CAP_13_FLAMBAGEM_DE_COLUNAS

Prévia do material em texto

Problem 13-1
Determine the critical buckling load for the column. The material can be assumed rigid.
Solution:
Equilibrium: The disturbing force F can be deternined by summing
moments about point A.
 ΣΜA=0; P L θ⋅( )⋅ F
L
2
⎛⎜
⎝
⎞
⎠
⋅− 0=
F 2P θ⋅=
Spring Formula :
The restoring spring force Fs can be deternined using
the spring formula Fs = kx.
Fs k
L
2
⋅ θ⋅=
k L⋅ θ⋅
2
=
Critical Buckling Load :
For the mechanism to be on the verge of buckling, the
disturbing force F must be equal to the spring force Fs.
Thus, 2Pcr θ⋅
k L⋅ θ⋅
2
=
Pcr
k L⋅
4
= Ans
Problem 13-2
The column consists of a rigid member 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.
Solution:
Equilibrium:
 ΣΜA=0; P L⋅ sin θ( )⋅ k L⋅ sin θ( )⋅ L cos θ( )⋅( )⋅− 0=
P k L⋅ cos θ( )⋅=
Since θ is small, cos θ( )=1.
Pcr k L⋅= Ans
Problem 13-3
An A-36 steel column has a length of 4 m and is pinned at both ends. If the cross sectional area has the
dimensions shown, determine the critical load.
Given: L 4m:= E 200 GPa⋅:= σY 250MPa:=
b 25mm:= a 10mm:=
Solution:
Section Property: bo 2 b⋅ a+:=
A 2 bo⋅ a⋅ a
2−:=
Ix
a bo
3⋅
12
2 b⋅ a3⋅
12
+:= Iy Ix:=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 22.7 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 20.66 MPa=
 < σY = 250 MPa (O.K.!)
Therefore, Euler's formula is valid.
Problem 13-4
Solve Prob. 13-3 if the column is fixed at its bottom and pinned at its top.
Given: L 4m:= E 200 GPa⋅:= σY 250MPa:=
b 25mm:= a 10mm:=
Solution:
Section Property: bo 2 b⋅ a+:=
A 2 bo⋅ a⋅ a
2−:=
Ix
a bo
3⋅
12
2 b⋅ a3⋅
12
+:= Iy Ix:=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For fixed-pinned ends, K 0.7:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 46.4 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 42.15 MPa=
 < σY = 250 MPa (O.K.!)
Therefore, Euler's formula is valid.
Problem 13-5
A square bar is made from PVC plastic that has a modulus of elasticity of E = 9 GPa and a yield strain
of εY = 0.001 mm/mm. Determine its smallest cross-sectional dimensions a so it does not fail from
elastic buckling. It is pinned at its ends and has a length of 1250 mm.
Given: L 1.250m:= E 9 GPa⋅:=
εY 0.001
mm
mm
:=
Solution:
Stress-strain Relationship: σY E εY⋅:=
σY 9.00 MPa=
Section Property:
A a2= I
a4
12
=
Buckling Load :
Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
σY A⋅
π
2 E⋅ I⋅
K L⋅( )2
= I
σY A⋅ K L⋅( )
2⋅
π
2 E⋅
=
For pinned ends, K 1.0:=
Thus, a
4
12
σY a
2⋅ 1.0L( )2⋅
π
2 E⋅
=
a
12σY L
2⋅
π
2 E⋅
:=
a 43.59 mm= Ans
Problem 13-6
The rod is made from an A-36 steel rod. Determine the smallest diameter of the rod, to the nearest
mm, that will support the load of P = 25 kN without buckling. The ends are roller-supported.
Given: L 500mm:= E 200 GPa⋅:=
Pcr 25kN:=
Solution:
Section Property:
I
π do
4⋅
64
= A
π do
2⋅
4
=
Buckling Load :
Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
I
Pcr K L⋅( )
2⋅
π
2 E⋅
=
For roller ends, K 1.0:=
Thus, π do
4⋅
64
Pcr 1.0 L⋅( )
2⋅
π
2 E⋅
=
do
4
64Pcr L
2⋅
π
3 E⋅
:=
do 15.94 mm= Use do 16mm:= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
= σcr
4Pcr
π do
2⋅
:=
σcr 124.34 MPa= < σY = 250 MPa (O.K.!)
Problem 13-7
The rod is made from a 25-mm-diameter steel rod. Determine the critical buckling load if the ends are
roller supported. Est = 200 GPa, σY = 350 MPa.
Given: L 500mm:= E 200 GPa⋅:=
do 25mm:= σY 350MPa:=
Solution:
Section Property:
I
π do
4⋅
64
:= A
π do
2⋅
4
:=
Buckling Load :
Applying Euler's formula,
For roller ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 151.4 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 308.43 MPa=
 < σY = 350 MPa (O.K.!)
Problem 13-8
An A-36 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.
Given: E 200 GPa⋅:= L 5m:= bo 100mm:= bi 80mm:=
σY 250MPa:= ho 50mm:= hi 30mm:=
Solution:
Section Property:
A bo ho⋅ bi hi⋅−:=
Ix
bo ho
3⋅
12
bi hi
3⋅
12
−:=
Iy
ho bo
3⋅
12
hi bi
3⋅
12
−:=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For fixed ends, K 0.5:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 272.1 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 104.67 MPa=
 < σY = 250 MPa (O.K.!)
Therefore, Euler's formula is valid.
Problem 13-9
An A-36 steel column has a length of 4.5 m and is pinned at both ends. If the cross-sectional area has
the dimensions shown, determine the critical load.
Given: E 200 GPa⋅:= L 4.5m:= bf 200mm:= df 12mm:=
σY 250MPa:= tw 12mm:= dw 150mm:=
Solution:
D dw 2df+:=Section Property:
A bf D⋅ dw bf tw−( )⋅−:=
Ix
bf D
3⋅
12
bf tw−( ) dw3⋅
12
−:=
Iy
2df bf
3⋅
12
dw tw
3⋅
12
+:=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 1561.7 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 236.63 MPa=
 < σY = 250 MPa (O.K.!)
Therefore, Euler's formula is valid.
Problem 13-10
The W250 X 67 is made of A-36 steel and is used as a column that has a length of 4.55 m. If its ends
are assumed pin supported, and it is subjected to an axial load of 500 kN, determine the factor of safety
with respect to buckling.
Given: E 200 GPa⋅:= L 4.55m:=
σY 250MPa:= P 500kN:=
Solution:
Use W 250x67 : Ix 104 10
6( )⋅ mm4:= A 8560mm2:=
Iy 22.2 10
6( )⋅ mm4:=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 2116.7 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 247.28 MPa=
 < σY = 250 MPa (O.K.!)
Factor of safety:
Fsafety
Pcr
P
:= Fsafety 4.23= Ans
Problem 13-11
The W250 X 67 is made of A-36 steel and is used as a column that has a length of 4.55 m. If the ends
of the column are fixed supported, can the column support the critical load without yielding?
Given: E 200 GPa⋅:= L 4.55m:=
σY 250MPa:= P 500kN:=
Solution:
Use W 250x67 : Ix 104 10
6( )⋅ mm4:= A 8560mm2:=
Iy 22.2 10
6( )⋅ mm4:=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For fixed ends, K 0.5:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 8466.8 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 989.11 MPa=
 > σY = 250 MPa (No !) Ans
The column will yield before the axial force achieves the critical load Pcr,
and the Euler's formula is not valid.
Problem 13-12
Determine the maximum force P that can be applied to the handle so that the A-36 steel control rod AB
does not buckle. The rod has a diameter of 30 mm. It is pin connected at its ends.
Given: L 0.9m:= E 200 GPa⋅:= a 0.6m:=
do 30mm:= σY 250MPa:= b 0.9m:=
Solution:
Section Property:
I
π do
4⋅
64
:= A
π do
2⋅
4
:=
Buckling Load : For rod AB,
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:=
Pcr 96.89 kN=
FAB Pcr:=
Equations of Equilibrium : 
 ΣΜh=0; FAB( )− a⋅ P b⋅+ 0=
P
a
b
⎛⎜
⎝
⎞
⎠
FAB:=
P 64.6 kN= Ans
Critical Stress check :
σcr
Pcr
A
:= σcr 137.08 MPa=
 < σY = 250 MPa (O.K.!)
Therefore, Euler's formula is valid.
Problem 13-13
The two steel channels are to be laced together to form a 9-m-long bridge column assumed to be pin
connected at its ends. Each channel has a cross-sectional area of A = 1950 mm2 and moments of
inertia Ix = 21.60(106) mm4, Iy = 0.15(106) mm4. 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 = 200 GPa, σY = 350 MPa.Given: E 200 GPa⋅:= L 9m:=
σY 350MPa:= A' 1950mm
2:=
I'x 21.60 10
6( )⋅ mm4:= c'1 6.5mm:=
I'y 0.15 10
6( )⋅ mm4:= c'2 30mm:=
Solution:
Section Property:
Ix 2I'x:= A 2A':=
Iy 2I'y 2A' 0.5d( )
2⋅+=
In order for the column to buckle about x-x and y-y axes
at the same time, Iy must be equal to Ix.
Iy Ix= 2I'y 2A' 0.5d( )
2⋅+ 2I'x=
d 2
I'x I'y−
A'
:= d 209.76 mm= Ans
 > 2c'2 = 60 mm (O.K. !)
Iy 2I'y 2A' 0.5d( )
2⋅+:=
Iy 43.20 10
6× mm4=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 1052.8 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 269.94 MPa=
 > σY = 350 MPa (O.K. !)
Therefore, Euler's formula is valid.
Problem 13-14
The W200 X 100 is used as a structural A-36 steel column that can be assumed fixed at its base and
pinned at its top. Determine the largest axial force P that can be applied without causing it to buckle.
Given: E 200 GPa⋅:= L 7.5m:=
σY 250MPa:=
Solution:
Use W 200x100 : Ix 133 10
6( )⋅ mm4:= A 12700mm2:=
Iy 36.6 10
6( )⋅ mm4:=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For fixed-pinned ends, K 0.7:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 2621.2 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 206.39 MPa=
 < σY = 250 MPa (O.K.!)
Problem 13-15
Solve Prob. 13-14 if the column is assumed fixed at its bottom and free at its top.
Given: E 200 GPa⋅:= L 7.5m:=
σY 250MPa:=
Solution:
Use W 200x100 : Ix 133 10
6( )⋅ mm4:= A 12700mm2:=
Iy 36.6 10
6( )⋅ mm4:=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For fixed-free ends, K 2.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 321.1 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 25.28 MPa=
 < σY = 250 MPa (O.K.!)
Problem 13-16
A steel column has a length of 9 m and is fixed at both ends. If the cross-sectional area has the
dimensions shown, determine the critical load. Est = 200 GPa, σY = 250 MPa.
Given: E 200 GPa⋅:= L 9m:= bf 200mm:= df 10mm:=
σY 250MPa:= tw 10mm:= dw 150mm:=
Solution:
D dw 2df+:=Section Property:
A bf D⋅ dw bf tw−( )⋅−:=
Ix
bf D
3⋅
12
bf tw−( ) dw3⋅
12
−:=
Iy
2df bf
3⋅
12
dw tw
3⋅
12
+:=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For fixed ends, K 0.5:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 1300.9 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 236.53 MPa=
 < σY = 250 MPa (O.K.!)
Therefore, Euler's formula is valid.
Problem 13-17
Solve Prob. 13-16 if the column is pinned at its top and bottom.
Given: E 200 GPa⋅:= L 9m:= bf 200mm:= df 10mm:=
σY 250MPa:= tw 10mm:= dw 150mm:=
Solution:
D dw 2df+:=Section Property:
A bf D⋅ dw bf tw−( )⋅−:=
Ix
bf D
3⋅
12
bf tw−( ) dw3⋅
12
−:=
Iy
2df bf
3⋅
12
dw tw
3⋅
12
+:=
Buckling Load :
Applying Euler's formula, I min Ix Iy,( ):=
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 325.2 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 59.13 MPa=
 < σY = 250 MPa (O.K.!)
Therefore, Euler's formula is valid.
Problem 13-18
The 3.6-m A-36 steel pipe column has an outer diameter of 75 mm and a thickness of 6 mm.
Determine the critical load if the ends are assumed to be pin connected.
Given: E 200 GPa⋅:= L 3.6m:=
σY 250MPa:= do 75mm:= t 6mm:=
Solution:
Section Property: di do 2t−:=
A
π
4
do
2 di
2−⎛⎝
⎞
⎠⋅:= I
π
64
do
4 di
4−⎛⎝
⎞
⎠⋅:=
Buckling Load :
Applying Euler's formula,
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 118.8 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 91.33 MPa=
 < σY = 250 MPa (O.K.!)
