow A = 17.315(11.7) = 203 kip 7 P = 200 kip
OK
Use W8 * 40
Ans.
1385
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–85. Determine the largest length of a W8 * 31
structural A992 steel section if it is pin supported and is
subjected to an axial load of 18 kip. Use the AISC equations.
Section properties: For W8 * 31
ry = 2.02 in.
A = 9.13 in2
Assume it is a long column:
sallow =
12p2E
2
23 A KL
r B
; a
KL 2
12 p2E
b =
r
23 sallow
KL
12p2E
=
A 23 sallow
r
Here sallow =
P
18
=
= 1.9715 ksi
A
9.13
KL
12p2(29)(103)
=
= 275.2 7 200
A 23(1.9715)
r
Thus use
KL
= 200
r
1.0 (L)
= 200
2.02
L = 404 in. = 33.7 ft
Ans.
Ans:
L = 33.7 ft
1386
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–86. Using the AISC equations, select from Appendix B
the lightest-weight structural A992 steel column that is 12 ft
long and supports an axial load of 40 kip. The ends
are fixed.
Try W6 * 9
a
A = 2.68 in2
ry = 0.905 in.
KL
2 p2 (29)(103)
2 p2E
b =
=
= 107
A
r c A sY
50
0.5(12)(12)
KL
=
= 79.56
ry
0.905
KL
KL
6 a
b
ry
r c
Intermediate column
2
2
c 1 - 12 a (KL>r)c b d sy
c1 - 12 a79.56
107 b d50
KL>r
sallow =
3
c 53 + 38 a (KL>r)c b - 18 a (KL>r)c b d
KL>r
KL>r
=
3
1 79.56
c 53 + 38 a 79.56
107 b - 8 a 107 b d
= 15.40 ksi
Pallow = sallow A
= 19.10 (2.68)
= 51.2 kip 7 40 kip
OK
Use W6 * 9
Ans.
Ans:
Use W6 * 9
1387
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–87. A 5-ft-long rod is used in a machine to transmit an
axial compressive load of 3 kip. Determine its smallest
diameter if it is pin connected at its ends and is made of a
2014-T6 aluminum alloy.
Section Properties:
p d 4
pd4
a b =
4 2
64
A =
p 2
d ;
4
r =
I
d
64
=
=
AA
C p4 d2 4
I =
pd4
sallow =
P
3
3.820
= p 2 =
A
d2
4 d
Assume long column:
1.0 (5)(12)
KL
240
=
=
d
r
d
4
sallow =
54 000
A
B
KL 2
r
;
54000
3.820
=
2
d
C 240 D 2
d
d = 1.42 in.
Ans.
KL
240
=
= 169 7 55
r
1.42
O.K.
Ans:
d = 1.42 in.
1388
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–88. Determine the largest length of a W8 * 31
structural A992 steel column if it is to support an axial load
of 10 kip. The ends are pinned.
W8 * 31
a
ry = 2.02 in.
A = 9.13 in2
KL
2 p2 (29)(103)
2 p2E
b =
=
= 126.1
A
r c A sY
36
KL
1.0 L
=
ry
2.02
Assume
sallow =
KL
KL
b
7 a
ry
r c
12p2E
A
B
2
23 KL
r
Here sallow =
;
KL
12 p2E
=
A 23 sallow
r
P
10
=
= 1.10
A
9.13
KL
KL
12p2(29)(103)
=
= 369.2 7 a
b
A 23 (1.10)
r
r c
Assumption
OK
1.0 (L)
= 369.2
2.02
L = 745.9 in. = 62.2 ft
Ans.
1389
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–89. Using the AISC equations, check if a column
having the cross section shown can support an axial force of
1500 kN. The column has a length of 4 m, is made from A992
steel, and its ends are pinned.
20 mm
350 mm
20 mm
300 mm
10 mm
Section Properties:
A = 0.3(0.35) - 0.29(0.31) = 0.0151 m2
Iy =
1
1
(0.04) A 0.33 B +
(0.31) A 0.013 B = 90.025833 A 10 - 6 B m4
12
12
ry =
Iy
90.02583(10 - 6)
=
= 0.077214 m
AA
A
0.0151
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus,
a
AISC
Column
1(4)
KL
= 51.80
b =
r y
0.077214
Formula:
For
A992
steel,
a
KL
2p2E
b =
r c A sg
KL
KL
2p2[200(109)]
= 107. Since
6 a
b , the column is an intermediate
A 345(106)
r
r c
column. Applying Eq. 13–23,
=
B1 sallow =
2(KL>r)2c
R sY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)3c
B1 =
(KL>r)2
(51.802)
2(1072)
R (345)(106)
3(51.80)
(51.803)
5
+
3
8(107)
8(1073)
= 166.1 MPa
The allowable load is
Pallow = sallowA
= 166.1 A 106 B (0.0151)
O.K.
= 2508 kN 7 P = 1500 kN
Thus, the column is adequate.
Ans.
Ans:
Yes.
1390
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–90. The A992-steel tube is pinned at both ends. If it is
subjected to an axial force of 150 kN, determine the
maximum length of the tube using the AISC column
design formulas.
100 mm
Section Properties.
80 mm
A = p A 0.052 - 0.042 B = 0.9 A 10 - 3 B p m2
I =
p
A 0.054 - 0.044 B = 0.9225 A 10 - 6 B p m4
4
0.9225 A 10 B p
I
=
= 0.03202 m
AA
C 0.9 A 10 - 3 B p
-6
r =
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus,
1(L)
KL
=
= 31.23L
r
0.03202
AISC Column Formulas.
sallow =
12p2E
23(KL>r)2
150 A 103 B
0.9 A 10 - 3 B p
=
12p2 C 200 A 109 B D
23(31.23L)2
L = 4.4607 m = 4.46 m
Here,
KL
= 31.23(4.4607) = 139.33.
r
2p2 C 200 A 109 B D
345 A 10 B
column is correct.
=
C
6
= 107. Since a
Ans.
For
A992
steel
a
KL
2p2E
b =
r c A sY
KL
KL
b 6
6 200, the assumption of a long
r c
r
Ans:
L = 4.46 m
1391
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
600 lb
13–91. The bar is made of a 2014-T6 aluminum alloy.
Determine its smallest thickness b if its width is 5b. Assume
that it is pin connected at its ends.
b
5b
8 ft
Section Properties:
A = b(5b) = 5b2
Iy =
1
5 4
(5b) A b3 B =
b
12
12
ry =
5 4
Iy
23
12 b
=
b
=
AA
C 5b2
6
600 lb
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus,
a
1(8)(12) 332.55
KL
b =
=
13
r y
b
6 b
Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply
Eq. 13–26.
sallow =
54 000
(KL>r)2
0.600
54 000
=
5b2
A 332.55 B 2
b
b = 0.7041 in.
Here,
KL
KL
332.55
=
= 472.3. Since
7 55, the assumption is correct. Thus,
r
r
0.7041
b = 0.704 in.
Ans.
Ans:
b = 0.704 in.
1392
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–92. The bar is made of a 2014-T6 aluminum alloy.
Determine its smallest thickness b if its width is 5b. Assume
that it is fixed connected at its ends.
600 lb
b
5b
8 ft
Section Properties:
600 lb
2
A = b(5b) = 5b
Iy =
1
5 4
(5b) A b3 B =
b
12
12
ry =
5 4
Iy
23
12 b
b
=
AA
C 5b2 = 6
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus,
a
0.5(8)(12)
166.28
KL
=
b =
r y
b
23
6 b
Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply
Eq. 13–26.
sallow =
54 000
(KL>r)2
0.600
54 000
=
2
5b
A 166.28 B 2
b
b = 0.4979 in.
Here,
KL
KL
166.28
=
= 334.0. Since
7 55, the assumption is correct.
r
r
0.4979
Thus,
b = 0.498 in.
Ans.
1393
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–93. The 2014-T6 aluminum column of 3-m length has
the cross section shown. If the column is pinned at both
ends and braced against the weak axis at its mid-height,
determine the allowable axial force P that can be safely
supported by the column.
