The formula used for structural steelwork is the Perry-Robertson formula that represented as the average end stress to cause yield in a strut.
where;
σp = stress based on Perry-Robertson formula. σE = Euler’s stress
σyield = yield stress depending on the yield strength of material
η = constant depending on the material. For a brittle material η = 0.015L/r, for a ductile material η = 0.3 (L/100r)2 5.7 Rankine-Gordon Formula 2 ) / ( 1 a L r y R + = σ σ where E a 2y π σ
= theoretically but is usually found by experiment. Typical values are :
a Material Compressive yield stress
(MN/m2) Pinned ends Fixed ends
Mild steel 315 1/7500 1/30000
Cast iron 540 1/1600 1/6400
Appendix A
TUTORIAL 5
1. A steel column has a length of 9m and is fixed at its bottom and free at its top. If the cross-sectional area has the dimensions shown;
a) determine the critical load.
b) also check whether the Euler’s Equation is appropriate or not. Given Est= 200GPa, σy= 250 MPa. 200mm 10mm 10mm 150mm 10mm [Ans: a) Pcr = 82.25kN, b) σcr = 14.95MPa]
2. a) The 4m steel column has an outer diameter of 80mm and a thickness of
6mm. Determines the critical load if the bottom is fixed and the top are pinned. Also check whether the Euler’s Equation is appropriate or not. Given E= 210GPa, σy= 250 MPa.
b) A steel strut is 6.5m long and constructed from circular tube with an outside diameter of 120mm as shown in figure below. The type of strut is fixed at the bottom and pinned at the top. The strut must resist an axial load of 150kN from buckling. Using a factor of safety of 2.0, determine the required tube thickness and the critical buckling stress. Given E = 200GPa.
3. The steel angle has a cross sectional area of 1550mm2. A radius of gyration about the x axis is 32mm and about y axis is 22mm. The smallest radius of gyration occurs about z axis is 16mm. If the angle is to be used as a pin-connected with 3m long column, determine the largest axial load that can be applied through its centroid, C without causing it to buckle, E= 200GPa.
y
z
x C x
y z [Ans: Pcr =87.03kN]
4. a) Define the following terms; i. short column ii. slender column iii. Euler buckling load
b) List 5 assumptions or limitation of Euler’s formula.
c) A column is 8m high and is constructed from circular hollow section as shown in figure. The column is fixed at one end and is free at the other end. Determine the critical buckling load. Given E = 2x105 N/mm2.
5. a) The 4m steel pipe column has an outer diameter of 100mm and a thickness
of 8mm. Determine the critical load if the ends are assumed to be pin connected. Also check whether the Euler’s Equation is appropriate or not. Given E=210GPa , σy=250MPa.
b) Determine the Euler’s buckling load for an I-section of a column as shown in figure with length of 10m and fixed at both ends. Given E = 200GPa.
[Ans: a) Pcr = 19.31kN, σcr = 138.8MPa, b) Pcr = 459kN] 6. For the column fixed at the base and the top,
Mx M P Cut section y L P M P L P
Cut any section vertically.
M The internal moment at arbitrary section is;
. M Py M M Py M x x + − = − + =
The critical load acted is 4 22
L EI Pcr
π
7. a) Write the relationship between Euler’s buckling stress with slenderness ratio. b) If a pin-ended solid circular strut is 3m long and 60mm diameter, determine the
strut’s slenderness ratio.
c) If the same material on 7(b) is reshaped into a square bar with the length and same width of 60mm, determine the percentage reduction in the slenderness ratio. [Ans: b) L/r = 200, c) 13.4%]
8. A steel strut with total length of 5m is constructed of circular tube. The dimension of tube is diameter, Doutside = 100mm and thickness t = 12mm. The type of strut is partially restrained which is fixed at the bottom and pinned at the top. Determine;
a) slenderness ratio. b) Euler buckling load.
c) ratio of axial stress under the action of the buckling load to the elastic strength of the material. Given E=200GPa , σy=250MPa.
d) safety load the column can support with a factor of safety of 1.5. [Ans: a) L/r = 111.46, b) Pcr = 527kN, c) 0.635, d) Pallow = 351.3kN]
9. a) For a pin-ended solid rectangular column with a dimension 120cm x 50cm and 5m in length. Given E = 200GPa. Determine;
i. slenderness ratios at both axes. ii. Euler’s buckling load.
iii. allowable load with a factor of safety of 2.5.
b) Determine the Euler’s buckling load for an I-section of length 6m when fixed at both ends as shown in figure.
[Ans: a) i) (L/r)x = 34.65, (L/r)y = 14.43, ii) Pcr = 987MN, iii) Pallow =394.5MN, b) Pcr =26.4MN]
10. a) State all Euler’s buckling load formulas for different end support conditions. b) Determine the Euler’s buckling load for a C-section of column as shown in figure
with length of 10m and fixed at both ends. Given E = 200GPa.
[Ans: Pcr =436.63kN]
11. a) The cross sectional area of column which is fixed at the bottom and pinned at the top is shown in figure. The length of column is 1.5 meter. Given E = 200GPa. Determine;
i) slenderness ratio. ii) Euler’s buckling load.
iii) the allowable load with a factor of safety of 2.5.
b) For a fixed and free ends column of length 2m with a cross section as shown in figure, determine;
i) slenderness ratio of the column. ii) Euler’s buckling load.
iii) allowable load that the column can carry by assuming a factor of safety is 2.5. Given E = 200GPa.
[Ans: a) i) L/r = 20.21, ii) Pcr = 200.13MN, iii) Pallow =80MN, b) i) L/r = 128, ii) Pcr = 414.5kN, iii) Pallow = 165.8kN]
12. a) A steel column has a hollow circular cross sectional area with an outside diameter of 120mm and inside diameter of 100mm. The column is 6.0m length and is pin ended at both ends. Given E = 200GPa. Determine;
i. slenderness ratio
ii. Euler’s Buckling load
iii. Axial stress in the column when the column is subjected to the Euler’s Buckling load.
b) The cross sectional area of the column given in 12(a) is reshaped with the new radius of gyration r = 39.9mm and area = 11 355mm2. Determine the current value of slenderness ratio, Euler Buckling load and axial stress. c) A column of 3m height has a cross section as shown in figure. Both of its
end are fixed. Determine the safe load when a factor of safety equals to four. Check whether the Euler’s Equation is appropriate or not. Given E= 200GPa, σy= 250 MPa.