Problem 13-19
The 3.6-m A-36 steel column has an outer diameter of 75 mm and a thickness of 6 mm. Determine the
critical load if the bottom is fixed and the top is pinned.
Given: E 200 GPa⋅:= L 3.6m:=
σY 250MPa:= do 75mm:= t 6mm:=
Solution:
Section Property: di do 2t−:=
A
π
4
do
2 di
2−⎛⎝
⎞
⎠⋅:= I
π
64
do
4 di
4−⎛⎝
⎞
⎠⋅:=
Buckling Load :
Applying Euler's formula,
For fixed-pinned ends, K 0.7:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 242.4 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 186.38 MPa=
 < σY = 250 MPa (O.K.!)
Problem 13-20
The 3-m wooden rectangular column has the dimensions shown. Determine the critical load if the ends
are assumed to be pin connected. Ew = 12 GPa, σY = 35 MPa.
Given: E 12 GPa⋅:= L 3m:=
σY 35MPa:= b 50mm:= h 100mm:=
Solution:
Section Property:
A b h⋅:= Ix
b h3⋅
12
:= Iy
h b3⋅
12
:=
Buckling Load : I min Ix Iy,( ):=
Applying Euler's formula,
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 13.7 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 2.74 MPa=
 < σY = 35 MPa (O.K.!)
Problem 13-21
The 3-m column has the dimensions shown. Determine the critical load if the bottom is fixed and the
top is pinned. Ew = 12 GPa, σY = 35 MPa.
Given: E 12 GPa⋅:= L 3m:=
σY 35MPa:= b 50mm:= h 100mm:=
Solution:
Section Property:
A b h⋅:= Ix
b h3⋅
12
:= Iy
h b3⋅
12
:=
Buckling Load : I min Ix Iy,( ):=
Applying Euler's formula,
For fixed-pinned ends, K 0.7:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 28 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 5.60 MPa=
 < σY = 35 MPa (O.K.!)
Problem 13-22
The members of the truss are assumed to be pin connected. If member BD is an A-36 steel rod of
radius 50 mm, determine the maximum load P that can be supported by the truss without causing the
member to buckle.
Given: a 4m:= E 200 GPa⋅:=
h 3m:= σY 250MPa:=
ro 50mm:=
Solution:
Section Property:
I
π ro
4⋅
4
:= A π ro
2⋅:=
Buckling Load : LBD a:=
Applying Euler's formula,
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K LBD⋅( )2
:=
Pcr 605.6 kN= FBD Pcr:=
Equations of Equilibrium : 
 
ΣΜC=0; R a⋅ FBD h⋅− 0= R
h
a
⎛⎜
⎝
⎞
⎠
FBD:=
Support Reactions : By symmetry, A=B=R
+ ΣFy=0; 2R 2P− 0= P R:=
P 454.19 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 77.11 MPa=
 < σY = 250 MPa (O.K.!)
Problem 13-23
Solve Prob. 13-22 in the case of member AB, which has a radius of 50 mm.
Given: a 4m:= E 200 GPa⋅:=
h 3m:= σY 250MPa:=
ro 50mm:=
Solution:
Section Property:
I
π ro
4⋅
4
:= A π ro
2⋅:=
Buckling Load : LAB a
2 h2+:=
Applying Euler's formula,
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K LAB⋅( )2
:=
Pcr 387.6 kN= FAB Pcr:=
At Joint A: Equations of Equilibrium : 
+ ΣFy=0; R FAB
h
LAB
⎛
⎜
⎝
⎞
⎠
⋅− 0= R FAB
h
LAB
⎛
⎜
⎝
⎞
⎠
⋅:=
Support Reactions : By symmetry, A=B=R
+ ΣFy=0; 2R 2P− 0= P R:=
P 232.55 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 49.35 MPa=
 < σY = 250 MPa (O.K.!)
Problem 13-24
The truss is made from A-36 steel bars, each of which has a circular cross section with a diameter of
40 mm. 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.
Given: a 1.2m:= E 200 GPa⋅:= do 40mm:=
h 0.9m:= σY 250MPa:=
Solution:
Section Property: I
π do
4⋅
64
:= A
π
4
do
2⋅:=
Buckling Load :
 Member AB and BC are in compression.
LAB 2a:= LBC a
2 h2+:= LAC LBC:=
Applying Euler's formula,
Assume failure of rod AB: Assume failure of rod BC:
For pinned ends, K 1.0:= For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K LAB⋅( )2
:= P'cr
π
2 E⋅ I⋅
K LBC⋅( )2
:=
Pcr 43.1 kN= P'cr 110.2 kN=
FAB Pcr:= FBC P'cr:=
At Joint A: Equations of Equilibrium : 
+ ΣFy=0; FAC
h
LAC
⋅ P− 0= FAC P
LAC
h
⋅=
+ ΣFx=0; FAC
a
LAC
⋅ FAB− 0= FAB FAC
a
LAC
⋅= FAB P
a
h
⋅=
Pcr P
a
h
⋅= P Pcr
h
a
⋅:= P 32.30 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 34.27 MPa= < σY = 250 MPa (O.K.!)
Check Rod BC :
Support Reactions :
 
ΣΜD=0; P− 2a( )⋅ Bh h⋅+ 0= Bh P
2a
h
⋅=
At Joint B: Equations of Equilibrium : 
+ ΣFx=0; FAB FBC
a
LAC
⋅+ Bh− 0=
P
a
h
⋅ P'cr
a
LAC⋅+ P
2a
h
⋅− 0=
P P'cr
h
LAC
⋅:= P 66.15 kN= (Not Control !)
h 0.9m:=Given: a 1.2m:= E 200 GPa⋅:=
Equations of Equilibrium : 
P 50kN:= σY 250MPa:=
Solution:
At Joint A:
(O.K.!)
+ ΣFy=0; FAC
h
LAC
⋅ P− 0= FAC P
LAC
h
⋅=
+ ΣFx=0; FAC
a
LAC
⋅ FAB− 0=
FAB FAC
a
LAC
⋅= FAB P
a
h
⋅:=
Buckling Load : LAB 2a:=
Applying Euler's formula,
For pinned ends, K 1.0:= Pcr
π
2 E⋅ I⋅
K LAB⋅( )2
=
I
1.0 2a( )⋅[ ]2
π
2 E⋅
Pcr⋅=
Pcr FAB:= I
4a2
π
2 E⋅
P
a
h
⋅⎛⎜
⎝
⎞
⎠
⋅=
Section Property: I
π do
4⋅
64
= A
π
4
do
2⋅=
π do
4⋅
64
4a2
π
2 E⋅
P
a
h
⋅⎛⎜
⎝
⎞
⎠
⋅=
do
4
256a2
π
3 E⋅
P
a
h
⋅⎛⎜
⎝
⎞
⎠
⋅:=
do 44.62 mm=
Use: do 45mm:= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
A
π
4
do
2⋅:= σcr
Pcr
A
:= σcr 41.92 MPa=
 < σY = 250 MPa
Problem 13-25
The truss is made from A-36 steel bars, each of which has a circular cross section. If the applied load
P = 50 kN, determine the diameter of member AB to the nearest multiples of 5mm that will prevent this
member from buckling. The members are pin supported at their ends.
Problem 13-26
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.
Given: E 200 GPa⋅:= L 0.6m:= Fsafety 1.75:=
σY 703MPa:= b 36mm:= t 12mm:=
Solution:
Section Property:
Ix
t b3⋅
12
:= Ix 46656.00 mm
4=
Iy
b t3⋅
12
:= Iy 5184.00 mm
4=
A b t⋅:= A 432.00 mm2=
Critical Buckling Load :
With respect to the x-x axis: for pinned ends, K 1.0:=
Applying Euler's formula,
Pcr
π
2 E⋅ Ix⋅
K L⋅( )2
:= Pcr 255.8 kN=
With respect to the y-y axis: for fixed ends, K 0.5:=
Applying Euler's formula,
P'cr
π
2 E⋅ Iy⋅
K L⋅( )2
:= P'cr 113.7 kN= (Controls !)
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
P'cr
A
:= σcr 263.19 MPa=
 < σY = 703 MPa (O.K.!)
Factor of safety:
Fsafety
P'cr
P
=
P
P'cr
Fsafety
:=
P 64.97 kN= Ans
σY 250MPa:=
Given: h 3.6m:= θAB 45deg:= E 200 GPa⋅:=
P 30kN:= θBC 30deg:=
(1)
Solution:
At Joint B: Equations of Equilibrium : Given
+ ΣFy=0; FAB cos θAB( )⋅ FBC cos θBC( )⋅+ P− 0=
(O.K.!)
+ ΣFx=0; FAB sin θAB( )⋅ FBC sin θBC( )⋅− 0= (2)
Solving Eqs. (1) and (2): Guess FAB 1kN:= FBC 1kN:=
FAB
FBC
⎛⎜
⎜⎝
⎞
⎠
Find FAB FBC,( ):=
FAB
FBC
⎛⎜
⎜⎝
⎞
⎠
15.53
21.96
⎛
⎜
⎝
⎞
⎠
kN=
Section Property: I
π do
4⋅
64
= A
π
4
do
2⋅=
Buckling Load : Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
For pinned ends, K 1.0:=
I
1.0 L( )⋅[ ]2
π
2 E⋅
Pcr⋅=
π do
4⋅
64
L2
π
2 E⋅
Pcr⋅= do
4
64L2
π
3 E⋅
Pcr( )⋅=
For Member AB: LAB
h
cos θAB( )
:=
Pcr FAB:= L LAB:= dAB
4
64L2
π
3 E⋅
Pcr( )⋅:= dAB 45.15 mm=
Use: dAB 50mm:= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
A
π
4
dAB
2⋅:= σcr
Pcr
A
:= σcr 7.91 MPa= < σY = 250 MPa (O.K.!)
For Member BC: LBC
h
cos θBC( )
:=
P'cr FBC:= L' LBC:= dBC
4
64L'2
π
3 E⋅
P'cr( )⋅:= dBC 44.49 mm=
Use: dBC 45mm:= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
A'
π
4
dBC
2⋅:= σ'cr
P'cr
A'
:= σ'cr 13.81 MPa= < σY = 250 MPa
Problem 13-27
The linkage is made using two A-36 steel rods, each having a circular cross section. Determine the
diameter of each rod to the nearest multiples of 5mm that will support a load of P = 30 kN. Assume
that the rods are pin connected at their ends. Use a factor of safety with respect to buckling of 1.8.
Problem 13-28
The linkage is made using two A-36 steel rods, each having a circular cross section. If each rod has a
diameter of 20 mm, determine the largest load it can support without causing any rod to buckle.
Assume that the rods are pin-connected at their ends.
Given: h 3.6m:= θAB 45deg:= E 200 GPa⋅:=
do 20mm:= θBC 30deg:= σY 250MPa:=
Solution: Set P' 1:=
At Joint B: Equations of Equilibrium : Given
+ ΣFy=0; F'AB cos θAB( )⋅ F'BC cos θBC( )⋅+ P'− 0= (1)
+ ΣFx=0; F'AB sin θAB( )⋅ F'BC sin θBC( )⋅− 0= (2)
Solving Eqs. (1) and (2): Guess F'AB 1:= F'BC 1:=
F'AB
F'BC
⎛⎜
⎜⎝
⎞
⎠
Find F'AB F'BC,( ):=
F'AB
F'BC
⎛⎜
⎜⎝
⎞
⎠
0.5176
0.7321
⎛
⎜
⎝
⎞
⎠
=
FAB
FBC
⎛⎜
⎜⎝
⎞
⎠
F'AB P⋅
F'BC P⋅
⎛⎜
⎜⎝
⎞
⎠
kN=
Section Property: I
π do
4⋅
64
:= A
π
4
do
2⋅:=
Buckling Load : Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
For pinned ends, K 1.0:= Pcr
π
2 E⋅ I⋅
L2
=
For Member AB: LAB
h
cos θAB( )
:=
Pcr FAB= L LAB:= F'AB P⋅
π
2 E⋅ I⋅
LAB
2
= P
π
2 E⋅ I⋅
F'AB LAB
2⋅
:=
P 1.16 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr F'AB P⋅:= σcr
Pcr
A
:= σcr 1.90 MPa= < σY = 250 MPa (O.K.!)
Check Member BC: LBC
h
cos θBC( )
:= FBC F'BC P⋅:= FBC 0.846 kN=
P'cr
π
2 E⋅ I⋅
LBC
2
:= P'cr 0.8972 kN= > FBC = 0.846 kN (O.K.!)