15 mm
170 mm
15 mm
15 mm
100 mm
Section Properties.
A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2
Ix =
1
1
(0.1) A 0.23 B (0.085) A 0.173 B = 31.86625 A 10 - 6 B m4
12
12
Iy = 2 c
1
1
(0.015) A 0.13 B d +
(0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4
12
12
rx =
31.86625 A 10
Ix
=
AA
C 5.55 A 10 - 3 B
ry =
Iy
2.5478 A 10 - 6 B = 0.02143 m
=
AA
C
5.55 A 10 - 3 B
-6
B
= 0.07577
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 3 m
and Ly = 1.5 m. Thus,
a
(1)(3)
KL
= 39.592
b =
r x
0.07577
a
(1)(1.5)
KL
b =
= 70.009 (controls)
r y
0.02143
2014-T6 Alumimum Alloy Column Formulas. Since a
KL
b 7 55, the column can
r y
be classified as a long column,
sallow = D
372.33 A 103 B
T MPa
= C
372.33 A 103 B
S MPa
a
KL 2
b
r
70.0092
= 75.966 MPa
Thus, the allowed force is
Pallow = sallowA = 75.966 A 106 B C 5.55 A 10 - 3 B D = 421.61 kN = 422 kN
Ans.
Ans:
Pallow = 422 kN
1394
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–94. The 2014-T6 aluminum column has the cross section
shown. If the column is pinned at both ends and subjected to
an axial force P = 100 kN, determine the maximum length
the column can have to safely support the loading.
15 mm
170 mm
15 mm
15 mm
100 mm
Section Properties.
A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2
Iy = 2 c
ry =
1
1
(0.015) A 0.13 B d +
(0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4
12
12
2.5478 A 10 - 6 B
Iy
=
= 0.02143 m
C 5.55 A 10 - 3 B
AA
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then,
a
1(L)
KL
b =
= 46.6727L
r y
0.02143
2014-T6 Alumimum Alloy Column Formulas. Assuming a long column,
sallow = D
372.33 A 103 B
100 A 103 B
= C
5.55 A 10 - 3 B
a
KL 2
b
r
T MPa
372.33 A 103 B
(46.672L)2
S A 106 B Pa
Ans.
L = 3.080 m = 3.08 m
Since a
KL
b = 46.6727(3.080) = 144 7 55, the assumption is correct.
r y
Ans:
L = 3.08 m
1395
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
13–95. The tube is 0.5 in. thick, is made of aluminum alloy
2014-T6, and is fixed connected at its ends. Determine the
largest axial load that it can support.
y
x
8 in.
x
8 in.
y
12 ft
Section Properties:
P
2
A = (8)(8) - (7)(7) = 15 in
Ix = Iy =
1
1
(8)(83) (7)(73) = 141.25 in4
12
12
rx = ry =
I
141.25
=
= 3.069 in.
AA
A 15
Allowable stress:
0.5(12)(12)
KL
KL
=
= 23.46, 12 6
6 55
r
3.069
r
Intermediate column
sallow = 30.7 - 0.23 a
KL
b
r
= 30.7 - 0.23(23.46) = 25.30 ksi
Pallow = sallow A
= 25.30(15) = 380 kip
Ans.
Ans:
Pallow = 380 kip
1396
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
*13–96. The tube is 0.5 in. thick, is made from aluminum
alloy 2014-T6, and is fixed at its bottom and pinned at its
top. Determine the largest axial load that it can support.
y
x
8 in.
x
8 in.
y
12 ft
Section Properties:
P
2
A = (8)(8) - (7)(7) = 15 in
Ix = Iy =
1
1
(8)(83) (7)(73) = 141.25 in4
12
12
rx = ry =
I
141.25
=
= 3.069 in.
AA
A 15
Allowable stress:
0.7(12)(12)
KL
KL
=
= 32.8446, 12 6
6 55
r
3.069
r
Intermediate column
sallow = 30.7 - 0.23 a
KL
b = 30.7 - 0.23(32.8446) = 23.15 ksi
r
Pallow = sallow A = 23.15(15) = 347 kip
Ans.
1397
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–97. The tube is 0.25 in. thick, is made of a 2014-T6
aluminum alloy, and is fixed at its bottom and pinned at its
top. Determine the largest axial load that it can support.
P
x
y
6 in.
x
6 in.
y
10 ft
P
Section Properties:
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
1
1
(6) A 63 B (5.5) A 5.53 B = 31.7448 in4
12
12
r =
31.7448
I
=
= 2.3496 in.
AA
A 5.75
Slenderness Ratio: For a column fixed at one end and pinned at the other end,
K = 0.7. Thus,
0.7(10)(12)
KL
=
= 35.75
r
2.3496
Aluminium (2014 – T6 alloy) Column Formulas: Since 12 6
KL
6 55, the column
r
is classified as an intermediate column. Applying Eq. 13–25,
sallow = c 30.7 - 0.23 a
KL
b d ksi
r
= [30.7 - 0.23(35.75)]
= 22.477 = 22.48 ksi
The allowable load is
Pallow = sallowA = 22.48(5.75) = 129 kip
Ans.
Ans:
Pallow = 129 kip
1398
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
13–98. The tube is 0.25 in. thick, is made of a 2014-T6
aluminum alloy, and is fixed connected at its ends.
Determine the largest axial load that it can support.
x
y
6 in.
x
6 in.
y
10 ft
P
Section Properties:
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
1
1
(6) A 63 B (5.5) A 5.53 B = 31.7448 in4
12
12
r =
I
31.7448
=
= 2.3496 in.
AA
A 5.75
Slenderness Ratio: For column fixed at both ends, K = 0.5. Thus,
0.5(10)(12)
KL
=
= 25.54
r
2.3496
Aluminium (2014 – T6 alloy) Column Formulas: Since 12 6
KL
6 55, the
r
column is classified as an intermediate column. Applying Eq. 13–25,
sallow = c 30.7 - 0.23a
KL
b d ksi
r
= [30.7 - 0.23(25.54)]
= 24.83 ksi
The allowable load is
Pallow = sallowA = 24.83(5.75) = 143 kip
Ans.
Ans:
Pallow = 143 kip
1399
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–99. The tube is 0.25 in. thick, is made of 2014-T6
aluminum alloy and is pin connected at its ends. Determine
the largest axial load it can support.
P
x
y
6 in.
x
6 in.
y
10 ft
P
Section Properties:
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
1
1
(6) A 63 B (5.5) A 5.53 B = 31.7448 in4
12
12
r =
I
31.7448
=
= 2.3496 in.
AA
A 5.75
Slenderness Ratio: For a column pinned as both ends, K = 1. Thus,
1(10)(12)
KL
=
= 51.07
r
2.3496
Aluminum (2014 – T6 alloy) Column Formulas: Since 12 6
KL
6 55, the
r
column is classified as an intermediate column. Applying Eq. 13–25,
sallow = c 30.7 - 0.23 a
KL
b d ksi
r
= [30.7 - 0.23(51.07)]
= 18.95 ksi
The allowable load is
Pallow = sallowA = 18.95(5.75) = 109 kip
Ans.
Ans:
Pallow = 109 kip
1400
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–100. A rectangular wooden column has the cross
section shown. If the column is 6 ft long and subjected to an
axial force of P = 15 kip, determine the required minimum
1
dimension a of its cross-sectional area to the nearest 16
in.
so that the column can safely support the loading. The
column is pinned at both ends.
a
2a
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then,
(1)(6)(12)
KL
72
=
=
a
a
d
NFPA Timber Column Formula. Assuming an intermediate column,
sallow = 1.20 c 1 -
1 KL>d 2
a
b d ksi
3 26.0
15
1 72>a 2
= 1.20 c 1 - a
b d
2a(a)
3 26.0
a = 2.968 in.
Use a = 3 in.
Ans.
KL
72
KL
=
= 24. Since 11 6
6 26, the assumption is correct.
d
3
d
1401
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–101. A rectangular wooden column has the cross
section shown. If a = 3 in. and the column is 12 ft long,
determine the allowable axial force P that can be safely
supported by the column if it is pinned at its top and fixed at
its base.
a
2a
Slenderness Ratio. For a column fixed at its base and pinned at its top K = 0.7.
