2.3 Rectangular steel tubing
2.3.1.4 Inner rectangular area
Integrate and evaluate the limits of Equation 2.93 from x1 to h1/2 to obtain
(2.94)
dV2a = [fy/(h − c)](htan θ + 2zo)[x + (c − h/2)]dx (2.95) Integrate and evaluate the limits of Equation 2.95 from −x1 to x1 to obtain V2a = −[K2/(h − c)]2x1(c − h/2) (2.96) dV3a = −[fy/(h − c)](cot θ + tan θ) {x2 +
[c − (1/2)(h − h1)]x + (1/2)h1(c − h/2)}dx (2.97) Integrate and evaluate the limits of Equation 2.97 from −h/2 to −x1 to obtain
Case 3: (h/2 − x1) < c < h/2. V1a is integrated with limits from (h/2 − c) to V3a is the same as above except that the denominator has been changed to obtain
Similarly, V3b is integrated with limits from −(c − h/2) to −x1 to obtain Integrate and evaluate the limits of Equation 2.117 from x1 to (h/2 − c) to obtain
(2.118)
dV2ax2a = [fy/(h − c)](htan θ + 2zo)[x2 + (c − h/2)x]dx (2.119) Integrate and evaluate the limits of Equation 2.119 from −x1 to x1 to obtain
(2.120)
Integrate and evaluate the limits of Equation 2.121 from −h1/2 to −x1 to
V2ax2a is integrated with limits from −x1 to (h/2 − c) to get
(2.130) V3ax3a = Equation 2.126.
M = V1ax1a + V2bx2b + V2ax2a + V3ax3a (2.131) Case 4: h/2 < c < (h/2 + x1). V1ax1a = Equation 2.128, and V3ax3a is the same as above except that the denominator has been changed to obtain
(2.132)
V2ax2a is integrated with limits from −x1 to −(c − h/2) to get
(2.133) V2bx2b is integrated with limits from −(c − h/2) to ξ1 to get
(2.134) M = V1ax1a + V2bx2b + V2ax2a + V3ax3a (2.135) Case 5: (h/2 + x1) < c < (h1/2 + h/2). V1ax1a = Equation 2.128, and V2bx2b is integrated with limits from −x1 to x1 to get
(2.136) V3bx3b is integrated with limits from −(c − h/2) to −x1 to get
(2.137)
V3ax3a is integrated with limits from −h1/2 to −(c − h/2) to get
(2.138) M = V1ax1a + V2bx2b − V3bx3b + V3ax3a (2.139) V x2a 2a= −[K2/ (6h c− )] ( /
{
h 2−c)3+x12[2x1+3(c h− / ))]2}
V x3a 3a=(K1/12c x)
{
12⎡⎣⎣3x12−4x c1( −0 50. h+0 50. h1) + 3h c h1( − / )2⎤⎦−( )
h13/ (4 c−0 25. h1−0 50. h)}}
V x2a 2a=(K2/ ) (6c
{
c h− / )23−x12[−2x1+3(c h− / )]2}
V x2b 2b=(K2/ )6c x
{
12[2x1+3(c h− / )] (2 − −c h/ )23}
V x2b 2b=(K2/ )3 2c x13
V x3b 3b =(K1/12c x)
{{
12[3x12−4x c1( −0 50. h+0 50. h1) + 3h c h1( − / )] (2 − −c h/ ) ( .23 0 50h h+ −1 c)}
V x3a 3a=(K1/12c) (
{
c h− / ) ( .230 50h h+ − −1 c)( )
h13/ ((4 c−0 25. h1−0 50. h)}
Case 6: c > (h/2 + h1/2). V1ax1a = Equation 2.128 and V2bx2b = Equation 2.136.
V3bx3b is integrated with limits from −h1/2 to −x1 to get
(2.140)
M = V1ax1a + V2bx2b − V3bx3b (2.141)
in which
K1 = (cot θ + tan θ) (2.142)
K2 = (htan θ + 2zo) (2.143) x2 = (1/2)(d sin θ − b cos θ) (2.144) x1 = (1/2)(d1 sin θ − b1cos θ) (2.145)
h = b cos θ + d sin θ (2.146)
h1 = b1 cos θ + d1 sin θ (2.147)
b1 = b − 2t (2.148)
d1 = d − 2t (2.149)
zo = (1/2)(b cos θ − d sin θ) (2.150) zo 1 = (1/2)(b1 cos θ − d1sin θ) (2.151) Equations 2.147 to 2.151 yield the capacity of the section along the capacity axis, measured as M (moment capacity) and P (axial capacity). The value of M is sufficient to compare with the component of the resultant external bending moment along this axis for equilibrium or safety in design.