Problem 13-29
The A-36 steel pipe has an outer diameter of 50 mm and a thickness of 12 mm. If it is held in place by
a guywire, determine the largest vertical force P that can be applied without causing the pipe to buckle.
Assume that the ends of the pipe are pin connected.
Given: L 4.2m:= θBC 30deg:= E 200 GPa⋅:=
do 50mm:= t 12mm:= σY 250MPa:=
Solution: Set P' 1:=
At Joint B: Equations of Equilibrium : Given
+ ΣFy=0; F'BC sin θBC( )⋅ P'− 0= (1)
+ ΣFx=0; F'BC cos θBC( )⋅ F'AB− 0= (2)
Solving Eqs. (1) and (2): Guess F'AB 1:= F'BC 1:=
F'AB
F'BC
⎛⎜
⎜⎝
⎞
⎠
Find F'AB F'BC,( ):=
F'AB
F'BC
⎛⎜
⎜⎝
⎞
⎠
1.7321
2.0000
⎛
⎜
⎝
⎞
⎠
=
FAB
FBC
⎛⎜
⎜⎝
⎞
⎠
F'AB P⋅
F'BC P⋅
⎛⎜
⎜⎝
⎞
⎠
kN=
Section Property: di do 2t−:=
A
π
4
do
2 di
2−⎛⎝
⎞
⎠⋅:= I
π do
4 di
4−⎛⎝
⎞
⎠⋅
64
:=
Buckling Load : Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
For pinned ends, K 1.0:= Pcr
π
2 E⋅ I⋅
L2
=
Pcr FAB= F'AB P⋅
π
2 E⋅ I⋅
L2
= P
π
2 E⋅ I⋅
F'AB L
2⋅
:=
P 18.37 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr F'AB P⋅:= σcr
Pcr
A
:= σcr 22.21 MPa= < σY = 250 MPa (O.K.!)
Problem 13-30
The A-36 steel pipe has an outer diameter of 55 mm. If it is held in place by a guywire, determine its
required inner diameter to the nearest multiples of 5mm, so that it can support a maximum vertical load
of P = 20 kN without causing the pipe to buckle. Assume the ends of the pipe are pin connected.
Given: L 4.2m:= θBC 30deg:= E 200 GPa⋅:=
do 55mm:= P 20kN:= σY 250MPa:=
Solution:
At Joint B: Equations of Equilibrium : Given
+ ΣFy=0; FBC sin θBC( )⋅ P− 0= (1)
+ ΣFx=0; FBC cos θBC( )⋅ FAB− 0= (2)
Solving Eqs. (1) and (2): Guess FAB 1kN:= FBC 1kN:=
FAB
FBC
⎛⎜
⎜⎝
⎞
⎠
Find FAB FBC,( ):=
FAB
FBC
⎛⎜
⎜⎝
⎞
⎠
34.64
40.00
⎛
⎜
⎝
⎞
⎠
kN=
Section Property: di do 2t−=
A
π
4
do
2 di
2−⎛⎝
⎞
⎠⋅= I
π do
4 di
4−⎛⎝
⎞
⎠⋅
64
=
Buckling Load : Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
For pinned ends, K 1.0:= Pcr FAB= FAB
π
2 E⋅ I⋅
L2
=
I
FAB L
2⋅
π
2 E⋅
:=
π do
4 di
4−⎛⎝
⎞
⎠⋅
64
FAB L
2⋅
π
2 E⋅
=
di
4
do
4 64 FAB⋅ L
2
π
3 E⋅
−:= di 41.07 mm=
Use: di 40mm:= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
A
π
4
do
2 di
2−⎛⎝
⎞
⎠⋅:= I
π do
4 di
4−⎛⎝
⎞
⎠⋅
64
:= Pcr
π
2 E⋅ I⋅
K L⋅( )2
:=
σcr
Pcr
A
:= σcr 32.35 MPa= < σY = 250 MPa (O.K.!)
Problem 13-31
The linkage is made using two A-36 steel rods, each having a circular cross section. Determine the
diameter of each rod to the nearest multiples of 5mm that will support the 4.5-kN load. Assume that
the rods are pin connected at their ends. Use a factor of safety with respect to buckling of F.S. = 1.8.
Given: a 4.8m:= h 3.6m:= E 200 GPa⋅:=
b 2.7m:= P 4.5kN:= σY 250MPa:=
Fsafety 1.8:=
Solution: LAB a
2 h2+:= LBC b
2 h2+:=
At Joint B: Equations of Equilibrium : Given
+ ΣFy=0; FAB( )
h
LAB
⋅ FBC( )
h
LBC
⋅+ P− 0= (1)
+ ΣFx=0; FAB( )
a
LAB
⋅ FBC( )
b
LBC
⋅− 0= (2)
Solving Eqs. (1) and (2): Guess FAB 1kN:= FBC 1kN:=
FAB
FBC
⎛⎜
⎜⎝
⎞
⎠
Find FAB FBC,( ):=
FAB
FBC
⎛⎜
⎜⎝
⎞
⎠
2.7
3.6
⎛
⎜
⎝
⎞
⎠
kN=
Section Property: I
π do
4⋅
64= A
π
4
do
2⋅=
Buckling Load : Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
For pinned ends, K 1.0:=
I
1.0 L( )⋅[ ]2
π
2 E⋅
Pcr⋅=
π do
4⋅
64
L2
π
2 E⋅
Pcr⋅= do
4
64L2
π
3 E⋅
Pcr( )⋅=
For Member AB:
Pcr Fsafety FAB⋅:= L LAB:= dAB
4
64L2
π
3 E⋅
Pcr( )⋅:= dAB 36.66 mm=
Use: dAB 40mm:= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
A
π
4
dAB
2⋅:= σcr
Pcr
A
:= σcr 3.87 MPa= < σY = 250 MPa (O.K.!)
For Member BC:
P'cr Fsafety FBC⋅:= L' LBC:= dBC
4
64L'2
π
3 E⋅
P'cr( )⋅:= dBC 34.11 mm=
Use: dBC 40mm:= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
A'
π
4
dBC
2⋅:= σ'cr
P'cr
A'
:= σ'cr 5.16 MPa= < σY = 250 MPa (O.K.!)
Problem 13-32
The linkage is made using two A-36 steel rods, each having a circular cross section. If each rod has a
diameter of 20 mm, determine the largest load it can support without causing any rod to buckle.
Assume that the rods are pin connected at their ends.
Given: a 4.8m:= h 3.6m:= E 200 GPa⋅:=
b 2.7m:= do 20mm:= σY 250MPa:=
Fsafety 1.8:=
Solution: LAB a
2 h2+:= LBC b
2 h2+:=
Set P' 1:=
At Joint B: Equations of Equilibrium : Given
+ ΣFy=0; F'AB( )
h
LAB
⋅ F'BC( )
h
LBC
⋅+ P'− 0= (1)
+ ΣFx=0; F'AB( )
a
LAB
⋅ F'BC( )
b
LBC
⋅− 0= (2)
Solving Eqs. (1) and (2): Guess F'AB 1:= F'BC 1:=
F'AB
F'BC
⎛⎜
⎜⎝
⎞
⎠
Find F'AB F'BC,( ):=
F'AB
F'BC
⎛⎜
⎜⎝
⎞
⎠
0.6
0.8
⎛
⎜
⎝
⎞
⎠
=
FAB
FBC
⎛⎜
⎜⎝
⎞
⎠
F'AB P⋅
F'BC P⋅
⎛⎜
⎜⎝
⎞
⎠
kN=
Section Property: I
π do
4⋅
64
:= A
π
4
do
2⋅:=
Buckling Load : Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
For pinned ends, K 1.0:= Pcr
π
2 E⋅ I⋅
L2
=
For Member AB:
Pcr FAB= L LAB:= F'AB P⋅
π
2 E⋅ I⋅
LAB
2
= P
π
2 E⋅ I⋅
F'AB LAB
2⋅
:=
P 0.718 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr F'AB P⋅:= σcr
Pcr
A
:= σcr 1.37 MPa= < σY = 250 MPa (O.K.!)
Check Member BC: FBC F'BC P⋅:= FBC 0.574 kN=
P'cr
π
2 E⋅ I⋅
LBC
2
:= P'cr 0.7656 kN= > FBC = 0.574 kN (O.K.!)
a'h
ah
LAC
:=
Problem 13-33
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, σY = 360 MPa.
Given: h 100mm:= av 4m:= E 200 GPa⋅:=
b 50mm:= ah 3m:= σY 360MPa:=
P 18kN:= LAB 6m:= LAC av
2 ah
2+:=
Solution: a'v
av
LAC
:=
At Joint B: Equations of Equilibrium : 
+ ΣFy=0; FAC− a'v( )⋅ FAB+ 0= (1)
+ ΣFx=0; FAC− a'h( )⋅ P+ 0= (2)
Solving Eqs. (1) and (2): FAC
P
a'h
:= FAB P
a'v
a'h
⋅:=
FAC 30 kN= FAB 24 kN=
Section Property: Iy
h b3⋅
12
:= A h b⋅:=
Buckling Load : Applying Euler's formula,
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ Iy⋅
K LAB⋅( )2
:= Pcr 57.12 kN=
Fsafety
Pcr
FAB
:= Fsafety 2.38= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 11.42 MPa= < σY = 360 MPa (O.K.!)
Problem 13-34
Determine the maximum load P the frame can support without buckling member AB. Assume that AB
is made of steel and is pinned at its ends for y-y axis buckling and fixed at its ends for x-x axis
buckling. Est = 200 GPa, σY = 360 MPa.
Given: h 100mm:= av 4m:= E 200 GPa⋅:=
b 50mm:= ah 3m:= σY 360MPa:=
P 18kN:= LAB 6m:= LAC av
2 ah
2+:=
Solution: a'v
av
LAC
:= a'h
ah
LAC
:=
At Joint B: Equations of Equilibrium : 
+ ΣFy=0; FAC− a'v( )⋅ FAB+ 0= (1)
+ ΣFx=0; FAC− a'h( )⋅ P+ 0= (2)
Solving Eqs. (1) and (2): FAC
P
a'h
:= FAB P
a'v
a'h
⋅:= (3)
FAC 30 kN= FAB 24 kN=
Section Property: A h b⋅:= Ix
b h3⋅
12
:= Iy
h b3⋅
12
:=
Buckling Load : Applying Euler's formula,
About x-x axix: About y-y axix:
For fixed ends, For pinned ends, 
P''cr
π
2 E⋅ Iy⋅
Ky LAB⋅( )2
:=
P'cr 913.85 kN= P''cr 57.12 kN=
Pcr min P'cr P''cr,( ):= Pcr 57.12 kN=
Substituting FAB=Pcr into Eq.(3), P Pcr
a'h
a'v
⋅:= P 42.84 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 11.42 MPa= < σY = 360 MPa (O.K.!)
P'cr
π
2 E⋅ Ix⋅
Kx LAB⋅( )2
:=
Kx 0.5:= Ky 1.0:=
Problem 13-35
Determine the maximum force P that can be applied to the handle so that the A-36 steel control rod BC
does not buckle. The rod has a diameter of 25 mm.
Given: a 350mm:= θ 45deg:= E 200 GPa⋅:=
b 250mm:= do 25mm:= σY 250MPa:=
c 800mm:= P kN:=
Solution:
Equations of Equilibrium: For the handle.
 ΣΜA=0; P a( )⋅ FBC b cos θ( )⋅( )⋅− 0=
FBC
P a⋅
b cos θ( )⋅
:= FBC 1.9799 P=
Section Property: I
π do
4⋅
64
:= A
π
4
do
2⋅:=
Buckling Load : Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
For pinned ends, K 1.0:= Pcr FBC:= L c:=
FBC
π
2 E⋅ I⋅
K c⋅( )2
=
P a⋅
b cos θ( )⋅
π
2 E⋅ I⋅
K c⋅( )2
=
P
π
2 E⋅ I⋅ b⋅ cos θ( )⋅
a K c⋅( )2⋅
:=
P 29.870 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr
P a⋅
b cos θ( )⋅
:= σcr
Pcr
A
:= σcr 120.48 MPa= < σY = 250 MPa (O.K.!)
Problem 13-36
Determine the maximum allowable load P that can be applied to member BC without causing member
AB to buckle. Assume that AB is made of steel and is pinned at its ends for x-x axis buckling and fixed
at its ends for y-y axis buckling. Use a factor of safety with respect to buckling of F.S. = 3. Est = 200
GPa, σY = 360 MPa.
Given: h 30mm:= L 2m:= E 200 GPa⋅:=
b 20mm:= Fsafety 3:= σY 360MPa:=
Solution: P kN:=
Support Reactions : For member BC.