Then,
0.7(12)(12)
KL
=
= 33.6
d
3
NFPA Timer Column Formula. Since 26 6
KL
6 50, the column can be classified
d
as a long column.
sallow =
540 ksi
540
=
= 0.4783 ksi
(KL>d)2
33.62
The allowable force is
Pallow = sallowA = 0.4783(3)(6) = 8.61 kip
Ans.
Ans:
Pallow = 8.61 kip
1402
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–102. A rectangular wooden column has the cross
section shown. If a = 3 in. and the column is subjected to an
axial force of P = 15 kip, determine the maximum length
the column can have to safely support the load. The column
is pinned at its top and fixed at its base.
a
2a
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7.
Then,
KL
0.7L
=
= 0.2333L
d
3
NFPA Timber Column Formula. Assuming an intermediate column,
sallow = 1.20 c 1 -
1 KL>d 2
a
b d ksi
3 26.0
15
1 0.2333L 2
= 1.20 c 1 - a
b d
3(6)
3
26.0
L = 106.68 in. = 8.89 ft
Ans.
KL
KL
= 0.2333(106.68) = 24.89. Since 11 6
6 26, the assumption is
d
d
correct.
Here,
Ans:
L = 8.89 ft
1403
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–103. The timber column has a square cross section and
is assumed to be pin connected at its top and bottom. If it
supports an axial load of 50 kip, determine its smallest side
dimension a to the nearest 12 in. Use the NFPA formulas.
14 ft
a
Section Properties:
A = a2
sallow = s =
50
P
= 2
A
a
Assume long column:
sallow =
540
2
A KL
d B
540
50
=
(1.0)(14)(12) 2
a2
C
D
a
a = 7.15 in.
(1.0)(14)(12)
KL
KL
=
= 23.5,
6 26
d
7.15
d
Assumption NG
Assume intermediate column:
sallow = 1.20 B 1 -
1 KL>d 2
a
b R
3 26.0
2
50
1
a
b R
= 1.20 B 1 - a
2
a
26.0
3
1.0(14)(12)
a = 7.46 in.
1.0(14)(12)
KL
KL
=
= 22.53, 11 6
6 26
d
7.46
d
Assumption O.K.
1
Use a = 7 in.
2
Ans.
Ans:
1
Use a = 7 in.
2
1404
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–104. The bar is made of aluminum alloy 2014-T6.
Determine its thickness b if its width is 1.5b. Assume that it
is fixed connected at its ends.
800 lb
b
1.5b
5 ft
800 lb
Section Properties:
A = 1.5 b2
Iy =
1
(1.5b)(b3) = 0.125 b4
12
ry =
Iy
0.125 b4
=
= 0.2887 b
AA
A 1.5 b2
sallow =
P
0.8
0.5333
=
=
A
1.5 b2
b2
Assume long column:
sallow =
54 000
(KL>r)2
54 000
0.5333
=
2
(0.5)(5)(12)
b
C
D2
0.2887 b
b = 0.571 in.
ry = 0.2887(0.571) = 0.1650 in.
(0.5)(12)(12)
KL
= 181.8,
=
ry
0.1650
KL
7 55
ry
Assumption OK
Use b = 0.571 in.
Ans.
1405
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–105. The column is made of wood. It is fixed at its
bottom and free at its top. Use the NFPA formulas to
determine its greatest allowable length if it supports an
axial load of P = 6 kip.
P
3 in.
y
x
y 6 in.
x
L
Assume long column:
sallow = s =
6
P
=
= 0.3333 ksi
A
6(3)
sallow =
540
(KL>d)2
0.3333 =
540
[2.0(L)>3]2
K = 2.0
d = 3 in.
Ans.
L = 60.37 in. = 5.03 ft
Check:
2.0(60.37)
KL
=
= 40.25,
d
3
26 6
KL
6 50
d
Assumption OK
Ans:
L = 5.03 ft
1406
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
13–106. The column is made of wood. It is fixed at its
bottom and free at its top. Use the NFPA formulas to
determine the largest allowable axial load P that it can
support if it has a length L = 6 ft.
3 in.
y
x
y 6 in.
x
L
K = 2.0
L = 6(12) = 72 in.
d = 3 in.
2.0(72)
KL
KL
=
= 48, 26 6
6 50
d
3
d
Long column
sallow =
540
540
=
= 0.2344 ksi
2
(KL>d)
(48)2
Pallow = sallow A
= 0.2344(6)(3) = 4.22 kip
Ans.
Ans:
Pallow = 4.22 kip
1407
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–107. The W8 * 15 structural A992 steel column is
assumed to be pinned at its top and bottom. Determine the
largest eccentric load P that can be applied using Eq. 13–30
and the AISC equations of Sec. 13.6. The load at the top
consists of a force P and a moment M = P (8 in.).
P
y
M
x
x
y
10 ft
Section Properties: For a W8 * 15 wide flange section,
A = 4.44 in2
d = 8.11 in.
Ix = 48.0 in4
ry = 0.876 in.
rx = 3.29 in.
Slenderness Ratio: By observation, the largest slenderness ratio is about y–y axis.
For a column pinned at both ends, K = 1. Thus,
a
1(10)(12)
KL
b =
= 137.0
r y
0.876
Allowable Stress: The allowable stress can be determined using AISC Column
2
KL
2p2[29(103)]
b = 2p E =
Formulas. For A992 steel, a
= 107. Since
A
r c A sr
50
KL
KL
a
b …
… 200, the column is a long column. Applying Eq. 13–21,
r c
r
sallow =
12 p2E
23(KL>r)2
12p2(29.0)(103)
=
23(137.02)
= 7.958 ksi
Maximum Stress: Bending is about x–x axis. Applying Eq. 13–30, we have
smax = sallow =
P
Mc
+
A
I
P(8) A 2 B
P
+
4.44
48
8.11
7.958 =
P = 8.83 kip
Ans.
Ans:
P = 8.83 kip
1408
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–108. Solve Prob. 13–107 if the column is fixed at its
top and bottom.
P
y
M
x
x
y
10 ft
Section Properties: For a W8 * 15 wide flange section,
A = 4.44 in2
d = 8.11 in.
Ix = 48.0 in4
rx = 3.29 in.
ry = 0.876 in.
Slenderness Ratio: By observation, the largest slenderness ratio is about y–y axis.
For a column fixed at both ends, K = 0.5. Thus,
a
0.5(10)(12)
KL
= 68.49
b =
r y
0.876
Allowable Stress: The allowable stress can be determined using AISC Column Formulas.
For A992 steel, a
KL
KL
KL
2p2 [29(103)]
2 p2E
= 107. Since
=
6 a
b ,
b =
A
50
r C A sr
r
r c
the column is an intermediate column. Applying Eq. 13–23,
sallow =
2(KL>r)2c
dsY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)3c
c1 -
=
(KL>r)2
c1 -
(68.492)
2(1072)
d (50)
3(68.49)
(68.493)
5
+
3
8(107)
8(1073)
= 21.215 ksi
Maximum Stress: Bending is about x–x axis. Applying Eq. 13–30, we have
smax = sallow =
P
Mc
+
A
I
P(8) A 2 B
P
+
4.44
48
8.11
21.215 =
Ans.
P = 23.5 kip
1409
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–109. Solve Prob. 13–107 if the column is fixed at its
bottom and pinned at its top.
P
y
M
x
x
y
10 ft
Section Properties: For a W8 * 15 wide flange section,
A = 4.44 in2
Ix = 48.0 in4
d = 8.11 in.
rx = 3.29 in.
ry = 0.876 in.
Slenderness Ratio: By observation, the largest slenderness ratio is about y–y axis.
