Assignment 2 Solutions

Problem 1)

True Length = SQRT of (diff x^2 + diff y^2 + diff z^2)

Column Length Beam Length
AE 7.00 AB 8.60 (root of 3^2+4^2+7^2)
BF 4.00 BC 4.00 (root of 0^2+0^2+4^2)
CG 4.00 CD 5.52 (root of 1.5^2+4^2+3.5^2)
DH 5.50 AD 12.59 (root of 1.5^2+12^2+3.5^2)

Although depicted in a top-over-front view, the text of the problem describes the drawing as actually the front view above and the top view below.
i.e. the lower drawing is a top view (this probelm only).

The areas are easiest solved using vector cross-products.
Divide the top into two triangles.

The problem becomes basic vector manipulation. An area of a triangle is one half the area of the parallelogram represented by two vectors.
( i.e. 1/2 (vector1 x vector2) )

Given the three points A(0,7,7) B(4,0,4) D(12,3.5,5.5)

Vector AB = (4-0,0-7,4-7) = (4,-7,-3)
Vector AD = (12-0,3.5-7,5.5-7) = (12,-3.5,-1.5)

The area is equal to the numerical value of 1/2 (AB x AD)

1/2 | i
| 4
| 12
j
-7
-3.5
k |
-3 |
-1.5 |
 = 1/2 |-30j + 70k| = 38.0788

Vector BC = (4,0,0)
Vector BD = (8,3.5,1.5) then, 1/2 (BC x BD) = 7.615773

Area of ABCD = Area of ABD + Area of BCD
Area of ABD = 38.07887 ft^2
Area of BCD = 7.615773 ft^2
Area of ABCD = 45.69464 ft^2

Thickness of ABCD = 0.020833 ft.
Volume of ABCD = 0.951972 ft^3
Density of Steel = 490 lb/ft^3 for ASTM A36 Structural Carbon Steel, Rolled

Total Weight of Top Sheet = 466.4661 lb.

Problem 2)

Again, this problem is solved quickly using vectors.

% Find relative Z values

Az = 8*tan(30) = 4.6188
Cz = 5*tan(45) = 5.0000

% Position of Vertices
B = [0 0 0]
A = [-8 0 -Az] = [-8 0 -4.6188]
C = [0 -5 -Cz] = [0 -5 -5]

% Create vectors
AB = B-A = [8 0 4.6188]
AC = C-A = [8 5 -0.3812]
CA = A-C = [-8 5 0.3812]
CB = B-C = [0 5 5]

% Find true length of AC (root of x^2+y^2+z^2)
X = norm(AC) = 9.4417

% Find vector angles
% ( angle = acos(dot product of two / multiplying norms of two )
AngA = acos(dot(AB,AC)/(norm(AB)*norm(AC))) = 44.4712
AngB = acos(dot(AB,CB)/(norm(AB)*norm(CB))) = 69.2952
AngC = acos(dot(CA,CB)/(norm(CA)*norm(CB))) = 66.2336

Problem 3)

This is a problem of Descriptive Geometry.

View P represents the Plan View (in our terms the "top")

View E1 represents the Elevation View (in our terms the "front")

The revolution method involves rotating a line, or part, so that the edge you are trying to measure is at a right angle to you as the viewer. Only then is its True Length (TL) shown.

The direct method here would mean to simply measure the lines shown to check your answers.

Slope can ONLY be determined in the elevation views (where there exists a horizontal plane) and only when an entity of true length is being referenced in the same view.

FIRST DIAGRAM

Here, the line AB is revolved around Axis A in the elevation view.
As shown, the length of AB is equal to ABr. After this clockwise revolution about Axis A, ABr is now inline with the horizontal plane. The projection of this into the Plan view (P) means ABr represents the true length in view P.

There is no slope of AB that may be indicated, because the line does not appear true length in an elevation view. If a line is revolved about a vertical axis, the angle between the line and a horizontal plane remains constant. In other words, the slope of the line is not changed when the line is revolved about a vertical axis.

SECOND DIAGRAM

Here, the line AB is revolved around Axis A in the Plan View.
As above, length ABr is equal to length AB. This revolution from B to Br moves AB, or now ABr, at a right angle when projected to the Elevation view.

In the Elevation view, point B moves parallel to the rotation line, or at a right angle to the axis of revolution.

The slope is taken directly from the sketch, at about 45 degrees.