+ ΣFy=0; FBA FC+ P− 0=
By symmetry, FBA FC=
Hence, FBA 0.5P:= (1)
FC 0.5P:=
Section Property: A h b⋅:=
Ix
b h3⋅
12
:= Iy
h b3⋅
12
:=
Buckling Load : Applying Euler's formula,
About x-x axix: About y-y axix:
For pinned ends, Kx 1.0:= For fixed ends, Ky 0.5:=
P'cr
π
2 E⋅ Ix⋅
Kx L⋅( )2
:= P''cr
π
2 E⋅ Iy⋅
Ky L⋅( )2
:=
P'cr 22.21 kN= P''cr 39.48 kN=
Pcr min P'cr P''cr,( ):= Pcr 22.207 kN=
FBA
Pcr
Fsafety
:= FBA 7.402 kN=
From Eq.(1), P 2FBA:= P 14.80 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 37.01 MPa= < σY = 360 MPa (O.K.!)
Problem 13-37
Determine the maximum allowable load P that can be applied to member BC without causing member
AB to buckle. Assume that AB is made of steel and is pinned at its ends for x-x axis buckling and fixed
at its ends for y-y axis buckling. Use a factor of safety with respect to buckling of F.S. = 3. Est = 200
GPa, σY = 360 MPa.
Given: h 30mm:= L 2m:= E 200 GPa⋅:=
b 20mm:= Fsafety 3:= σY 360MPa:=
Solution: P kN:=
Support Reactions : For member BC.
+ ΣFy=0; FBA FC+ P− 0=
By symmetry, FBA FC=
Hence, FBA 0.5P:= (1)
FC 0.5P:=
Section Property: A h b⋅:=
Ix
b h3⋅
12
:= Iy
h b3⋅
12
:=
Buckling Load : Applying Euler's formula,
About x-x axix: About y-y axix:
For pinned ends, Kx 1.0:= For fixed ends, Ky 0.5:=
P'cr
π
2 E⋅ Ix⋅
Kx L⋅( )2
:= P''cr
π
2 E⋅ Iy⋅
Ky L⋅( )2
:=
P'cr 22.21 kN= P''cr 39.48 kN=
Pcr min P'cr P''cr,( ):= Pcr 22.207 kN=
FBA
Pcr
Fsafety
:= FBA 7.402 kN=
From Eq.(1), P 2FBA:= P 14.80 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 37.01 MPa= < σY = 360 MPa (O.K.!)
Problem 13-38
Determine if the frame can support a load of P = 20 kN if the factor of safety with respect to buckling
of member AB is F.S. = 3. Assume that AB is made of steel and is pinned at its ends for x-x axis
buckling and fixed at its ends for y-y axis buckling. Est = 200 GPa, σY = 360 MPa.
Given: h 30mm:= L 2m:= E 200 GPa⋅:=
b 20mm:= Fsafety 3:= σY 360MPa:=
P 20kN:=
Solution:
Support Reactions : For member BC.
+ ΣFy=0; FBA FC+ P− 0=
By symmetry, FBA FC=
Hence, FBA 0.5P:= (1)
FC 0.5P:=
Section Property: A h b⋅:=
Ix
b h3⋅
12
:= Iy
h b3⋅
12
:=
Buckling Load : Applying Euler's formula,
About x-x axix: About y-y axix:
For pinned ends, Kx 1.0:= For fixed ends, Ky 0.5:=
P'cr
π
2 E⋅ Ix⋅
Kx L⋅( )2
:= P''cr
π
2 E⋅ Iy⋅
Ky L⋅( )2
:=
P'cr 22.21 kN= P''cr 39.48 kN=
Pcr min P'cr P''cr,( ):= Pcr 22.207 kN=
Factor of Safety :
FS
Pcr
FBA
:=
FS 2.221= < Fsafety
Hence, the frame cannot support the load with the required factor of safety. Ans
Problem 13-39
The members of the truss are assumed to be pinconnected. If member GF is an A-36 steel rod having
a diameter of 50 mm, determine the greatest magnitude of load P that can be supported by the truss
without causing this member to buckle.
Given: a 4m:= E 200 GPa⋅:=
h 3m:= σY 250MPa:=
do 50mm:=
Solution:
Section Property:
I
π do
4⋅
64
:= A
π
4
do
2⋅:=
Buckling Load : LGF a:=
Applying Euler's formula,
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K LGF⋅( )2
:=
Pcr 37.8 kN= FGF Pcr:=
Equations of Equilibrium : Use method of sections.
 
ΣΜB=0; R a⋅ FGF h⋅− 0= R
h
a
⎛⎜
⎝
⎞
⎠
FGF:=
Support Reactions : By symmetry, A=B=R
+ ΣFy=0; 2R 2P− 0= P R:=
P 28.39 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 19.28 MPa=
 < σY = 250 MPa (O.K.!)
Problem 13-40
The members of the truss are assumed to be pin connected. If member AG is an A-36 steel rod having
a diameter of 50 mm, determine the greatest magnitude of load P that can be supported by the truss
without causing this member to buckle.
Given: a 4m:= E 200 GPa⋅:=
h 3m:= σY 250MPa:=
do 50mm:=
Solution:
Section Property:
I
π do
4⋅
64
:= A
π
4
do
2⋅:=
Buckling Load : LAG a
2 h2+:=
Applying Euler's formula,
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ I⋅
K LAG⋅( )2
:=
Pcr 24.2 kN= FAG Pcr:=
Equations of Equilibrium : Use method of joints at Joint A.
+ ΣFy=0; R FAG( )
h
LAG
⋅− 0= R
h
LAG
⎛
⎜
⎝
⎞
⎠
FAG:=
Support Reactions : By symmetry, A=B=R
+ ΣFy=0; 2R 2P− 0= P R:=
P 14.53 kN= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 12.34 MPa=
 < σY = 250 MPa (O.K.!)
w
kN
m
:=
Problem 13-41
Determine the maximum distributed loading that can be applied to the wide-flange beam so that the
brace CD does not buckle. The brace is an A-36 steel rod having a diameter of 50 mm.
Given: b 2m:= do 50mm:= E 200 GPa⋅:=
Lc 4m:= σY 250MPa:=
Solution: Lb 2 b⋅:=
Equations of Equilibrium: For the beam.
 ΣΜA=0; w Lb⋅ 0.5Lb( )⋅ FCD b( )⋅− 0=
FCD
w Lb
2⋅
2 b⋅
:= FCD 4.00 w m⋅=
Section Property: I
π do
4⋅
64
:= A
π
4
do
2⋅:=
Buckling Load : Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
For pinned ends, K 1.0:= Pcr FCD:= L Lc:=
FCD
π
2 E⋅ I⋅
K Lc⋅( )2
=
w Lb
2⋅
2 b( )⋅
π
2 E⋅ I⋅
K Lc⋅( )2
=
w
2π2 E⋅ I⋅ b⋅
Lb
2 K Lc⋅( )2⋅
:=
w 9.462
kN
m
= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr
π
2 E⋅ I⋅
K Lc⋅( )2
:= σcr
Pcr
A
:= σcr 19.28 MPa= < σY = 250 MPa (O.K.!)
Problem 13-42
The 50-mm diameter C86100 bronze rod is fixed supported at A and has a gap of 2 mm from the wall
at B. Determine the increase in temperature ∆T that will cause the rod to buckle. Assume that the
contact at B acts as a pin.
Unit used: °C deg:=
Given: L 1m:= do 50mm:= go 2mm:=
α 17 10 6−( )⋅ 1
°C
:= E 103GPa:= σY 345MPa:=
Solution:
Section Property: I
π do
4⋅
64
:= A
π
4
do
2⋅:=
Compatibity Condition : This requires,
+ go δT δF+=
go α ∆T( )⋅ L⋅
F L⋅
A E⋅
−=
F A E⋅ α ∆T⋅
go
L
−
⎛
⎜
⎝
⎞
⎠
⋅=
Buckling Load : Applying Euler's formula, Pcr
π
2 E⋅ I⋅
K L⋅( )2
=
For fixed-pinned ends, K 0.7:= Pcr F:=
A E⋅ α ∆T⋅
go
L
−
⎛
⎜
⎝
⎞
⎠
⋅
π
2 E⋅ I⋅
K L⋅( )2
=
∆T
1
α
π
2 I⋅
A K L⋅( )2
go
L
+
⎡⎢
⎢⎣
⎤⎥
⎥⎦
:=
∆T 302.78 °C= Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= σcr
Pcr
A
:= σcr 324.16 MPa= < σY = 345 MPa (O.K.!)
Problem 13-43
Consider an ideal column as in Fig. 13-12c, having both ends fixed. Show that the critical load on the
column is given by Pcr = 4π
2EI/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 ./PM')()( ++= xEIPCxEIPCv cossin 21
Problem 13-44
Consider an ideal column as in Fig. 13-12d, having one end fixed and the other pinned.Show that the
critical load on the column is given by Pcr = 20.19 EI/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 + (P/EI) v = (R'/EI) (L-x). The
solution is of the form
After application of the boundary conditions show that . Solve by trial and
error for the smallest root.
).()/()()( xLPR'xEIPCxEIPCv −++= cossin 21
.)( LEIPLEIP =tan
Problem 13-45
The column is supported at B by a support that does not permit rotation but allows vertical deflection.
Determine the critical load Pcr . EI is constant.
Problem 13-46
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 = A sin kx + B cos kx - c2x/k2, where c2 = F/2EI,
k2 = P/EI.
Problem 13-47
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 = A sin kx + B cos kx + (w/(2P)) x2 - (wL/(2P)) x - (wEI/P2) where k2 = P/EI.
Problem 13-48
Determine the load P required to cause the A-36 steel W200 X 22 column to fail either by buckling or
by yielding. The column is fixed at its base and free at its top.
Given: E 200 GPa⋅:= L 2.4m:=
σY 250MPa:= ey 25mm:=
Solution:
Use W 200x22 : Ix 20.0 10
6( )⋅ mm4:= A 2860mm2:=
Iy 1.42 10
6( )⋅ mm4:= d 206mm:= rx 83.6mm:=
Buckling about y-y axis :
For fixed-free ends, K 2.0:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 121.66 kN=
P Pcr:= P 121.66 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 42.54 MPa=
 < σY = 250 MPa (O.K.!)
Check yielding about x-x axis :
Applying the secant formula: c 0.5d:=
σmax
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅:=
σmax 59.69 MPa= < σY = 250 MPa (O.K.!)
Problem 13-49
The wood column is assumed fixed at its base and pinned at its top. Determine the maximum eccentric
load P that can be applied without causing the column to buckle or yield. Ew = 12 GPa, σY = 56 MPa.
Given: E 12 GPa⋅:= L 3.6m:= b 38mm:=
σY 56MPa:= ey 44mm:= h 88mm:=
Solution:
Section Property:
Ix
b h3⋅
12
:= Iy
h b3⋅
12
:=
A b h⋅:= rx
Ix
A
:=
Buckling about y-y axis :
For fixed-pinned ends, K 0.7:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 7.5 kN=
P Pcr:= P 7.50 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 2.24 MPa=
 < σY = 56 MPa (O.K.!)
Check yielding about x-x axis :
Applying the secant formula: c 0.5h:=
σmax
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅:=
σmax 10.89 MPa= < σY = 56 MPa (O.K.!)
Problem 13-50
The wood column is fixed at its base and fixed at its top. Determine the maximum eccentric load P that
can be applied at its top without causing the column to buckle or yield. Ew = 12 GPa, σY = 56 MPa.
Given: E 12 GPa⋅:= L 3.6m:= b 38mm:=
σY 56MPa:= ey 44mm:= h 88mm:=
Solution:
Section Property:
Ix
b h3⋅
12
:= Iy
h b3⋅
12
:=
A b h⋅:= rx
Ix
A
:=
Buckling about y-y axis :
For fixed-fixed ends, K 0.5:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 14.71 kN=
P Pcr:= P 14.71 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 4.40 MPa=
 < σY = 56 MPa (O.K.!)
Check yielding about x-x axis :
Applying the secant formula: c 0.5h:=
σmax
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅:=
σmax 21.35 MPa= < σY = 56 MPa (O.K.!)
Problem 13-51
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, σY = 55 MPa.
Given: E12 GPa⋅:= L 2m:=
σY 55MPa:= e 120mm:= do 100mm:=
Solution:
Section Property:
I
do
4
12
:= A do
2:= r
I
A
:=
Buckling :
For fixed-free ends, K 2.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 61.69 kN=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 6.17 MPa=
 < σY = 55 MPa (O.K.!)