For a column fixed at one end and pinned at the other, K = 0.7. Thus,
a
0.7(10)(12)
KL
b =
= 95.89
r y
0.876
Allowable Stress: The allowable stress can be determined using AISC Column
2p2 [29(103)]
KL
2p2E
= 107. Since
=
Formulas. For A–36 steel, a
b =
A
50
A sr
r
y
KL
KL
6 a
b . the column is a intermediate column. Applying Eq. 13–23,
r
r y
sallow =
2(KL>r)2c
d sY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)3c
c1 -
=
(KL>r)2
c1 -
(95.892)
2(126.12)
d (50)
3(95.89)
(95.893)
5
+
3
8(126.1)
8(126.13)
= 15.643 ksi
Maximum Stress: Bending is about x–x axis. Applying Eq. 13–30, we have
smax = sallow =
P
Mc
+
A
I
P(8) A 2 B
P
+
4.44
48
8.11
15.643 =
P = 17.4 kip
Ans.
Ans:
P = 17.4 kip
1410
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–110. The W10 * 19 structural A992 steel column is
assumed to be pinned at its top and bottom. Determine the
largest eccentric load P that can be applied using Eq. 13–30
and the AISC equations of Sec. 13.6.
P
20 kip
y
6 in.
x
x
y
12 ft
Section Properties for W10 * 19:
A = 5.62 in2
d = 10.24 in.
rx = 4.14 in.
ry = 0.874 in.
Ix = 96.3 in4
1.0(12)(12)
KL
=
= 164.76
ry
0.874
a
2
3
KL
2p2E = 2p (29)(10 ) = 107, KL 7 a KL b
b =
ry
r c
50
A
r c A sY
(sa)allow =
12p2(29)(103)
12p2E
=
= 5.501 ksi
23(KL>r)2
23(164.76)2
smax = (sA)allow =
Mxc
P
+
A
Ix
P(6) A 2
P + 20
+
5.62
96.3
10.24
5.501 =
B
P = 3.91 kip
Ans.
Ans:
P = 3.91 kip
1411
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–111. The W8 * 15 structural A992 steel column is
fixed at its top and bottom. If it supports end moments of
M = 5 kip·ft, determine the axial force P that can be
applied. Bending is about the x–x axis. Use the AISC
equations of Sec. 13.6 and Eq. 13–30.
P
x
M
y
y
x
16 ft
M
Section Properties for W8 * 15:
A = 4.44 in2
Ix = 48.0 in4
P
ry = 0.876 in.
d = 8.11 in.
0.5(16)(12)
KL
=
= 109.59
ry
0.876
a
KL
2p2(29)(103)
2 p2E
= 107
=
b =
A
50
r c A sY
KL
KL
>a
b
ry
r c
sallow =
12p2(29)(103)
12p2E
=
2
23(KL>r)
23(109.59)2
= 12.434 ksi
smax =
P
Mc
P
Mc
+
=
+
A
I
A
I
5(12) A 2 B
P
+
4.44
48
8.11
12.434 =
= 32.7 kip
Ans.
Ans:
P = 29.6 kip
1412
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–112. The W8 * 15 structural A992 steel column is
fixed at its top and bottom. If it supports end moments of
M = 23 kip·ft, determine the axial force P that can be
applied. Bending is about the x–x axis. Use the interaction
formula with (sb) = 24 ksi.
P
x
M
y
y
x
16 ft
M
Section Properties for W8 * 15:
A = 4.44 in2
Ix = 48.0 in4
P
ry = 0.876 in.
d = 8.11 in.
Interaction Method:
0.5(16)(12)
KL
=
= 109.59
ry
0.876
a
KL
KL
KL
2p2(29)(103)
2p2E
= 107,
7 a
b
=
b =
ry
r c
A
50
r c A sY
(sa)allow =
12p2(29)(103)
12p2E
=
2
23(KL>r)
23(109.59)2
= 12.434 ksi
sa =
P
P
=
= 0.2252P
A
4.44
23(12) A 2 B
Mc
=
= 23.316
I
48
8.11
sb =
sa
sb
+
= 1
(sa)allow
(sb)allow
0.2252P
23.316
+
= 1
12.434
24
P = 1.57 kip
Note:
Ans.
0.2252(1.52)
sa
= 0.0284 6 0.15
=
(sa)allow
12.434
Therefore the method is allowed.
1413
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–113. The A992-steel W10 * 45 column is fixed at its
base. Its top is constrained so that it cannot move along the
x–x axis but it is free to rotate about and move along the
y–y axis. Determine the maximum eccentric force P that
can be safely supported by the column using the allowable
stress method.
12 in.
P
y
x
y
x
24 ft
Section Properties. From the table listed in the appendix, the section properties for
a W10 * 45 are
A = 13.3 in2
bf = 8.02 in.
rx = 4.32 in.
Iy = 53.4 in4
ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base
and free at its top, Kx = 2. Thus,
a
2(288)
KL
= 133.33 (controls)
b =
r x
4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Then,
a
0.7(288)
KL
b =
= 100.30
r y
2.01
Allowable Stress. The allowable stress will be determined using the AISC column
formulas. For A992 steel,
2p2 C 29 A 103 B D
KL
2p2E
KL
KL
b =
=
= 107. Since a
b 6 a
b 6 200,
r c
r c
r x
C
50
B sY
the column is classified as a long column.
a
sallow =
=
12p2E
23(KL>r)2
12p2 C 29 A 103 B D
23(133.332)
= 8.400 ksi
Maximum Stress. Bending is about the weak axis. Since M = P(12) and
bf
8.02
=
= 4.01 in,
c =
2
2
sallow =
P
Mc
+
A
I
8.400 =
[P(12)](4.01)
P
+
13.3
53.4
P = 8.604 kip = 8.60 kip
Ans.
Ans:
P = 8.60 kip
1414
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–114. The A992-steel W10 * 45 column is fixed at its
base. Its top is constrained so that it cannot move along the
x–x axis but it is free to rotate about and move along the
y–y axis. Determine the maximum eccentric force P that can
be safely supported by the column using an interaction
formula. The allowable bending stress is (sb) allow = 15 ksi.
12 in.
P
y
x
y
x
24 ft
Section Properties. From the table listed in the appendix, the section properties for
a W10 * 45 are
A = 13.3 in2
bf = 8.02 in.
rx = 4.32 in.
Iy = 53.4 in4
ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base
and free at its top, Kx = 2. Thus,
a
2(288)
KL
= 133.33 (controls)
b =
r x
4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Then,
a
0.7(288)
KL
b =
= 100.30
r y
2.01
Allowable Stress. The allowable stress will be determined using the AISC column
formulas. For A992 steel, a
a
2p C 29 A 10 B D
KL
2p2E
b =
= C
= 107. Since
r c
B sY
50
2
3
KL
KL
b 6 a
b 6 200, the column is classified as a long column.
r c
r x
sallow =
=
12p2E
23(KL>r)2
12p2 C 29 A 103 B D
23 A 133.332 B
= 8.400 ksi
Interaction Formula. Bending is about the weak axis. Here, M = P(12) and
bf
8.02
=
= 4.01 in.
c =
2
2
P>A
Mc>Ar2
+
= 1
(sa)allow
(sb)allow
P>13.3
+
8.400
P(12)(4.01) n C 13.3 A 2.012 B D
15
= 1
P = 14.57 kip = 14.6 kip
Ans.
14.57>13.3
sa
=
= 0.1304 6 0.15
(sa)allow
8.400
O.K.
Ans:
P = 14.6 kip
1415
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–115. The A-36-steel W12 * 50 column is fixed at its
base. Its top is constrained so that it cannot move along the
x–x axis but it is free to rotate about and move along the
y–y axis. If the eccentric force P = 15 kip is applied to
the column, investigate if the column is adequate to support
the loading. Use the allowable stress method.
12 in.
P
y
x
y
x
24 ft
Section Properties. From the table listed in the appendix, the section properties for
a W12 * 50 are
A = 14.7 in2
bf = 8.08 in.
rx = 5.18 in.
