Steel Design Civil Engineering

Steel Design Civil Engineering
Part I (30 points) – Tension Test. A tensile test was conducted on a solid, circular-section
specimen with diameter 1.25 inch. The “gage length” (initial, longitudinal length of specimen) is
L0=5.50 inches. The deformation was measured using this length. The test data are shown below.
Deformation of
the coupon, ?L
(inches)
External force
applied to the
coupon, F (kips)
Please :
a. Draw the stress-strain curve from
force/deformation table shown to the left
(please, use engineering stress and
engineering strain).
.
b. Estimate the proportional limit
c. Determine best-fit (least squares) line in the
linear elastic range. Use the slope of the
best-fit line to estimate the Modulus of
Elasticity
d. Estimate the 0.2% offset yield strength
e. Estimate the ultimate stress.
f. If a load of 50.0 kips is applied and then
removed, estimate the permanent
deformation in inches (Hint: Is the
elastic/plastic limit going to be exceeded?).
0.00000 0.00000
0.00528 13.49903
0.00924 26.99806
0.01540 41.72428
0.01892 51.54175
0.01980 55.22331
0.02112 57.67768
0.02200 58.90486
0.02640 61.35923
0.03344 62.58642
0.04400 63.81360
0.10560 68.72234
0.13200 72.40389
0.18040 73.63108
0.35200 79.76700
0.52800 83.44855
0.70400 84.67574
0.79200 84.67574
0.88000 83.44855
Note (for Part I):Use MS Excel to plot data and to determine the best-fit line.
in paper format, only. Submission should include the stress/strain data (tabular form), any
plots that you have used, a summary of your results, etc. Submission of the Excel file is not
needed.
Part II (30 points) – Review of Mechanics & Statics. For each “built-up” section shown below:
a) Compute the Total Area, A (in2
);
b) Indicate the strong and weak axes of inertia of the built-up section (strong = principal axis
with maximum moment of inertia; weak = principal axis with minimum moment of inertia);
c) Compute the Strong-Axis Moment of Inertia of the built-up section, Ix (in4
), about the
centroid of the built-up section;
d) Compute the Weak-Axis Moment of Inertia of the built-up section, Iy (in4
), about the centroid
of the built-up section;
Part II.1
A W14 x 82 with a 8 in. x 1.0 in.
cover plate on each flange
Part II.2
Two side-by-side L6x6x1:
Note 1:
An example of “W-shape with cover plates” is shown
in Figure 1.8 of the textbook by Segui.
Note 2:
For Areas, Moments of Inertia of:
? Rectangular plates (part II.1), please check AISC,
Part 17, Table 17-27;
? For W14x82 (part II.1), L6x6x1 (part II.2), please
determine moment of inertias and areas from the
tables in Part 1 of the AISC manual.
Hint (to solve Part II.2): Please, recall the following properties of moments of inertias:
? Please, review “the parallel-axis theorem” to transfer moment of inertias along parallel axes;
x’
y’
CG
? I suggest that you also possibly review the chapter on inertias from the textbook by Gere and Goodno,
utilized in Analysis 1 and Statics. In particular, you may review the methods to determine the
centroidal moment of inertia about an inclined axis, given Ix’ and Iy’. Also, please recall the following
property for plane sections:
constant
(for any , pair of axes, with axes being mutually orthogonal, i.e., )
x y I I
x y xy
? ?
?
? The AISC manual has information on inertias Ix’ and Iy’ of each L shape (about centroid CG), and the
principal-minimum moment of inertia only, the direction of which is inclined with respect to x’ (or y’). In
the manual, there is no information on the maximum moment of inertia, which must be determined in
question c;
? If the built-up section has an axis of symmetry, one of the two principal axes of inertia will coincide
with the direction of this axis.
Part III (40 points) – LRFD Load Combinations
A steel column is subjected to an axial force, derived from the combination of various external
loads. The column is designed for an ultimate axial force, factored according to LRFD in AISC,
equal to Pu=700 kips, and caused by the most unfavorable combination of loads described in 1)
through 5) below. The nominal (unfactored) values of internal axial force, derived from the load
analysis of the column, are:
1) PD = (compression, unknown value) caused by dead load;
2) PL = 250 kips (compression) caused by live load of the floors (“L” in LRFD);
3) PLr = 50 kips (compression) caused by roof live load (“Lr” in LRFD);
4) Nominal wind loads (both a tension or compression forces may occur in the column
depending on wind direction on the building; both cases must be verified): PW,1=128 kips
(internal compression force) or PW,2=104 kips (internal tension force);
5) Nominal Earthquake loads (either tension or compression; both cases must be verified):
PE,1 = 60 kips (causing compression) or PE,2 = 60 kips (causing tension).
Determine the nominal axial force PD, which can be carried by the element, based on Pu above.
Please note that all LRFD load combinations apply.

FOR YOUR ASSIGNMENTS TO BE DONE AT A CHEAPER PRICE PLACE THIS ORDER OR A SIMILAR ORDER WITH US NOW