Check yielding :
Applying the secant formula: c 0.5do:=
Given
σY
Pmax
A
1
e c⋅
r2
sec
K L⋅
2r
Pmax
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess Pmax 1kN:=
Pmax Find Pmax( ):= Pmax 31.37 kN= (Controls !)
Hence,
Pallow min Pcr Pmax,( ):=
Pallow 31.37 kN= Ans
Problem 13-52
The W200 X 71 structural A-36 steel column is fixed at its bottom and free at its top. If it is subjected
to the eccentric load of 375 kN, determine the factor of safety with respect to either the initiation of
buckling or yielding.
Given: E 200 GPa⋅:= L 3.6m:= P 375kN:=
σY 250MPa:= ey 200mm:=
Solution:
Use W 200x71 : Ix 76.6 10
6( )⋅ mm4:= A 9100mm2:=
Iy 25.4 10
6( )⋅ mm4:= d 216mm:= rx 91.7mm:=
Buckling about y-y axis :
For fixed-free ends, K 2.0:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 967.16 kN=
Pallow Pcr:= Pallow 967.16 kN=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 106.28 MPa=
 < σY = 250 MPa (O.K.!)
Check yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σY
Pmax
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
Pmax
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess Pmax 1kN:=
Pmax Find Pmax( ):= Pmax 531.59 kN= (Controls !)
 < Pallow = 967.16 kN
Factor of Safety :
Fsafety
Pmax
P
:= Fsafety 1.42= Ans
Problem 13-53
The W200 X 71 structural A-36 steel column is fixed at its bottom and pinned at its top. If it is
subjected to the eccentric load of 375 kN, determine if the column fails by yielding. The column is
braced so that it does not buckle about the y-y axis.
Given: E 200 GPa⋅:= L 3.6m:= P 375kN:=
σY 250MPa:= ey 200mm:=
Solution:
Use W 200x71 : Ix 76.6 10
6( )⋅ mm4:= A 9100mm2:=
Iy 25.4 10
6( )⋅ mm4:= d 216mm:= rx 91.7mm:=
For fixed-pined ends, K 0.7:=
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σmax
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅:=
σmax 149.16 MPa=
 < σY = 250 MPa (O.K.!)
Hence, the column does not fail by yielding. Ans
Problem 13-54
The brass rod is fixed at one end and free at the other end. 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. Ebr = 101 GPa,
σY = 69 MPa.
Given: E 101 GPa⋅:= do 100mm:= P 200kN:=
σY 69MPa:= e 10mm:= L m:=
Solution:
Section Property:
I
πdo
4
64
:= A
πdo
2
4
:= r
I
A
:=
Buckling :
For fixed-free ends, K 2.0:= Pcr P:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
= Lcr
π
2 E⋅ I⋅
P K2⋅
:= Lcr 2.473 m=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 25.46 MPa=
 < σY = 69 MPa (O.K.!)
Check yielding :
Applying the secant formula: c 0.5do:=
Given
σY
P
A
1
e c⋅
r2
sec
K Lmax⋅
2r
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess Lmax 1m:=
Lmax Find Lmax( ):= Lmax 1.706 m= (Controls !)
Hence,
Lallow min Lcr Lmax,( ):=
Lallow 1.706 m= Ans
Problem 13-55
The brass rod is fixed at one end and free at the other end. 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. Ebr = 101 GPa, σY = 69
MPa.
Given: E 101 GPa⋅:= do 100mm:= L 2m:=
σY 69MPa:= e 10mm:=
Solution:
Section Property:
I
πdo
4
64
:= A
πdo
2
4
:= r
I
A
:=
Buckling :
For fixed-free ends, K 2.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 305.82 kN=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 38.94 MPa=
 < σY = 69 MPa (O.K.!)
Check yielding :
Applying the secant formula: c 0.5do:=
Given
σY
Pmax
A
1
e c⋅
r2
sec
K L⋅
2r
Pmax
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess Pmax 1kN:=
Pmax Find Pmax( ):= Pmax 173.70 kN= (Controls !)
Hence,
Pallow min Pcr Pmax,( ):=
Pallow 173.70 kN= Ans
Maximum Deflection :
∆max e sec
K L⋅
2
Pallow
E I⋅
⋅
⎛
⎜
⎝
⎞
⎠
1−
⎛
⎜
⎝
⎞
⎠
⋅:=
∆max 16.50 mm= Ans
Problem 13-56
A W310 X 39 structural A-36 steel column is fixed connected at its ends and has a length L = 6.9 m.
Determine the maximum eccentric load P that can be applied so the column does not buckle or yeild.
Compare this value with an axial critical load P' applied through the centroid of the column.
Given: E 200 GPa⋅:= L 6.9m:=
σY 250MPa:= ey 150mm:=
Solution:
Use W 310x39 : Ix 84.8 10
6( )⋅ mm4:= A 4930mm2:=
Iy 7.23 10
6( )⋅ mm4:= d 310mm:= rx 131mm:=
Buckling about y-y axis :
For fixed-ends, K 0.5:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 1199.03 kN=
P' Pcr:= P' 1199.03 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 243.21 MPa=
 < σY = 250 MPa (O.K.!)
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 509.74 kN= (Controls !)
 < P' =1199 kN
Hence, Pmax min P' P,( ):=
Pmax 509.74 kN= Ans
Problem 13-57
A W360 X 45 structural A-36 steel column is fixed connected at its ends and has a length L = 6 m.
Determine the maximum eccentric load P that can be applied so the column does not buckle or yeild.
Compare this value with an axial critical load P' applied through the centroid of the column.
Given: E 200 GPa⋅:= L 6.9m:=
σY 250MPa:= ey 150mm:=
Solution:
Use W 360x45 : Ix 121 10
6( )⋅ mm4:= A 5710mm2:=
Iy 8.16 10
6( )⋅ mm4:= d 352mm:= rx 146mm:=
Buckling about y-y axis :
For fixed ends, K 0.5:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 1353.26 kN=
P' Pcr:= P' 1353.26 kN= Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 237.00 MPa=
 < σY = 250 MPa (O.K.!)
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 624.10 kN= (Controls !)
 < P' =1353 kN
Hence, Pmax min P' P,( ):=
Pmax 624.10 kN= Ans
Problem 13-58
Solve Prob. 13-57 if the column is fixed at its bottom and free at its top.
Given: E 200 GPa⋅:= L 6.9m:=
σY 250MPa:= ey 150mm:=
Solution:
Use W 360x45 : Ix 121 10
6( )⋅ mm4:= A 5710mm2:=
Iy 8.16 10
6( )⋅ mm4:= d 352mm:= rx 146mm:=
Buckling about y-y axis :
For fixed-free ends, K 2.0:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 84.58 kN=
P' Pcr:= P' 84.58 kN= (Controls !) Ans
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 14.81 MPa=
 < σY = 250 MPa (O.K.!)
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 457.81 kN=
 > P' =85 kN
Hence, Pmax min P' P,( ):=
Pmax 84.58 kN= Ans
Problem 13-59
The wood column is fixed at its base and can be assumed pin connected at its top. Determine the
maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 12
GPa, σY = 56 MPa.
Given: E 12 GPa⋅:= L 3m:= b 100mm:=
σY 56MPa:= ey 125mm:= h 250mm:=
Solution:
Section Property:
Ix
b h3⋅
12
:= Iy
h b3⋅
12
:=
A b h⋅:= rx
Ix
A
:=
Buckling about y-y axis :
For fixed-pinned ends, K 0.7:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 559.5 kN=
P' Pcr:= P' 559.50 kN=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 22.38 MPa=
 < σY = 56 MPa (O.K.!)
Check yielding about x-x axis :
Applying the secant formula: c 0.5h:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
SolvingEq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 320.08 kN= (Controls !)
 < P' =559.5 kN
Hence, Pmax min P' P,( ):=
Pmax 320.08 kN= Ans
Problem 13-60
The wood column is fixed at its base and can be assumed fixed connected at its top. Determine the
maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 12
GPa, σY = 56 MPa.
Given: E 12 GPa⋅:= L 3m:= b 100mm:=
σY 56MPa:= ey 125mm:= h 250mm:=
Solution:
Section Property:
Ix
b h3⋅
12
:= Iy
h b3⋅
12
:=
A b h⋅:= rx
Ix
A
:=
Buckling about y-y axis :
For fixed-fixed ends, K 0.5:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 1096.62 kN=
P' Pcr:= P' 1096.62 kN=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 43.86 MPa=
 < σY = 56 MPa (O.K.!)
Check yielding about x-x axis :
Applying the secant formula: c 0.5h:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 334.13 kN= (Controls !)
 < P' =1096.6 kN
Hence, Pmax min P' P,( ):=
Pmax 334.13 kN= Ans
Problem 13-61
The aluminum 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. Eal = 70 GPa, σY = 95 MPa.
Given: bf 160mm:= dw 150mm:= L 5m:=
tf 10mm:= tw 10mm:= e 5mm:=
E 70 GPa⋅:= σY 95MPa:= Fsafety 3:=
Solution:
Section Property : D dw tf+:=
xc
0.5tf bf tf⋅( ) 0.5dw tf+( ) dw tw⋅( )+
bf tf⋅ dw tw⋅+
:=
xc 43.71 mm=
Iy
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf xc−( )2⋅+
1
12
tw⋅ dw
3⋅ dw tw⋅( ) 0.5dw tf+ xc−( )2⋅++
...:=
Iy 7780672.04 mm
4=
Ix
1
12
tf⋅ bf
3⋅
1
12
dw⋅ tw
3⋅+:= Ix 3425833.33 mm
4=
A D bf⋅ dw bf tw−( )⋅−:= A 3100 mm2= ry
Iy
A
:=
Buckling about x-x axis :
For fixed-free ends, K 2.0:=
Pcr
π
2 E⋅ Ix⋅
K L⋅( )2
:= Pcr 23.67 kN=
P' Pcr:= P' 23.67 kN= (Controls !)
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 7.63 MPa=
 < σY = 95 MPa (O.K.!)
Yielding about y-y axis :
Applying the secant formula: c xc:= ey xc e−:=
Given
σY
P
A
1
ey c⋅
ry
2
sec
K L⋅
2ry
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 45.61 kN=
Hence, Pmax min P' P,( ):= Pmax 23.67 kN=
Pallow
Pmax
Fsafety
:= Pallow 7.89 kN= Ans
Problem 13-62
A W250 X 22 structural A-36 steel member is used as a fixed-connected column. 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.
Given: E 200 GPa⋅:= L 7.5m:=
σY 250MPa:= ey 200mm:=
Solution:
Use W 250x22 : Ix 28.8 10
6( )⋅ mm4:= A 2850mm2:=
Iy 1.22 10
6( )⋅ mm4:= d 254mm:= rx 101mm:=
Buckling about y-y axis :
For fixed ends, K 0.5:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 171.25 kN=
P' Pcr:= P' 171.25 kN= (Controls !)
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 60.09 MPa=
 < σY = 250 MPa (O.K.!)
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 195.49 kN=
 > P' =171 kN
Hence, Pmax min P' P,( ):= Pmax 171.25 kN= Ans
Problem 13-63
Solve Prob. 13-62 if the column is pin-connected at its ends.
Given: E 200 GPa⋅:= L 7.5m:=
σY 250MPa:= ey 200mm:=
Solution:
Use W 250x22 : Ix 28.8 10
6( )⋅ mm4:= A 2850mm2:=
Iy 1.22 10
6( )⋅ mm4:= d 254mm:= rx 101mm:=
Buckling about y-y axis :
For pinned ends, K 1.0:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 42.81 kN=
P' Pcr:= P' 42.81 kN= (Controls !)
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 15.02 MPa=
 < σY = 250 MPa (O.K.!)
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 172.94 kN=
 > P' =43 kN
Hence, Pmax min P' P,( ):= Pmax 42.81 kN= Ans
Problem 13-64
Solve Prob. 13-62 if the column is fixed at its bottom and pinned at its top.
Given: E 200 GPa⋅:= L 7.5m:=
σY 250MPa:= ey 200mm:=
Solution:
Use W 250x22 : Ix 28.8 10
6( )⋅ mm4:= A 2850mm2:=
Iy 1.22 10
6( )⋅ mm4:= d 254mm:= rx 101mm:=
Buckling about y-y axis :
For fixed-pinned ends, K 0.7:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 87.37 kN=
P' Pcr:= P' 87.37 kN= (Controls !)
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 30.66 MPa=
 < σY = 250 MPa (O.K.!)