Iy = 56.3 in4
ry = 1.96 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base
and free at its top, Kx = 2. Thus,
a
2(288)
KL
= 111.20 (controls)
b =
r x
5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Thus,
a
0.7(288)
KL
b =
= 102.86
r y
1.96
Allowable Stress. The allowable stress will be determined using the AISC column
formulas. For A-36 steel, a
a
2p C 29 A 10 B D
KL
2p2E
= 126.10. Since
b =
=
C
r c A sY
36
2
3
KL
KL
b 6 a
b , the column can be classified as an intermediate column.
r x
r c
B1 sallow =
2(KL>r)C 2
R sY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)C
8(KL>r)C 3
C1 -
=
(KL>r)2
111.202
2 A 126.102 B
S (36)
3(111.20)
5
111.203
+
3
8(126.10)
8 A 126.103 B
= 11.51 ksi
Maximum Stress. Bending is about the weak axis. Since, M = 15(12) = 180 kip # in.
bf
8.08
=
= 4.04 in.,
and c =
2
2
smax =
180(4.04)
P
Mc
15
+
=
+
= 13.94 ksi
A
I
14.7
56.3
Since smax 7 sallow, the W12 * 50 column is inadequate according to the allowable
stress method.
Ans:
No.
1416
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–116. The A-36-steel W12 * 50 column is fixed at its
base. Its top is constrained so that it cannot move along the
x–x axis but it is free to rotate about and move along the
y–y axis. If the eccentric force P = 15 kip is applied to
the column, investigate if the column is adequate to support
the loading. Use the interaction formula. The allowable
bending stress is (sb)allow = 15 ksi.
12 in.
bf = 8.08 in.
rx = 5.18 in.
y
24 ft
ry = 1.96 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base
and free its top, Kx = 2. Thus,
a
2(288)
KL
= 111.20 (controls)
b =
r x
5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Then,
a
Allowable
=
Axial
0.7(288)
KL
= 102.86
b =
r y
1.96
Stress.
For
A-36
steel,
a
KL
2p2E
b =
r c A sY
2p2 C 29 A 103 B D
KL
KL
b 6 a
b , the column can be
= 126.10 . Since a
r x
r c
36
C
classified as an intermediate column.
C1 sallow =
2(KL>r)c 2
S sY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)c 3
C1 -
=
(KL>r)2
111.202
2 A 126.102 B
S (36)
3(111.20)
5
111.203
+
3
8(126.10)
8 A 126.103 B
= 11.51 ksi
Interaction Formula. Bending is about the weak axis. Here, M = 15(12)
bf
8.08
= 180 kip # in . and c =
=
= 4.04 in.
2
2
P>A
Mc>Ar2
15>14.7
+
=
+
(sa)allow
(sb)allow
11.51
180(4.04) n C 14.7 A 1.962 B D
15
y
x
Section Properties. From the table listed in the appendix, the section properties for
a W12 * 50 are
A = 14.7 in2
P
= 0.9471 6 1
15>14.7
sa
=
= 0.089 6 0.15
(sa)allow
11.51
O.K.
Thus, a W12 * 50 column is adequate according to the interaction formula.
1417
x
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
13–117. A 16-ft-long column is made of aluminum alloy
2014-T6. If it is fixed at its top and bottom, and a concentric
load P and a moment M = P (4.5 in.) are applied at point A,
determine the maximum allowable magnitude of P using
the equations of Sec. 13.6 and Eq. 13–30.
y
M
x
0.5 in.
y
8 in.
A
x 8 in.
0.5 in.
0.5 in.
Section Properties:
A = 2(0.5)(8) + 8(0.5) = 12 in2
Ix =
1
1
(8)(93) (7.5)(83) = 166 in4
12
12
Iy = 2 a
ry =
1
1
b(0.5)(83) +
(8)(0.53) = 42.75 in4
12
12
42.75
Iy
= 1.8875 in.
=
A
12
AA
Allowable Stress Method:
0.5(16)(12)
KL
KL
=
= 50.86, 12 6
6 55
ry
ry
1.8875
sallow = c 30.7 - 0.23 a
KL
bd
r
= [30.7 - 0.23(50.86)] = 19.00 ksi
smax = sallow =
19.00 =
Mx c
P
+
A
Ix
P(4.25)(4.5)
P
+
12
166
P = 95.7 kip
Ans.
Ans:
P = 95.7 kip
1418
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–118. A 16-ft-long column is made of aluminum alloy
2014-T6. If it is fixed at its top and bottom, and a concentric
load P and a moment M = P (4.5 in.) are applied at point A,
determine the maximum allowable magnitude of P using
the equations of Sec. 13.6 and the interaction formula with
(sb)allow = 20 ksi.
P
y
M
x
0.5 in.
y
8 in.
A
x 8 in.
0.5 in.
0.5 in.
Section Properties:
A = 2(0.5)(8) + 8(0.5) = 12 in2
Ix =
1
1
(8)(93) (7.5)(83) = 166 in4
12
12
Iy = 2 a
ry =
1
1
b(0.5)(83) +
(8)(0.53) = 42.75 in4
12
12
Iy
42.75
=
= 1.8875 in.
AA
A 12
Interaction Method:
0.5(16)(12)
KL
KL
=
= 50.86, 12 6
6 55
ry
ry
1.8875
sallow = c 30.7 - 0.23 a
KL
bd
r
= [30.7 - 0.23(50.86)]
= 19.00 ksi
sa =
P
P
=
= 0.08333P
A
12
sb =
P(4.25)(4.50)
Mc
= 0.1152P
=
Ix
166
sb
sa
+
= 1.0
(sa)allow
(sb)allow
0.08333P
0.1152P
+
= 1
19.00
20
P = 98.6 kip
Ans.
Ans:
P = 98.6 kip
1419
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–119. The 2014-T6 aluminum hollow column is fixed at
its base and free at its top. Determine the maximum
eccentric force P that can be safely supported by the
column. Use the allowable stress method. The thickness of
the wall for the section is t = 0.5 in.
6 in.
P
3 in.
6 in.
8 ft
Section Properties.
A = 6(3) - 5(2) = 8 in2
Ix =
1
1
(3) A 63 B (2) A 53 B = 33.1667 in4
12
12
rx =
33.1667
Ix
=
= 2.036 in.
AA
A
8
Iy =
1
1
(6) A 33 B (5) A 23 B = 10.1667 in4
12
12
ry =
10.1667
Iy
= 1.127 in.
=
A
8
AA
Slenderness Ratio. For a column fixed at its base and free at its top, K = 2. Thus,
a
2(8)(12)
KL
= 170.32
b =
r y
1.127
Allowable Stress. Since a
KL
b 7 55, the column can be classified as a long
r y
column.
sallow =
54 000 ksi
54 000 ksi
=
= 1.862 ksi
(KL>r)2
170.322
Maximum Stress. Bending occurs about the strong axis so that M = P(6) and
6
c = = 3 in.
2
sallow =
1.862 =
P
Mc
+
A
I
C P(6) D (3)
P
+
8
33.1667
P = 2.788 kip = 2.79 kip
Ans.
Ans:
P = 2.79 kip
1420
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–120. The 2014-T6 aluminum hollow column is fixed
at its base and free at its top. Determine the maximum
eccentric force P that can be safely supported by the
column. Use the interaction formula. The allowable bending
stress is (sb)allow = 30 ksi. The thickness of the wall for the
section is t = 0.5 in.
6 in.
P
3 in.
6 in.
8 ft
Section Properties.
A = 6(3) - 5(2) = 8 in2
Ix =
1
1
(3) A 63 B (2) A 53 B = 33.1667 in4
12
12
rx =
33.1667
Ix
=
= 2.036 in.
8
AA
A
Iy =
1
1
(6) A 33 B (5) A 23 B = 10.1667 in4
12
12
ry =
Iy
10.1667
=
= 1.127 in.
AA
A
8
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 2. Thus,
a
2(8)(12)
KL
= 170.32
b =
r y
1.127
KL
b 7 55, the column can be classified as the column is
r y
classified as a long column.
Allowable Stress. Since a
sallow =
54000 ksi
54000 ksi
=
= 1.862 ksi
2
(KL>r)
170.322
Interaction Formula. Bending is about the strong axis. Since M = P(6) and
6
c = = 3 in,
2
P>A
Mc>Ar2
+
= 1
(sa)allow
(sb)allow
P>8
+
1.862
[P(6)](3) n C 8 A 2.0362 B D
30
= 1
P = 11.73 kip = 11.7 kip
Ans.
