Assignment No 4 Ch 7.2 Tension
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DESIGN OF WOOD STRUCTURES
ASSIGNMENT No.
4, Chapt 7.2
Design of Tension Members
Problem No. 1
For the roof truss elevation shown in the figure below, determine if a 2 x 10 spruce-pine-
fir No. 1 member is adequate for the bottom chord, assuming that the roof dead load is 15
psf, the snow load is 45 psf, and the ceiling dead load is 10 psf of the horizontal plan
area. Trusses are spaced 24-in. on center. Assume normal temperature and dry service
conditions apply and that the members are connected with a single row of 3/4-in.-
diameter bolts with 1/16-in. oversized holes. The joints of the truss are assumed to be
pinned and the all gravity loads can be assumed to be applied as concentrated loads at the
joints of the truss.
Roof Dead Load = 15 psf; Snow load = 45 psf
Ceiling Dead Load = 10 psf
9 ft
3 ft
6 ft
6 ft
3 ft
9 ft
1/4
Problem No. 2
A 5-1/8 x 15 DF combination 5 glulam is used as the tension member in a large roof
truss. A single row of 1 in.-diameter bolts occurs at the net section of the member. It can
be assumed that the hole diameters are 1/8 in. larger than the bolt size. Loads are a
combination of dead and snow. Joints are assumed to be pin-connected.
.
Determine the factored axial capacity of the member assuming:
a)
Dry conditions (MC
16%).
b)
Wet conditions (MC > 16%).
Problem No. 3
The truss shown in the figure below has a 2 x 4 lower chord of Select Structural Spruce-
Pine-Fir (South). Trusses are space 24 in o.c. The uniform dead load on a horizontal
plane is 20 psf, while the snow load is 55 psf. There is no reduction of area for fasteners
and the truss is to be untreated and used in a dry normal building environment.
Verify the design in tension of the 2 x 4 lower chord.
2/4
Problem No. 4
Design the tension chord for the roof diaphragm of the warehouse building shown below.
The warehouse is 50 ft x 140 ft x 20 ft high. The roof is to be constructed of joists spaced
at 2 ft on centre supported on the top of glulam beams spaced at 10 ft on centre supported
by wood stud walls. The specified (unfactored) wind pressure on the long (140 ft) side of
the building is 24 psf. All framing lumber is S-P-F No. 2 grade and is used in a normal
building environment.
The roof diaphragm can be assumed to act as a simply supported beam and its moment
can be calculated from simple beam theory. Assume that each chord member is made of
two 2 x members splice together. Assume that a splice may occur at any location along
the 140 ft length of the chord member.
20 ft
50 ft
10 ft
14 @ 10 ft
Joists at
2 ft on center
3/4
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Problem No. 5
The elevation of a stud wall 16 feet long by 10 feet tall in an industrial building is shown
below. The wall is made of 2 x Hem Fir No. 3 studs spaced at 16 in. on center and of ½”
thick plywood. The unfactored design lateral force caused by wind and the unfactored
dead load resisted by the stud wall are also shown in the figure. For design purposes, it is
assumed that the overturning moment due to the wind load is resisted by two reactions
only at both end studs, while the gravity load is resisted in compression by all the studs
based on their tributary width. The moisture content is less than 19% and normal
temperatures apply.
a)
Calculate the factored tension force T
u
to be used for the design of the end studs. Note
that because the dead load causes compression (i.e. 0.5 C
D
) and the wind load causes
tension (i.e. T
w
) in the end studs, the overturning load combination (0.9D+1.0W)
must be considered.
b)
Determine the required width of the 2 x end studs to resist the factored tension force
calculated above.
4/4
16 ft
Wind Load
P
w
= 6.4 kips
Dead Load w
D
= 250 lbs/ft
Dead Load
Reactions
Wind Load Reactions
16 in typ.
10 ft
T
w
C
w
C
D
0.5C
D
0.5C
D
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Consider the truss shown in (Figure 1).
Part A
Identify the zero-force members in the truss.
Check all that apply.
• Vlew Available Hint(s)
O AB
O BC
O BD
O DE
O AE
O CD
BE
Figure
< 1 of 1
Submit
Previous Answers
3 kN
X Incorrect; Try Again; 5 attempts remaining
-2 m
2m
You have checked at least one member that is not a zerc
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session.masteringengineering.com
Homework #5
Grades for Sean E Stiner: (MA 244-02) (FA24) INTRO TO L...
M MasteringEngineering Mastering ComputerScience: Homew...
b Home | bartleby
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frameworks, solve the following:
Details for the bridge truss design:
8KN
6KN
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If one shoe is attached two-fifths of the way up the beam and another shoe is attached and three-fifths of the way up the beam, with θc = 16.5° and θb = 33.6° as shown in the figure, what is the tension in the cable, in newtons?
What is the x-component of the force, in newtons, that the pivot exerts on the bottom of the beam? Use the coordinate system specified in the figure.
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Homework No. 3:
4. A truss of 5 m span and 2.5 m height is subjected to wind load as shown in Fig. 12.39.
10 kN
20 kN
2.5 m
10 kN
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Example 8.6 A Weighted Forearm
Goal Apply the equilibrium conditions to the
human body.
-Humerus
Problem A W = 47.3 N (11 lb) weight is held
in a person's hand with the forearm horizontal,
as in Figure 8.11. The biceps muscle is
attached d = 0.0302 m from the joint, and the
weight is / = 0.351 m from the joint. Find the
upward force F exerted by the biceps on the
forearm (the ulna) and the downward force R
exerted by the humerus on the forearm, acting
at the joint. Neglect the weight of the forearm.
W
Biceps
Ulna
W
(a)
(b)
Figure 8.11 (a) A weight held with the forearm horizontal.
(b) The mechanical model for the system.
Strategy The forces acting on the forearm are equivalent to those acting on a bar of length 0.351 m, as shown
in Figure 8.1lb. Choose the usual x- and y-coordinates as shown and the axis O on the left end. (This
completes Steps 1 and 2.) Use the conditions of equilibrium to generate equations for the unknowns, and solve.
Solution
ET; = TR + TF + TBB = 0
R(0) + F(0.0302…
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1.5 m
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Ec
1 m
D.
1 m
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Figure 8
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10K
12'
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A 12' long beam in supported by an aluminum hanger at
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Esteel = 29,000,000 psi
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P16.52
Determine the maximum and minimum axial forces in mem-
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Solve for Stability and Determinacy for the truss frames shown.
Truss 1: b:
r =
Stable
Yes,
No
Truss 1
Truss 2: b =
r =
%3D
Stable
Yes,
No
Truss 2
Truss 3: b =
r =
%3D
Stable
Yes,
No
Truss 3
Truss 4: b =
r =
j=
%3D
Stable
Yes,
No
Truss 4
II
II
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B, C, and D. Find the axial forces in bar AF, CF, and BC.
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100 KN
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