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 187.74 kN=
 > P' =87.4 kN
Hence, Pmax min P' P,( ):= Pmax 87.37 kN= Ans
Problem 13-65
Determine the load P required to cause the steel W310 X 74structural A-36 steel column to fail either
by buckling or by yielding. The column is fixed at its bottom and the cables at its top act as a pin to
hold it.
Given: E 200 GPa⋅:= L 7.5m:=
σY 250MPa:= ey 50mm:=
Solution:
Use W 310x74 : Ix 165 10
6( )⋅ mm4:= A 9480mm2:=
Iy 23.4 10
6( )⋅ mm4:= d 310mm:= rx 132mm:=
Buckling about y-y axis :
For fixed-pinned ends, K 0.7:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 1675.82 kN=
P' Pcr:= P' 1675.82 kN=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 176.77 MPa=
 < σY = 250 MPa (O.K.!)
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 1551.14 kN= (Controls !)
 < P' =1676 kN
Hence, Pmax min P' P,( ):= Pmax 1551.14 kN= Ans
Problem 13-66
Solve Prob. 13-65 if the column is an A-36 steel W310 X 24 section.
Given: E 200 GPa⋅:= L 7.5m:=
σY 250MPa:= ey 50mm:=
Solution:
Use W 310x24 : Ix 42.8 10
6( )⋅ mm4:= A 3040mm2:=
Iy 1.16 10
6( )⋅ mm4:= d 305mm:= rx 119mm:=
Buckling about y-y axis :
For fixed-pinned ends, K 0.7:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 83.07 kN=
P' Pcr:= P' 83.07 kN= (Controls !)
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 27.33 MPa=
 < σY = 250 MPa (O.K.!)
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 459.16 kN=
 > P' =83 kN
Hence, Pmax min P' P,( ):= Pmax 83.07 kN= Ans
Problem 13-67
The W360 X 79 structural A-36 steel column is fixed at its base and free at its top. If P = 375 kN,
determine the sidesway deflection at its top and the maximum stress in the column.
Given: E 200 GPa⋅:= L 5.5m:= P 375kN:=
σY 250MPa:= ey 250mm:=
Solution:
Use W 360x79 : Ix 227 10
6( )⋅ mm4:= A 10100mm2:=
Iy 24.2 10
6( )⋅ mm4:= d 354mm:= rx 150mm:=
Buckling about y-y axis :
For fixed-free ends, K 2.0:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 394.78 kN= > P =375 kN (O.K.!)
Hence, the column does not buckle about the y-y axis.
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 39.09 MPa=
 < σY = 250 MPa (O.K.!)
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
σmax
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅:=
σmax 120.32 MPa= Ans
Since σmax < σY =250 MPa, the column does not yield.
Maximum Displacement :
vmax ey sec
K L⋅
2
P
E Ix⋅
⋅
⎛
⎜
⎝
⎞
⎠
1−⎛⎜
⎝
⎞
⎠
⋅:=
vmax 34.85 mm= Ans
Problem 13-68
The W360 X 79 steel 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 either buckle or yield. Est = 200 GPa, σY = 350 MPa.
Given: E 200 GPa⋅:= L5.5m:=
σY 350MPa:= ey 250mm:=
Solution:
Use W 360x79 : Ix 227 10
6( )⋅ mm4:= A 10100mm2:=
Iy 24.2 10
6( )⋅ mm4:= d 354mm:= rx 150mm:=
Buckling about y-y axis :
For fixed-free ends, K 2.0:=
Pcr
π
2 E⋅ Iy⋅
K L⋅( )2
:= Pcr 394.78 kN=
P' Pcr:= P' 394.78 kN= (Controls !)
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 39.09 MPa=
 < σY = 350 MPa (O.K.!)
Yielding about x-x axis :
Applying the secant formula: c 0.5d:=
Given
σY
P
A
1
ey c⋅
rx
2
sec
K L⋅
2rx
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess P 1kN:=
P Find P( ):= P 933.05 kN=
 > P' =395 kN
Hence, Pmax min P' P,( ):= Pmax 394.78 kN= Ans
Problem 13-69
The aluminum rod is fixed at its base and 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,
σY = 410 MPa.
Given: E 72 GPa⋅:= do 200mm:= P 200kN:=
σY 410MPa:= e 5mm:= L m:=
Solution:
Section Property:
I
πdo
4
64
:= A
πdo
2
4
:= r
I
A
:=
Buckling :
For fixed-free ends, K 2.0:= Pcr P:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
= Lcr
π
2 E⋅ I⋅
P K2⋅
:= Lcr 8.352 m=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 6.37 MPa=
 < σY = 410 MPa (O.K.!)
Check yielding :
Applying the secant formula: c 0.5do:=
Given
σY
P
A
1
e c⋅
r2
sec
K Lmax⋅
2r
P
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛
⎜
⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess Lmax 1m:=
Lmax Find Lmax( ):= Lmax 8.336 m= (Controls !)
Hence,
Lallow min Lcr Lmax,( ):=
Lallow 8.336 m= Ans
Problem 13-70
The aluminum rod is fixed at its base and 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, σY = 410 MPa.
Given: E 72 GPa⋅:= do 200mm:= L 2m:=
σY 410MPa:= e 5mm:=
Solution:
Section Property:
I
πdo
4
64
:= A
πdo
2
4
:= r
I
A
:=
Buckling :
For fixed-free ends, K 2.0:=
Pcr
π
2 E⋅ I⋅
K L⋅( )2
:= Pcr 3488.21 kN=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 111.03 MPa=
 < σY = 410 MPa (O.K.!)
Check yielding :
Applying the secant formula: c 0.5do:=
Given
σY
Pmax
A
1
e c⋅
r2
sec
K L⋅
2r
Pmax
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜⎝
⎞
⎠
⋅= (1)
Solving Eq. (1) by trial and error: Guess Pmax 1kN:=
Pmax Find Pmax( ):= Pmax 3200.5 kN= (Controls !)
Hence,
Pallow min Pcr Pmax,( ):=
Pallow 3200.5 kN= Ans
Maximum Deflection :
∆max e sec
K L⋅
2
Pallow
E I⋅
⋅
⎛
⎜
⎝
⎞
⎠
1−
⎛
⎜
⎝
⎞
⎠
⋅:=
∆max 70.61 mm= Ans
Problem 13-71
The steel column supports the two eccentric loadings. If it is assumed to be pinned at its top, fixed at
the bottom, and fully braced against buckling about the y-y axis, determine the maximum deflection of
the column and the maximum stress in the column. Est = 200 GPa, σY = 360 MPa.
Given: E 200 GPa⋅:= bf 100mm:= df 10mm:= e1 120mm:= P1 130kN:=
σY 360MPa:= tw 10mm:= dw 100mm:= e2 80mm:= P2 50kN:=
L 6m:=
Solution:
Section Property: D dw 2df+:=
A bf D⋅ dw bf tw−( )⋅−:=
Ix
bf D
3⋅
12
bf tw−( ) dw3⋅
12
−:= rx
Ix
A
:=
Buckling :
For fixed-pinned ends, K 0.7:=
Pcr
π
2 E⋅ Ix⋅
K L⋅( )2
:= Pcr 772.11 kN=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 257.37 MPa=
 < σY = 360 MPa (O.K.!)
Yielding about x-x axis :
Eccentricity of the two applied loasds is e
e1 P1⋅ e2 P2⋅−
P1 P2+
:= e 64.44 mm=
Applying the secant formula: c 0.5D:=
σmax
P1 P2+
A
1
e c⋅
rx
2
sec
K L⋅
2rx
P1 P2+
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅:= σmax 199.0 MPa= Ans
Since σmax < σY = 360MPa, the column does not yield.
Maximum Deflection :
∆max e sec
K L⋅
2
P1 P2+
E Ix⋅
⋅
⎛
⎜
⎝
⎞
⎠
1−
⎛
⎜
⎝
⎞
⎠
⋅:=
∆max 24.33 mm= Ans
Problem 13-72
The steel column supports the two eccentric loadings. If it is assumed to be fixed at its top and
bottom, and braced against buckling about the y-y axis, determine the maximum deflection of the
column and the maximum stress in the column. Est = 200 GPa, σY = 360 MPa.
Given: E 200 GPa⋅:= bf 100mm:= df 10mm:= e1 120mm:= P1 130kN:=
σY 360MPa:= tw 10mm:= dw 100mm:= e2 80mm:= P2 50kN:=
L 6m:=
Solution:
Section Property: D dw 2df+:=
A bf D⋅ dw bf tw−( )⋅−:=
Ix
bf D
3⋅
12
bf tw−( ) dw3⋅
12
−:= rx
Ix
A
:=
Buckling :
For fixed ends, K 0.5:=
Pcr
π
2 E⋅ Ix⋅
K L⋅( )2
:= Pcr 1513.34 kN=
Critical Stress : Euler's formula is only valid if σcr < σY.
σcr
Pcr
A
:= σcr 504.45 MPa=
 < σY = 360 MPa (O.K.!)
Yielding about x-x axis :
Eccentricity of the two applied loasds is e
e1 P1⋅ e2 P2⋅−
P1 P2+
:= e 64.44 mm=
Applying the secant formula: c 0.5D:=
σmax
P1 P2+
A
1
e c⋅
rx
2
sec
K L⋅
2rx
P1 P2+
E A⋅
⋅
⎛
⎜
⎝
⎞
⎠
⋅+
⎛⎜
⎜
⎝
⎞
⎠
⋅:= σmax 177.7 MPa= Ans
Since σmax < σY = 360MPa, the column does not yield.
Maximum Deflection :
∆max e sec
K L⋅
2
P1 P2+
E Ix⋅
⋅
⎛
⎜
⎝
⎞
⎠
1−
⎛
⎜
⎝
⎞
⎠
⋅:=
∆max 10.77 mm= Ans
Problem 13-73
A column of intermediate length buckles when the compressive strength is 280 MPa. If the slenderness
ratio is 60, determine the tangent modulus.
Given: σcr 280MPa:= λ' 60:=
Solution:
σcr
π
2 Et⋅
λ'2
=
Et
σcr λ'
2
⋅
π
2
:=
Et 102.13 GPa= Ans
Problem 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.
Given: σ1 91MPa:= ε1 0.002:=
σ2 126MPa:= ε2 0.005:=
Solution: Let λ
L
r
= σ
P
A
=
From the graph, 
E1
σ1
ε1
:= E1 45.50 GPa=
E2
σ2 σ1−
ε2 ε1−
:= E2 11.67 GPa=
σ - λ relationship: σcr
π
2 Et⋅
λ
2
=
For Et = E1 : σ σcr= σ
π
2 E1⋅
λ
2
= When σ σ1:= λ1 π
E1
σ1
:= λ1 70.25=
For Et = E2 : σ σcr= σ
π
2 E2⋅
λ
2
= When σ σ1:= λ'1 π
E2
σ1
:= λ'1 35.57=
Buckling Curve:
x1 10 1.01 10( )⋅, λ'1..:= x2 λ'1 1.01 λ'1⋅, λ1..:= x3 λ1 1.01 λ1⋅, 130..:=
σ1 x1( )
π
2 E2⋅
x1
2
⎛
⎜
⎜
⎝
⎞
⎠
1
MPa
⋅:= σ2 x2( )
π
2 E2⋅
λ'1
2
⎛
⎜
⎜
⎝
⎞
⎠
1
MPa
⋅:= σ3 x3( )
π
2 E1⋅
x3
2
⎛
⎜
⎜
⎝
⎞
⎠
1
MPa
⋅:=
50 100
0
200
400
Slenderless Ratio
St
re
ss
 (M
Pa
) σ1 x1( )
σ2 x2( )
σ3 x3( )
x1 x2, x3,
Problem 13-75
Construct the buckling curve, P/A versus L/r, for a column that has a bilinear stress-strain curve in
compression as shown.
Given: σ1 175MPa:= ε1 0.001:=
σ2 350MPa:= ε2 0.005:=
Solution: Let λ
L
r
= σ
P
A
=
From the graph, 
E1
σ1
ε1
:= E1 175.00 GPa=
E2
σ2 σ1−
ε2 ε1−
:= E2 43.75 GPa=
σ - λ relationship: σcr
π
2 Et⋅
λ
2
=
For Et = E1 : σ σcr= σ
π
2 E1⋅
λ
2
= When σ σ1:= λ1 π
E1
σ1
:= λ1 99.35=
For Et = E2 : σ σcr= σ
π
2 E2⋅
λ
2
= When σ σ1:= λ'1 π
E2
σ1
:= λ'1 49.67=
Buckling Curve:
x1 10 1.01 10( )⋅, λ'1..:= x2 λ'1 1.01 λ'1⋅, λ1..:= x3 λ1 1.01 λ1⋅, 200..:=
σ1 x1( )
π
2 E2⋅
x1
2
⎛
⎜
⎜
⎝
⎞
⎠
1
MPa
⋅:= σ2 x2( )
π
2 E2⋅
λ'1
2
⎛
⎜
⎜
⎝
⎞
⎠
1
MPa
⋅:= σ3 x3( )
π
2 E1⋅
x3
2
⎛
⎜
⎜
⎝
⎞
⎠
1
MPa
⋅:=
50 100 150 200
0
200
400
Slenderless Ratio
St
re
ss
 (M
Pa
) σ1 x1( )
σ2 x2( )
σ3 x3( )
x1 x2, x3,
Problem 13-76
Construct the buckling curve, P/A versus L/r, for a column that has a bilinear stress-strain curve in
compression as shown. The column is pinned at its ends.