1421
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
13–121. The 10-ft-long bar is made of aluminum alloy
2014-T6. If it is fixed at its bottom and pinned at the top,
determine the maximum allowable eccentric load P that can
be applied using the formulas in Sec. 13.6 and Eq. 13–30.
x
1.5 in.
1.5 in.
x
y
2 in.
y
2 in.
3 in.
Section Properties:
A = 6(4) = 24.0 in2
Ix =
1
(4) A 63 B = 72.0 in4
12
Iy =
1
(6) A 43 B = 32.0 in4
12
ry =
Iy
32.0
=
= 1.155 in.
AA
A 24
Slenderness Ratio: The largest slenderness ratio is about y -y axis. For a column
pinned at one end and fixed at the other end, K = 0.7. Thus,
a
0.7(10)(12)
KL
b =
= 72.75
r y
1.155
Allowable Stress: The allowable stress can be determined using aluminum
KL
7 55, the column is classified as a long
(2014 –T6 alloy) column formulas. Since
r
column. Applying Eq. 13–26,
sallow = c
=
54 000
d ksi
(KL>r)2
54 000
72.752
= 10.204 ksi
Maximum Stress: Bending is about x - x axis. Applying Eq. 13–30, we have
smax = sallow =
10.204 =
P
Mc
+
A
I
P(1.5)(3)
P
+
24.0
72.0
P = 98.0 kip
Ans.
Ans:
P = 98.0 kip
1422
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
13–122. The 10-ft-long bar is made of aluminum alloy
2014-T6. If it is fixed at its bottom and pinned at the top,
determine the maximum allowable eccentric load P that
can be applied using the equations of Sec. 13.6 and the
interaction formula with (sb)allow = 18 ksi.
x
1.5 in.
1.5 in.
x
y
2 in.
y
2 in.
3 in.
Section Properties:
A = 6(4) = 24.0 in2
Ix =
1
(4) A 63 B = 72.0 in4
12
Iy =
1
(6) A 43 B = 32.0 in4
12
rx =
Ix
72.0
=
= 1.732 in.
AA
A 24.0
ry =
32.0
Iy
=
= 1.155 in.
AA
A 24.0
Slenderness Ratio: The largest slenderness ratio is about y -y axis. For a column
pinned at one end and fixed at the other end, K = 0.7. Thus
a
0.7(10)(12)
KL
b =
= 72.75
r y
1.155
Allowable Stress: The allowable stress can be determined using aluminum
KL
(2014 –T6 alloy) column formulas. Since
7 55, the column is classified as a long
r
column. Applying Eq. 13–26,
(sa)allow = c
=
54 000
d ksi
(KL>r)2
54 000
72.752
= 10.204 ksi
Interaction Formula: Bending is about x -x axis. Applying Eq. 13–31, we have
Mc>Ar2
P>A
+
= 1
(sa)allow
(sb)allow
P(1.5)(3)>24.0(1.7322)
P>24.0
+
= 1
10.204
18
P = 132 kip
Ans.
Ans:
P = 132 kip
1423
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–123. Determine if the column can support the eccentric
compressive load of 1.5 kip. Assume that the ends
are pin connected. Use the NFPA equations in Sec. 13.6 and
Eq. 13–30.
1.5 kip
12 in.
3 in.
1.5 in.
6 ft
1.5 kip
A = 12 (1.5) = 18 in2;
Ix =
1
(1.5)(12)3 = 216 in4
12
d = 1.5 in.
1.0(6)(12)
KL
= 48
=
d
1.5
26 6
KL
6 50
d
(sa)allow =
smax =
=
A
540
B
KL 2
d
=
540
= 0.2344
(48)2
Mx c
P
+
A
Ix
1.5(3)(6)
1.5
+
= 0.208 ksi
18
216
(sa)allow 7 smax
The column is adequate.
Yes.
Ans.
Ans:
Yes.
1424
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1.5 kip
*13–124. Determine if the column can support the
eccentric compressive load of 1.5 kip. Assume that the
bottom is fixed and the top is pinned. Use the NFPA
equations in Sec. 13.6 and Eq. 13–30.
12 in.
3 in.
1.5 in.
6 ft
1.5 kip
A = 12 (1.5) = 18 in2;
Ix =
1
(1.5)(12)3 = 216 in4
12
d = 1.5 in.
0.7 (6)(12)
KL
=
= 33.6
d
1.5
26 6
KL
6 50
d
(sa)allow =
smax =
A
540
B
KL 2
d
=
540
= 0.4783
(33.6)2
1.5(3)(6)
Mxc
P
1.5
+
=
+
= 0.208 ksi
A
Ix
18
216
(sa)allow 7 smax
The column is adequate.
Yes.
Ans.
1425
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–125. The 10-in.-diameter utility pole supports the
transformer that has a weight of 600 lb and center of gravity
at G. If the pole is fixed to the ground and free at its top,
determine if it is adequate according to the NFPA equations
of Sec. 13.6 and Eq. 13–30.
G
15 in.
18 ft
2(18)(12)
KL
=
= 43.2 in.
d
10
26 6 43.2 … 50
Use Eq. 13–29,
sallow =
540
540
=
= 0.2894 ksi
(KL>d)
(43.2)2
smax =
Mc
P
+
A
I
smax =
(600)(15)(5)
600
+
p (5)2
A p B (5)4
4
smax = 99.31 psi 6 0.289 ksi
O.K.
Yes.
Ans.
Ans:
Yes.
1426
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
13–126. Using the NFPA equations of Sec. 13–6 and
Eq. 13–30, determine the maximum allowable eccentric
load P that can be applied to the wood column. Assume
that the column is pinned at both its top and bottom.
0.75 in.
6 in.
3 in.
12 ft
Section Properties:
A = 6(3) = 18.0 in2
Iy =
1
(6) A 33 B = 13.5 in4
12
Slenderness Ratio: For a column pinned at both ends, K = 1.0. Thus,
a
1.0(12)(12)
KL
b =
= 48.0
d y
3
Allowable Stress: The allowable stress can be determined using NFPA timber
KL
column formulas. Since 26 6
6 50, it is a long column. Applying Eq. 13–29,
d
sallow =
=
540
ksi
(KL>d)2
540
= 0.234375 ksi
48.02
Maximum Stress: Bending is about y - y axis. Applying Eq. 13–30, we have
smax = sallow =
0.234375 =
P
Mc
+
A
I
P(0.75)(1.5)
P
+
18.0
13.5
P = 1.69 kip
Ans.
Ans:
P = 1.69 kip
1427
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–127. Using the NFPA equations of Sec. 13.6 and
Eq. 13–30, determine the maximum allowable eccentric
load P that can be applied to the wood column. Assume
that the column is pinned at the top and fixed at the bottom.
P
0.75 in.
6 in.
3 in.
12 ft
Section Properties:
A = 6(3) = 18.0 in2
Iy =
1
(6) A 33 B = 13.5 in4
12
Slenderness Ratio: For a column pinned at one end and fixed at the other end,
K = 0.7. Thus,
a
0.7(12)(12)
KL
b =
= 33.6
d y
3
Allowable Stress: The allowable stress can be determined using NFPA timber
KL
column formulas. Since 26 6
6 50, it is a long column. Applying Eq. 13–29,
d
sallow =
=
540
ksi
(KL>d)2
540
= 0.4783 ksi
33.62
Maximum Stress: Bending is about y - y axis. Applying Eq. 13–30, we have
smax = sallow =
0.4783 =
P
Mc
+
A
I
P(0.75)(1.5)
P
+
18.0
13.5
P = 3.44 kip
Ans.
Ans:
P = 3.44 kip
1428
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–128. The wood column has a thickness of 4 in. and a
width of 6 in. Using the NFPA equations of Sec. 13.6 and
Eq. 13–30, determine the maximum allowable eccentric
load P that can be applied. Assume that the column is
pinned at both its top and bottom.
P
1 in.
6 in.