To obtain MR (the resultant moment capacity), Mz (the moment perpendicular to the capacity axis) should be included such that
(2.152) V x3b 3b=(K1/12c x)
{
12⎡⎣⎣3x12−4x c1( −0 50. h+0 50. h1)+ 3h c h1( − / )2⎤⎦−
( )
h13/ (4 c−0 25. h1−0 50. h)}}
MR=
{
M2+Mz2}
1 2/in which
Mz = (V1z1 + V2z2 + V3z3) (2.153) V1z1 = 0.50(cot θ − tan θ)V1x1 + {zo − 0.25h(cot θ − tan θ)}V1 (2.154) V2z2 = V2x2 tan θ (2.155) V3z3 = −0.50(cot θ − tan θ)V3x3 + {0.25h(cot θ − tan θ) − zo}V3 (2.156) P = (V1 + V2 + V3) (2.157) Note: When α < θ < (π/2 − α), the value of K2 in Equation 2.143 is changed to K2 = (h tanθ − 2zo) (2.158) in which
α = arctan(b/d) (2.159)
for the outer rectangle, and
α = arctan(b1/d1) (2.160) for the inner rectangle. When θ > (π/2 − α), Equation 2.155 is changed to
V2z2 = V2x2/tanθ (2.161) and Equation 2.143 is changed to
K2 = [(h/tanθ) + 2zo] (2.162) Equations 2.144, 2.145, 2.150, and 2.151 are changed to the following:
x2 = (1/2)(b sin θ − d cos θ) (2.163) x1 = (1/2)(b1 sin θ − d1 cos θ) (2.164) zo = (1/2)(d sin θ − b cos θ) (2.165) zo1 = (1/2)(d1 sin θ − b1 cos θ) (2.166) The pair of values of P and MR define the yield capacity of the steel section. The plot of these values will trace the yield capacity curve of the
steel section as determined from the above equations for rectangular tubing. For a square steel tubular section, set b = d in all the above equations.
2.3.1.5 Capacity curves
Capacity curves are obtained from Microsoft Excel worksheets where the above-listed equations (2.47 to 2.166) are programmed to solve the capaci-ties for any rectangular tubing. Figure 2.14 is the yield capacity curve of a rectangular steel tubing that is 8 × 4 in. (203.2 × 101.6 mm), for which t = 0.250 in. (6.35 mm) and fy = 36 ksi (248 MPa). The figure shows the plot of the yield capacity of this rectangular steel tubing at all envelopes of the compressive depth c of the section using the diagonal of the outer rectan-gular section as the capacity axis.
Figure 2.14 Rectangular steel tubing capacity curve.
203.2 × 101.6 tubing, t = 6.35 mm., fy = 248 MPa
–600 –400 –200 0 200 400 600 800 1000
0 10 20 30 40
Moment capacity in kN-m.
Axial capacity in kN
50
Note that values below the zero line are indicated for conditions of compressive depth less than the balanced condition. The balanced condition is where the axial capacity is zero and the moment value is a maximum.
This is the case for beams subjected to bending loads only.
For sections without tension, the values above the zero line are compared with the external loads the designer is considering. Figure 2.15 shows the column capacity curve for this example of the analytical method in predicting the yield capacity of rectangular tubing for column design.
To use the capacity curve, all that is needed is to determine the external loads to which the pipe will be subjected. Plot the set or sets of external axial and bending moment loads on the capacity curve. Then determine the locations of the plotted values with respect to the capacity curve. When these external loads are within the envelope of the capacity
Figure 2.15 Rectangular tubing column capacity curve.
Column Capacity Curve
0 100 200 300 400 500 600 700 800 900 1000
0 10 20 30 40
Moment capacity in kN-m
Axial capacity in kN
50
curve, the chosen tubular section is considered sufficient to support the external loads.
The decision as to where inside the curve to limit the external loads is the responsibility of the civil or structural engineer. Every civil/structural engineer can choose all the multipliers to the external loads such as moment magnification factor for slender columns, unanticipated eccen-tricities of the applied axial load, multipliers to live loads, code require-ments, safety factors, and other factors deemed applicable to a particular application.
2.3.1.6 Key points
The schematic locations of the key points on the capacity curve are similar to Figure 2.6. The following descriptions of these key points are for rectan-gular tubing. Three important key points on the yield capacity curve of rectangular steel tubing are Q, S, and U.
Key point U is the position of the compressive depth c at the center of the rectangular tubing section. It is the so-called balanced condition of load-ing when the compressive and tensile yield stress of the steel is developed at the same time. The value of the axial capacity is zero.
Key point Q is the condition when the compressive depth c approaches infinity (or a very large number). At that instant, the distribution of com-pressive stress becomes uniform and the value of the moment capacity approaches zero. The axial capacity is then simply the area of the rectangular tubing times the yield stress of steel.
Key point S is when the compressive depth is equal to h, the start of full compressive stress on the rectangular tubing.
The values of the axial and moment capacities at this point are recom-mended for minimum eccentricity criteria of the applied external loads.
For columns, key point Q is difficult, if not impossible, to attain because it is inherently out-of-plumb from the vertical axis of any column and because of buckling at mid-height of column length. Because of the unavoid-able eccentric loading of external loads on a column section, the minimum axial capacity may be limited at key point S. The value of P at key point S can also be set as the limit of the total compressive force to be applied at the bearing plates for columns.
Key point V is the limit of the capacity curve in the tension zone. This is attained when the compressive depth becomes very small, that is, less than t. When the compressive depth approaches zero, the numerical value of the axial yield capacity approaches that at key point S.
To use the capacity curve, plot sets of external axial and bending moment loads applied to the tubular section. These sets of external loads can represent any combination of loadings required by building codes or any applicable codes in structural design. For safety in design, these sets of points should be inside the capacity curve.
The column capacity curve of the example tubing is shown in Figure 2.15.
For columns with predominant tensile capacities, change the label
“Compression Zone” to “Tension Zone” and use the same plot of the column capacity curve. The values of the yield capacities at key points are shown in Table 2.8.