Given: σ1 140MPa:= ε1 0.001:=
σ2 260MPa:= ε2 0.004:=
Solution: Let λ
L
r
= σ
P
A
=
From the graph, 
E1
σ1
ε1
:= E1 140.00 GPa=
E2
σ2 σ1−
ε2 ε1−
:= E2 40.00 GPa=
σ - λ relationship: σcr
π
2 Et⋅
λ
2
=
For Et = E1 : σ σcr= σ
π
2 E1⋅
λ
2
= When σ σ1:= λ1 π
E1
σ1
:= λ1 99.35=
For Et = E2 : σ σcr= σ
π
2 E2⋅
λ
2
= When σ σ1:= λ'1 π
E2
σ1
:= λ'1 53.10=
Buckling Curve:
x1 10 1.01 10( )⋅, λ'1..:= x2 λ'1 1.01 λ'1⋅, λ1..:= x3 λ1 1.01 λ1⋅, 160..:=
σ1 x1( )
π
2 E2⋅
x1
2
⎛
⎜
⎜
⎝
⎞
⎠
1
MPa
⋅:= σ2 x2( )
π
2 E2⋅
λ'1
2
⎛
⎜
⎜
⎝
⎞
⎠
1
MPa
⋅:= σ3 x3( )
π
2 E1⋅
x3
2
⎛
⎜
⎜
⎝
⎞
⎠
1
MPa
⋅:=
50 100 150
0
200
400
Slenderless Ratio
St
re
ss
 (M
Pa
) σ1 x1( )
σ2 x2( )
σ3 x3( )
x1 x2, x3,
Problem 13-77
Determine the largest length of a structural A-36 steel rod if it is fixed supported and subjectedto an
axial load of 100 kN. The rod has a diameter of 50 mm. Use the AISC equations.
Given: E 200 GPa⋅:= P 100kN:=
σY 250MPa:= do 50mm:=
Solution:
Section Property :
A
π do
2⋅
4
:= I
π do
4⋅
64
:= r
I
A
:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 125.66=
Assume a long column, using Eq. 13.21 :
σ
P
A
:= σ 50.93 MPa=
σallow
12π2 E⋅
23λ'2
= λ'
12π2 E⋅
23σ
:= λ' 142.20=
> λ'c (Assumption is correct.)
Slenderness Ratio :
For fixed ends, K 0.5:=
λ'
K L⋅
r
= L
λ' r⋅
K
:= L 3.555 m= Ans
Problem 13-78
Determine the largest length of a W250 X 18 structural A-36 steel section if it is fixed supported and is
subjected to an axial load of 140 kN. Use the AISC equations.
Given: E 200 GPa⋅:= P 140kN:=
σY 250MPa:=
Solution:
Section Property :
W 250x18 : A 2280mm2:= ry 20.1mm:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 125.66=
Assume a long column, using Eq. 13.21 :
σ
P
A
:= σ 61.40 MPa=
σallow
12π2 E⋅
23λ'2
= λ'
12π2 E⋅
23σ
:= λ' 129.51=
> λ'c (Assumption is correct.)
Slenderness Ratio :
For fixed ends, K 0.5:=
λ'
K L⋅
r
= L
λ' ry⋅
K
:= L 5.206 m= Ans
Problem 13-79
Determine the largest length of a W310 X 67 structural A-36 steel column if it is pin supported and
subjected to an axial load of 1000 kN. Use the AISC equations.
Given: E 200 GPa⋅:= P 1000kN:=
σY 250MPa:=
Solution:
Section Property :
W 310x67 : A 8530mm2:= ry 49.3mm:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 125.66=
Assume a long column, using Eq. 13.21 :
σ
P
A
:= σ 117.23 MPa=
σallow
12π2 E⋅
23λ'2
= λ'
12π2 E⋅
23σ
:= λ' 93.73=
< λ'c (Assumption is not correct !)
Thus, it is an intermediate column. Using Eq. 13.23 :
Given
σ
1
λ'2
2λ'c
2
−
⎛
⎜
⎜
⎝
⎞
⎠
σY⋅
5
3
3λ'
8λ'c
+
λ'3
8λ'c
3
−
= (1)
Solving Eq. (1) by trial and error: Guess λ' 1:=
λ' Find λ'( ):= λ' 65.35=
For pinned ends, K 1.0:=
λ'
K L⋅
r
= L
λ' ry⋅
K
:= L 3.222 m= Ans
Problem 13-80
Determine the largest length of a W200 X 46 structural A-36 steel section if it is pin supported and is
subjected to an axial load of 380 kN. Use the AISC equations.
Given: E 200 GPa⋅:= P 380kN:=
σY 250MPa:=
Solution:
Section Property :
W 200x46 : A 5890mm2:= ry 51.0mm:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 125.66=
Assume a long column, using Eq. 13.21 :
σ
P
A
:= σ 64.52 MPa=
σallow
12π2 E⋅
23λ'2
= λ'
12π2 E⋅
23σ
:= λ' 126.34=
> λ'c (Assumption is correct.)
Slenderness Ratio :
For pinned ends, K 1.0:=
λ'
K L⋅
r
= L
λ' ry⋅
K
:= L 6.444 m= Ans
Problem 13-81
Using the AISC equations, check if a W150 X 14 structural A-36 steel column that is 3 m long can
support an axial load of 200 kN. The ends are fixed.
Given: E 200 GPa⋅:= P 200kN:=
σY 250MPa:= L 3m:=
Solution:
Section Property :
W 150x14 : A 1730mm2:= ry 23.0mm:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 125.66=
For fixed ends, K 0.5:=
λ'
K L⋅
ry
:= λ' 65.217= < λ'c 
Thus, it is an intermediate column. Using Eq. 13.23 :
σallow
1
λ'2
2λ'c
2
−
⎛
⎜
⎜
⎝
⎞
⎠
σY⋅
5
3
3λ'
8λ'c
+
λ'3
8λ'c
3
−
:= σallow 117.3 MPa=
Pallow A σallow⋅:=
Pallow 203.0 kN= > P = 200 kN (O.K.!)
Thus, the column is adequate. Ans
Problem 13-82
Using the AISC equations, select from Appendix B the lightest-weight structural steel column that is
4.2 m long and supports an axial load of 200 kN. The ends are pinned. Take σY = 350 MPa.
Given: E 200 GPa⋅:= P 200kN:=
σY 350MPa:= L 4.2m:=
Solution:
Section Property :
Try W 150x24 : A 2860mm2:= ry 36.8mm:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 106.21=
For pinned ends, K 1.0:= λ'
K L⋅
ry
:= λ' 114.130= > λ'c 
Thus, it is an long column. Using Eq. 13.21 :
σallow
12π2 E⋅
23λ'2
:= σallow 79.1 MPa=
Pallow A σallow⋅:=
Pallow 226.1 kN= > P = 200 kN (O.K.!)
Use W 150x24 Ans
Problem 13-83
Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that
is 3.6 m long and supports an axial load of 200 kN. The ends are fixed. Take σY = 350 MPa.
Given: E 200 GPa⋅:= P 200kN:=
σY 350MPa:= L 3.6m:=
Solution:
Section Property :
Try W 150x14 : A 1730mm2:= ry 23.0mm:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 106.21=
For fixed ends, K 0.5:=
λ'
K L⋅
ry
:= λ' 78.261= < λ'c 
Thus, it is an intermediate column. Using Eq. 13.23 :
σallow
1
λ'2
2λ'c
2
−
⎛
⎜
⎜
⎝
⎞
⎠
σY⋅
5
3
3λ'
8λ'c
+
λ'3
8λ'c
3
−
:= σallow 134.7 MPa=
Pallow A σallow⋅:=
Pallow 233.0 kN= > P = 200 kN (O.K.!)
Use W 150x14 Ans
Problem 13-84
Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that
is 4.2 m long and supports an axial load of 200 kN. The ends are fixed. 
Given: E 200 GPa⋅:= P 200kN:=
σY 250MPa:= L 4.2m:=
Solution:
Section Property :
Try W 150x18 : A 2290mm2:= ry 23.5mm:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 125.66=
For fixed ends, K 0.5:=
λ'
K L⋅
ry
:= λ' 89.362= < λ'c 
Thus, it is an intermediate column. Using Eq. 13.23 :
σallow
1
λ'2
2λ'c
2
−
⎛
⎜
⎜
⎝
⎞
⎠
σY⋅
5
3
3λ'
8λ'c
+
λ'3
8λ'c
3
−
:= σallow 98.9 MPa=
Pallow A σallow⋅:=
Pallow 226.5 kN= > P = 200 kN (O.K.!)
Use W 150x18 Ans
Problem 13-85
Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that
is 9 m long and supports an axial load of 1000 kN. The ends are fixed.
Given: E 200 GPa⋅:= P 1000kN:=
σY 250MPa:= L 9m:=
Solution:
Section Property :
Try W 250x80 : A 10200mm2:= ry 65.0mm:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 125.66=
For fixed ends, K 0.5:=
λ'
K L⋅
ry
:= λ' 69.231= < λ'c 
Thus, it is an intermediate column. Using Eq. 13.23 :
σallow
1
λ'2
2λ'c
2
−
⎛
⎜
⎜
⎝
⎞
⎠
σY⋅
5
3
3λ'
8λ'c
+
λ'3
8λ'c
3
−
:= σallow 114.5 MPa=
Pallow A σallow⋅:=
Pallow 1167.7 kN= > P = 1000 kN (O.K.!)
Use W 250x80 Ans
Problem 13-86
Determine the largest length of a W200 X 46 structural A-36 steel section if it is pin supported and is
subjected to an axial load of 90 kN. Use the AISC equations.
Given: E 200 GPa⋅:= P 90kN:=
σY 250MPa:=
Solution:
Section Property :
W 200x46 : A 5890mm2:= ry 51.0mm:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 125.66=
Assume a long column, using Eq. 13.21 :
σ
P
A
:= σ 15.28 MPa=
σallow
12π2 E⋅
23λ'2
= λ'
12π2 E⋅
23σ
:= λ' 259.61=
> λ'c (Assumption is correct.)
Slenderness Ratio : λ'max 200:=
For pinned ends, K 1.0:=
λ'max
K L⋅
r
= L
λ'max ry⋅
K
:= L 10.20 m= Ans
Problem 13-87
Determine the largest length of a W150 X 22 structural A-36 steel section if it is pin supported and
subjected to an axial load of 350 kN. Use the AISC equations.
Given: E 200 GPa⋅:= P 350kN:=
σY 250MPa:=
Solution:
Section Property :
W 150x22 : A 2860mm2:= ry 36.8mm:=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 125.66=
Assume a long column, using Eq. 13.21 :
σ
P
A
:= σ 122.38 MPa=
σallow
12π2 E⋅
23λ'2
= λ'
12π2 E⋅
23σ
:= λ' 91.74=
< λ'c (Assumption is not correct !)
Thus, it is an intermediate column. Using Eq. 13.23 :
Given
σ
1
λ'2
2λ'c
2
−
⎛
⎜
⎜
⎝
⎞
⎠
σY⋅
5
3
3λ'
8λ'c
+
λ'3
8λ'c
3
−
= (1)
Solving Eq. (1) by trial and error: Guess λ' 1:=
λ' Find λ'( ):= λ' 57.78=
For pinned ends, K 1.0:=
λ'
K L⋅
r
= L
λ' ry⋅
K
:= L 2.126 m= Ans
Problem 13-88
The bar is made of a 2014-T6 aluminum alloy. Determine its thickness b if its width is 5b. Assume that
it is pin connected at its ends.