4 in.
10 ft
Section properties:
A = 6(4) = 24 in2
Iy =
1
(6)(43) = 32 in4
12
d = 4 in.
Allowable Stress Method:
1.0(10)(12)
KL
=
= 30 in.
d
4
26 6
(sa)allow =
KL
6 50
d
540
540
=
= 0.6 ksi
(KL>d)2
302
smax = (sa)allow =
0.6 =
My c
P
+
A
Iy
P(1)(2)
P
+
24
32
P = 5.76 kip
Ans.
1429
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–129. The wood column has a thickness of 4 in. and a
width of 6 in. Using the NFPA equations of Sec. 13.6 and
Eq. 13–30, determine the maximum allowable eccentric
load P that can be applied. Assume that the column is
pinned at the top and fixed at the bottom.
P
1 in.
6 in.
4 in.
10 ft
Section Properties:
A = 6(4) = 24 in2
Iy =
1
(6)(43) = 32 in4
12
d = 4 in.
Allowable Stress Method:
0.7(10)(12)
KL
=
= 21
d
4
11 6
KL
6 26
d
(sa)allow = 1.20 c1 -
smax = (sa)allow =
0.9391 =
1 KL>d 2
1 21 2
a
b d = 1.20 c1 - a b d = 0.9391 ksi
3
26
3 26
Myc
P
+
A
Iy
P(1)(2)
P
+
24
32
P = 9.01 kip
Ans.
Ans:
P = 9.01 kip
1430
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10 mm
13–130. A steel column has a length of 5 m and is free at
one end and fixed at the other end. If the cross-sectional
area has the dimensions shown, determine the critical load.
Est = 200 GPa, sY = 360 MPa.
10 mm
60 mm
80 mm
Section Properties:
A = 0.06 (0.01) + 2 (0.06)(0.01) = 1.80(10 - 3) m2
y =
0.005 (0.06)(0.01) + 2[0.03 (0.06)(0.01)]
©yA
=
= 0.02167 m
©A
0.06 (0.01) + 2 (0.06)(0.01)
Ix =
1
(0.06)(0.01)3 + 0.06 (0.01)(0.02167 - 0.005)2
12
+ [
Iy =
1
(0.01)(0.06)3 + 0.01 (0.06)(0.03 - 0.02167)2] = 0.615 (10 - 6) m4
12
(controls)
1
1
(0.06)(0.08)3 (0.05)(0.06)3 = 1.66 (10 - 6) m4
12
12
Critical Load:
Pcr =
p2EI
;
(KL)2
K = 2.0
p2 (200)(109)(0.615)(10 - 6)
=
[2.0 (5)]2
= 12140 N = 12.1 kN
Ans.
Check Stress:
scr =
Pcr
12140
= 6.74 MPa 6 sY = 360 MPa
=
A
1.80 (10 - 3)
Hence, Euler’s equation is still valid.
Ans:
Pcr = 12.1 kN
1431
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–131. The square structural A992 steel tubing has outer
dimensions of 8 in. by 8 in. Its cross-sectional area is 14.40 in2
and its moments of inertia are Ix = Iy = 131 in4. Determine
the maximum load P it can support. The column can be
assumed fixed at its base and free at its top.
3 in.
x
P
y
12 ft
Section Properties:
A = 14.4 in2;
r =
Ix = Iy = 131 in4
I
131
=
= 3.01616 in.
AA
A 14.4
Yielding:
smax =
P
ec
KL
P
c 1 + 2 sec a
bd;
A
2 r AE A
r
K = 2.0
3 (4)
ec
=
= 1.319084
2
r
(3.01616)2
2(12)(12)
KL
P
P
=
= 0.073880 2P
2 r AE A
2(3.01616)A 29 (103)(14.40)
50 (14.4) = P[1 + 1.319084 sec (0.073880 2P)]
By trial and error:
P = 199 kip
(controls)
Ans.
Buckling:
P = Pcr =
scr =
p2(29)(103)(131)
p2E I
=
= 452 kip
2
(K L)
[2(12)(12)]2
Pcr
452
=
= 31.4 ksi 6 sY = 50 ksi
A
14.4
(OK)
Ans:
P = 161 kip
1432
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4.5 in.
*13–132. If the A-36 steel solid circular rod BD has a
diameter of 2 in., determine the allowable maximum force P
that can be supported by the frame without causing the rod
to buckle. Use F.S. = 2 against buckling.
A
B
C
4 ft
3 ft
3 ft
D
P
Equilibrium. The compressive force developed in BD can be determined by
considering the equilibrium of the free-body diagram of member ABC, Fig. a.
a + ©MA = 0;
4
FBD a b(3) + P(0.375) - P(6.375) = 0
5
FBD = 2.5P
Section Properties. The cross-sectional area and moment of inertia of BD are
A = p(12) = pin2
I =
p 4
(1 ) = 0.25p in4
4
Critical Buckling Load. Since BD is pinned at both of its ends, K = 1. The critical
buckling load is
Pcr = FBD(F.S.) = 2.5P(2) = 5P
The length of BD is L = 232 + 42 = 5 ft. Applying Euler’s formula,
Pcr =
5P =
p2EI
(KL)2
p2[29(103)](0.25p)
[1(5)(12)]2
P = 12.49 kip = 12.5 kip
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
5(12.49)
Pcr
=
= 19.88 ksi 6 sY = 36 ksi
p
A
(O.K.)
1433
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4.5 in.
13–133. If P = 15 kip, determine the required minimum
diameter of the A992 steel solid circular rod BD to the
1
nearest 16
in. Use F.S. = 2 against buckling.
A
B
C
4 ft
3 ft
3 ft
D
Equilibrium. The compressive force developed in BD can be determined by
considering the equilibrium of the free-body diagram of member ABC, Fig. a,
4
FBD a b (3) + 15(0.375) - 15(6.375) = 0
5
a + ©MA = 0;
P
FBC = 37.5 kip
Section Properties. The cross-sectional area and moment of inertia of BD are
A =
p 2
d
4
I =
p 4
p d 4
a b =
d
4 2
64
Critical Buckling Load. Since BD is pinned at both of its ends, K = 1. The critical
buckling load is
Pcr = FBD(F.S.) = 37.5(2) = 75 kip
The length of BD is L = 232 + 42 = 5 ft. Applying Euler’s formula,
Pcr =
75 =
p2EI
(KL)2
p
p2 c 29(103)d a d4b
64
[1(5)(12)]2
d = 2.094 in.
1
Use d = 2 in.
8
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
Pcr
75
=
= 21.15 ksi 6 sY = 50 ksi
p
A
(2.1252)
4
(O.K.)
Ans:
Use d = 2
1434
1
in.
8
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–134. The steel pipe is fixed supported at its ends. If
it is 4 m long and has an outer diameter of 50 mm,
determine its required thickness so that it can support an
axial load of P = 100 kN without buckling. Est = 200 GPa,
sY = 250 MPa.
P
4m
I =
p
(0.0254 - ri 4 )
4
Critical Load:
P
2
Pcr =
p EI
;
(K L)2
100(103) =
K = 0.5
p2(200)(109)[p4 (0.0254 - ri4)]
[0.5(4)]2
ri = 0.01908 m = 19.1 mm
t = 25 mm - 19.1 mm = 5.92 mm
Ans.
Check Stress:
s =
100(103)
Pcr
= 122 MPa 6 sY = 345 MPa
=
A
p(0.0252 - 0.01912)
(OK)
Ans:
t = 5.92 mm
1435
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–135. The W200 * 46 A992-steel column can be
considered pinned at its top and fixed at its base. Also, the
column is braced at its mid-height against the weak axis.
Determine the maximum axial load the column can support
without causing it to buckle.
6m
Section Properties. From the table listed in the appendix, the section properties for
a W200 * 46 are
A = 5890 mm2 = 5.89 A 10 - 3 B m2
Iy = 15.3 A 106 B mm4 = 15.3 A 10 - 6 B m4
6m
Ix = 45.5 A 106 B mm4 = 45.5 A 10 - 6 B m4
Critical Buckling Load. For buckling about the strong axis, Kx = 0.7 and Lx = 12 m.
Since the column is fixed at its base and pinned at its top,
Pcr =
p2EIx
(KL)x 2
=
p2 c 200 A 109 B d c 45.5 A 10 - 6 B d
[0.7(12)]2
= 1.273 A 106 B N = 1.27 MN
For buckling about the weak axis, Ky = 1 and Ly = 6 m since the bracing provides
a support equivalent to a pin. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y 2
=
p2 c 200 A 109 B d c 15.3 A 10 - 6 B d
[1(6)]2
= 838.92 kN = 839 kN (controls)Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY .
838.92 A 10 B
Pcr
=
= 142.43 MPa 6 sY = 345 MPa
A
5.89 A 10 - 3 B
3
scr =
O.K.
Ans:
Pcr = 839 kN
1436
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
*13–136. The structural A992 steel column has the cross
section shown. If it is fixed at the bottom and free at the
top, determine the maximum force P that can be applied at
A without causing it to buckle or yield. Use a factor of
safety of 3 with respect to buckling and yielding.
20 mm
A
10 mm
Section properties:
100 mm
-3
2
4m
©A = 0.2(0.01) + 0.15 (0.01) + 0.1(0.01) = 4.5(10 ) m
= 0.06722 m
1
(0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2
12
+
1
(0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2
12
+
1
(0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2
12
= 20.615278 (10 - 6) m4
1
1
1
(0.01)(0.23) +
(0.15)(0.013) +
(0.01)(0.13)
12
12
12
Ix =
= 7.5125 (10 - 6) m4
ry =
Iy
20.615278(10 - 6)
=
= 0.0676844
AA
A
4.5 (10 - 3)
Buckling about x–x axis:
Pcr =
p2(200)(109)(7.5125)(10 - 6)
p2 EI
=
2
(KL)
[2.0(4)]2
= 231.70 kN
scr =
(controls)
231.7 (103)
Pcr
= 51.5 MPa 6 sg = 345 MPa
=
A
4.5 (10 - 3)
Yielding about y–y axis:
smax =
ec
P
KL P
c 1 + 2 sec a
b d;
A
2r A EA
r
e = 0.06722 - 0.02 = 0.04722 m
0.04722 (0.06722)
ec
=
= 0.692919
r2
0.06768442
2.0 (4)
P
KL
P
=
= 1.96992 (10 - 3) 2P
2r A EA
2(0.0676844) A 200 (109)(4.5)(10 - 3)
345(106)(4.5)(10 - 3) = P[1 + 0.692919 sec (1.96992 (10 - 3) 2P)]
By trial and error:
P = 434.342 kN
Hence,
Pallow =
231.70
= 77.2 kN
3
Ans.
1437
A
100 mm
10 mm
~A
0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01)
©x
x =
=
©A
4.5(10 - 3)
Iy =
150 mm
10 mm
100 mm
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–137. The structural A992 steel column has the cross
section shown. If it is fixed at the bottom and free at the top,
determine if the column will buckle or yield when the load
P = 10 kN is applied. Use a factor of safety of 3 with respect
to buckling and yielding.
P
20 mm
A
10 mm
100 mm
Section properties:
4m
©A = 0.2 (0.01) + 0.15 (0.01) + 0.1 (0.01) = 4.5 (10 - 3) m2
150 mm
A
10 mm
100 mm
100 mm
10 mm
0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01)
©x~A
= 0.06722 m
x =
=
©A
4.5 (10 - 3)
Iy =
Ix =
ry =
1
(0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2
12
+
1
(0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2
12
+
1
(0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 = 20.615278 (10 - 6) m4
12
1
1
1
(0.01)(0.23) +
(0.15)(0.013) +
(0.01)(0.13) = 7.5125 (10 - 6) m4
12
12
12
Iy
BA
20.615278 (10 - 6)
=
B
4.5 (10 - 3)
= 0.0676844 m
Buckling about x–x axis:
Pcr =
p2(200)(109)(7.5125)(10 - 6)
p2 EI
=
= 231.70 kN
(KL)2
[2.0(4)]2
scr =
231.7 (103)
Pcr
= 51.5 MPa 6 sg = 345 MPa
=
A
4.5 (10 - 3)
Pallow =
O.K.
Pcr
231.7
=
= 77.2 kN 7 P = 10 kN
FS
3
Hence the column does not buckle.
Yielding about y–y axis:
smax =
P =
ec
P
KL
P
c 1 + 2 sec a
bd
A
2r A EA
r
e = 0.06722 - 0.02 = 0.04722 m
10
= 3.333 kN
3
3.333 (103)
P
= 0.7407 MPa
=
A
4.5 (10 - 3)
0.04722 (0.06722)
ec
=
= 0.692919
2
r
(0.067844)2
2.0 (4)
P
KL
3.333 (103)
=
= 0.113734
2 r AE A
2(0.0676844) A 200 (109)(4.5)(10 - 3)
smax = 0.7407 [1 + 0.692919 sec (0.113734)] = 1.26 MPa 6 sg = 345 MPa
Hence the column does not yield!
No.
Ans.
1438
Ans:
No, it does not buckle or yield.
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
sy
14–1. A material is subjected to a general state of plane
stress. Express the strain energy density in terms of the
elastic constants E, G, and n and the stress components sx,
sy, and txy.
txy
sx
Strain Energy Due to Normal Stresses: We will consider the application of normal
stresses on the element in two successive stages. For the first stage, we apply only sx
on the element. Since sx is a constant,
(Ui)1 =
s2x
s2x V
dV =
2E
Lv 2E
When sy is applied in the second stage, the normal strain Px will be strained by
Px ¿ = - vPy = -
vsy
E
. Therefore, the strain energy for the second stage is
(Ui)2 =
=
¢
s2y
+ sx Px ¿ ≤ dV
Lv 2E
B
s2y
Lv 2E
+ sx a -
vsy
E
b R dV
Since sx and sy are constants,
(Ui)2 =
V
(s2y - 2vsx sy)
2E
Strain Energy Due to Shear Stresses: The application of txy does not strain the
element in a normal direction. Thus, from Eq. 14–11, we have
(Ui)3 =
t2xy
Lv 2G
dV =
t2xy V
2G
The total strain energy is
Ui = (Ui)1 + (Ui)2 + (Ui)3
=
t2xy V
s2x V
V
+
(s2y - 2vsx sy) +
2E
2E
2G
t2xy V
V
2
2
=
(s + sy - 2vsx sy) +
2E x
2G
and the strain energy density is
t2xy
Ui
1
2
2
=
(s + sy - 2vsx sy) +
V
2E x
2G
Ans.
Ans:
t2xy
Ui
1 2
2
=
(sx + sy - 2nsxsy) +
V
2E
2G
1439
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–2. The strain-energy density must be the same whether
the state of stress is represented by sx, sy, and txy, or by the
principal stresses s1 and s2. This being the case, equate the
strain–energy expressions for each of these two cases and
show that G = E>[2(1 + n)].
U =
1
v
1 2
(s2x + s2y) - sxsy +
t R dV
E
2 G xy
Lv 2 E
U =
1
v
(s21 + s22) s s R dV
B
E 1 2
Lv 2 E
B
Equating the above two equations yields.
1
v
1 2
1
v
(s2 + s2y) ss +
t =
(s2 + s22) s s
2E x
E x y
2 G xy
2E 1
E 1 2
However, s1, 2 =
sx + sy
2
;
A
a
sx - sy
2
(1)
2
b + txy
2
Thus, A s21 + s22 B = s2x + s2y + 2 t2xy
and also
s1 s2 = sxsy - t2xy
Substitute into Eq. (1)
1
v
1 2
1
v
v 2
t =
(s2 + s2y + 2t2xy) ss +
t
A s2 + s2y B - sxsy +
2E x
E
2 G xy
2E x
E x y
E xy
2
txy
v 2
1 2
t =
+
txy
2 G xy
E
E
1
v
1
=
+
2G
E
E
1
1
=
(1 + v)
2G
E
G =
E
2(1 + v)
QED
1440