Given: E 73.1 GPa⋅:= P 3kN:= t b=
σY 414MPa:= L 2.4m:= h 5 b⋅=
Solution:
Section Property :
A t h⋅= A b 5 b⋅( )⋅= A 5 b2⋅=
Iy
h t3⋅
12
= Iy
5 b⋅( ) b3⋅
12
= Iy
5 b4⋅
12
=
ry
Iy
A
= ry
5 b4⋅
12 5 b2⋅( )⋅
= ry
b
12
=
Slenderness Ratio :
For pinned ends, K 1.0:=
Aluminum Association Formula : λ'
K L⋅
ry
= λ'
12 K⋅ L⋅
b
=
σ
P
A
= σ
P
5 b2⋅
=
Assume a long column, using Eq. 13.26 :
σallow378125MPa
λ'2
= λ'2
378125MPa
σ
=
12 K⋅ L⋅
b
⎛⎜
⎝
⎞
⎠
2
378125MPa 5 b2⋅( )⋅
P
=
b
4
12 K2⋅ L2⋅ P⋅
378125MPa 5( )⋅
:=
b 18.20 mm=
Hence, ry
b
12
:= λ'
K L⋅
ry
:=
λ' 456.85= > 55 (Assumption is correct.)
Thus, b 18.20 mm= Ans
Problem 13-89
The bar is made of a 2014-T6 aluminum alloy. Determine its thickness b if its width is 5b. Assume that
it isfixed connected at its ends.
Given: E 73.1 GPa⋅:= P 3kN:= t b=
σY 414MPa:= L 2.4m:= h 5 b⋅=
Solution:
Section Property :
A t h⋅= A b 5 b⋅( )⋅= A 5 b2⋅=
Iy
h t3⋅
12
= Iy
5 b⋅( ) b3⋅
12
= Iy
5 b4⋅
12
=
ry
Iy
A
= ry
5 b4⋅
12 5 b2⋅( )⋅
= ry
b
12
=
Slenderness Ratio :
For fixed ends, K 0.5:=
Aluminum Association Formula : λ'
K L⋅
ry
= λ'
12 K⋅ L⋅
b
=
σ
P
A
= σ
P
5 b2⋅
=
Assume a long column, using Eq. 13.26 :
σallow
378125MPa
λ'2
= λ'2
378125MPa
σ
=
12 K⋅ L⋅
b
⎛⎜
⎝
⎞
⎠
2
378125MPa 5 b2⋅( )⋅
P
=
b
4
12 K2⋅ L2⋅ P⋅
378125MPa 5( )⋅
:=
b 12.87 mm=
Hence, ry
b
12
:= λ'
K L⋅
ry
:=
λ' 323.04= > 55 (Assumption is correct.)
Thus, b 12.87 mm= Ans
Problem 13-90
The 50-mm-diameter rod is used to support an axial load of 40 kN. Determine its greatest allowable
length L if it is made of 2014-T6 aluminum. Assume that the ends are pin connected.
Given: E 73.1 GPa⋅:= do 50mm:=
σY 414MPa:= P 40kN:=
Solution:
Section Property :
A
π do
2⋅
4
:= A 1963.50 mm2=
I
π do
4⋅
64
:= I 306796.16 mm4=
ry
I
A
:= ry 12.50 mm=
Slenderness Ratio :
For pinned ends, K 1.0:=
Aluminum Association Formula : λ'
K L⋅
ry
=
σ
P
A
:=
Assume a long column, using Eq. 13.26 :
σallow
378125MPa
λ'2
= λ'
378125MPa
σ
:=
λ' 136.24= > 55 (Assumption is correct.)
Hence,
λ'
K L⋅
ry
= L
λ' ry⋅
K
:= L 1.703 m= Ans
Problem 13-91
The 50-mm-diameter rod is used to support an axial load of 40 kN. Determine its greatest allowable
length L if it is made of 2014-T6 aluminum. Assume that the ends are fixed connected.
Given: E 73.1 GPa⋅:= do 50mm:=
σY 414MPa:= P 40kN:=
Solution:
Section Property :
A
π do
2⋅
4
:= A 1963.50 mm2=
I
π do
4⋅
64
:= I 306796.16 mm4=
ry
I
A
:= ry 12.50 mm=
Slenderness Ratio :
For fixed ends, K 0.5:=
Aluminum Association Formula : λ'
K L⋅
ry
=
σ
P
A
:=
Assume a long column, using Eq. 13.26 :
σallow
378125MPa
λ'2
= λ'
378125MPa
σ
:=
λ' 136.24= > 55 (Assumption is correct.)
Hence,
λ'
K L⋅
ry
= L
λ' ry⋅
K
:= L 3.406 m= Ans
Problem 13-92
The tube is 6 mm 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.
Given: E 73.1 GPa⋅:= ao 150mm:= t 6mm:=
σY 414MPa:= L 3m:=
Solution:
Section Property : ai ao 2t−:=
A ao
2 ai
2−:= A 3456 mm2=
I
ao
4 ai
4−
12
:= I 11964672 mm4=
r
I
A
:= r 58.84 mm=
Slenderness Ratio :
For fixed-pinned ends, K 0.7:=
Aluminum Association Formula : λ'
K L⋅
r
:= λ' 35.69=
Since 12 < λ' < 55, it is an intermediate column. Applying Eq. 13.25 :
σallow 214.5 1.628λ'−( )MPa:=
σallow 156.4 MPa=
Hence, the allowable load is
Pallow σallow A⋅:=
Pallow 540.5 kN= Ans
Problem 13-93
The tube is 6 mm 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.
Given: E 73.1 GPa⋅:= ao 150mm:= t 6mm:=
σY 414MPa:= L 3m:=
Solution:
Section Property : ai ao 2t−:=
A ao
2 ai
2−:= A 3456.00 mm2=
I
ao
4 ai
4−
12
:= I 11964672.00 mm4=
r
I
A
:= r 58.84 mm=
Slenderness Ratio :
For fixed ends, K 0.5:=
Aluminum Association Formula : λ'
K L⋅
r
:= λ' 25.49=
Since 12 < λ' < 55, it is an intermediate column. Applying Eq. 13.25 :
σallow 214.5 1.628λ'−( )MPa:=
σallow 173.00 MPa=
Hence, the allowable load is
Pallow σallow A⋅:=
Pallow 597.9 kN= Ans
Problem 13-94
The tube is 6 mm thick, is made of a 2014-T6 aluminum alloy, and is pin connected at its ends.
Determine the largest axial load that it can support.
Given: E 73.1 GPa⋅:= ao 150mm:= t 6mm:=
σY 414MPa:= L 3m:=
Solution:
Section Property : ai ao 2t−:=
A ao
2 ai
2−:= A 3456.00 mm2=
I
ao
4 ai
4−
12
:= I 11964672.00 mm4=
r
I
A
:= r 58.84 mm=
Slenderness Ratio :
For pinned ends, K 1.0:=
Aluminum Association Formula : λ'
K L⋅
r
:= λ' 50.99=
Since 12 < λ' < 55, it is an intermediate column. Applying Eq. 13.25 :
σallow 214.5 1.628λ'−( )MPa:=
σallow 131.49 MPa=
Hence, the allowable load is
Pallow σallow A⋅:=
Pallow 454.4 kN= Ans
Problem 13-95
The bar is made of aluminum alloy 2014-T6. Determine its thickness b if its width is 1.5b. Assume that
it is pin connected at its ends.
Given: E 73.1 GPa⋅:= P 4kN:= t b=
σY 414MPa:= L 1.5m:= h 1.5 b⋅=
Solution:
Section Property :
A t h⋅= A b 1.5 b⋅( )⋅= A 1.5 b2⋅=
Iy
h t3⋅
12
= Iy
1.5 b⋅( ) b3⋅
12
= Iy
b4
8
=
ry
Iy
A
= ry
b4
8 1.5 b2⋅( )⋅
= ry
b
12
=
Slenderness Ratio :
For pinned ends, K 1.0:=
Aluminum Association Formula : λ'
K L⋅
ry
= λ'
12 K⋅ L⋅
b
=
σ
P
A
= σ
P
1.5 b2⋅
=
Assume a long column, using Eq. 13.26 :
σallow
378125MPa
λ'2
= λ'2
378125MPa
σ
=
12 K⋅ L⋅
b
⎛⎜
⎝
⎞
⎠
2
378125MPa 1.5 b2⋅( )⋅
P
=
b
4
12 K2⋅ L2⋅ P⋅
378125MPa 1.5( )⋅
:=
b 20.89 mm=
Hence, ry
b
12
:= λ'
K L⋅
ry
:=
λ' 248.75= > 55 (Assumption is correct.)
Thus, b 20.89 mm= Ans
Problem 13-96
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 A-36 steel, and its ends are pinned.
Given: bf 300mm:= df 20mm:= P 1500kN:=
tw 10mm:= dw 310mm:= L 4m:=
E 200 GPa⋅:= σY 250MPa:=
Solution:
Section Property: D dw 2df+:=
A bf D⋅ dw bf tw−( )⋅−:= A 15100 mm2=
Iy
2df( ) bf3⋅
12
dw tw
3⋅
12
+:= Iy 90025833.33 mm
4=
ry
Iy
A
:= ry 77.21 mm=
AISC Column Formula : λ'
K L⋅
r
=
From Eq. 13.22, λ'c
2π2 E⋅
σY
:= λ'c 125.66=
For pinnd ends, K 1.0:=
λ'
K L⋅
ry
:= λ' 51.804= < λ'c 
Thus, it is an intermediate column. Using Eq. 13.23 :
σallow
1
λ'2
2λ'c
2
−
⎛
⎜
⎜
⎝
⎞
⎠
σY⋅
5
3
3λ'
8λ'c
+
λ'3
8λ'c
3
−
:= σallow 126.2 MPa=
Pallow A σallow⋅:=
Pallow 1905.8 kN= > P = 1500 kN (O.K.!)
Thus, the column is adequate. Ans
Problem 13-97
A 1.5-m-long rod is used in a machine to transmit an axial compressive load of 15 kN. Determine its
diameter if it is pin connected at its ends and is made of a 2014-T6 aluminum alloy.
Given: E 73.1 GPa⋅:= L 1.5m:=
σY 414MPa:= P 15kN:=
Solution:
Section Property :
A
π do
2⋅
4
= I
π do
4⋅
64
=
ry
I
A
= ry
π do
4⋅
64
4
π do
2⋅
⋅= ry
do
4
=
Slenderness Ratio :
For pinned ends, K 1.0:=
Aluminum Association Formula : λ'
K L⋅
ry
= λ'
4 K⋅ L⋅
do
=
σ
P
A
= σ
4P
π do
2⋅
=
Assume a long column, using Eq. 13.26 :
σallow
378125MPa
λ'2
= λ'2
378125MPa
σ
=
4 K⋅ L⋅
do
⎛
⎜
⎝
⎞
⎠
2 378125MPa π do
2⋅⎛⎝
⎞
⎠⋅
4P
=
do
4
64 K2⋅ L2⋅ P⋅
378125MPa π( )⋅
:=
do 36.72 mm=
Hence, ry
do
4
:= λ'
K L⋅
ry
:=
λ' 163.39= > 55 (Assumption is correct.)
Thus, do 36.72 mm= Ans
Problem 13-98
Solve Prob. 13-97 if the rod is fixed connected at ends.
Given: E 73.1 GPa⋅:= L 1.5m:=
σY 414MPa:= P 15kN:=
Solution:
Section Property :
A
π do
2⋅
4
= I
π do
4⋅
64
=
ry
I
A
= ry
π do
4⋅
64
4
π do
2⋅
⋅= ry
do
4
=
Slenderness Ratio :
For fixed ends, K 0.5:=
Aluminum Association Formula : λ'
K L⋅
ry
= λ'
4 K⋅ L⋅
do
=
σ
P
A
= σ
4P
π do
2⋅
=
Assume a long column, using Eq. 13.26 :
σallow
378125MPa
λ'2
= λ'2
378125MPa
σ
=
4 K⋅ L⋅
do
⎛
⎜
⎝
⎞
⎠
2 378125MPa π do
2⋅⎛⎝
⎞
⎠⋅
4P
=
do
4
64 K2⋅ L2⋅ P⋅
378125MPa π( )⋅
:=
do 25.97 mm=
Hence, ry
do
4
:= λ'
K L⋅
ry
:=
λ' 115.54= > 55 (Assumption is correct.)
Thus, do 25.97 mm= Ans
Problem 13-99
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 250 kN, determine its side dimensions a to the nearest multiples
of 10 mm. Use the NFPA formulas.
Given: L 4.2m:= P 250kN:=
Solution:
Section Property : A a2=
Slenderness Ratio :
For pinned ends, K 1.0:=
NFPA Timber Column Formula : λ'
K L⋅
a
=
σ
P
A
= σ
P
a2
=
Assume a long column, using Eq. 13.29 :
σallow
3718MPa