1. Find: Ae [in2] dbolt=1/2 [in] tdamaged=1/32 [in] 2 [in] A) 0.3 plate
C) 0.5
D) 0.6
plate
bolts P
Find the effective area of the L beam, in inches squared. [pause] In this probelm, ---
L 2 x 2 x 1/8

2. Find: Ae [in2] dbolt=1/2 [in] tdamaged=1/32 [in] 2 [in] A) 0.3 plate
C) 0.5
D) 0.6
plate
bolts P
a 2 by 2 by 1/8 L beam is bolted to a plate and subjected to a force P. The bolts are spaced 2 inches apart ---
L 2 x 2 x 1/8

3. Find: Ae [in2] dbolt=1/2 [in] tdamaged=1/32 [in] 2 [in] A) 0.3 plate
C) 0.5
D) 0.6
plate
bolts P
with a given bolt diamter, d bolt. The effective area of a steel member, A e, ---
L 2 x 2 x 1/8

4. Find: Ae [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] 2 [in]
plate
equals, the net area of that member, A net, times, ---
L 2 x 2 x 1/8

5. Find: Ae [in2] effective area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
net
reduction
2 [in]
area
plate
coefficient
the reduction coefficient, U. In this equation, the net area, equals, ---
L 2 x 2 x 1/8

6. Find: Ae [in2] net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
the gross area of the member, minus, the area from the holes. From a standard steel reference manual, ---
L 2 x 2 x 1/8

7. Find: Ae [in2] net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
we can lookup that a 2 by 2 by 1/8 L beam has gross area of ----
L 2 x 2 x 1/8

8. Find: Ae [in2] net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
Agross=0.484 [in2]
0.484 inches squared. [pause] The area from the holes, equals, the quantity of holes ---
L 2 x 2 x 1/8

9. Find: Ae [in2] net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2]
Aholes = (# of holes)
(dhole+2*tdamaged)
for a given facture path, times the sum of, the diamter of a single hole plus 2 times the damaged thickness around the hole, times the thickness of beam. *
beam thickness *
L 2 x 2 x 1/8

10. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2] 1
Aholes = (# of holes)
(dhole+2*tdamaged)
The number of holes, equals, 1, because the beam will fracture through ---- *
beam thickness *
L 2 x 2 x 1/8

11. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2] 1
Aholes = (# of holes)
(dhole+2*tdamaged)
just one of the bolt holes. [pause] As a typical value, we’ll use, 1/32 of an inch --- *
beam thickness *
fracture
path
L 2 x 2 x 1/8

12. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2] 1
Aholes = (# of holes)
(dhole+2*tdamaged)
as the damagaed thickness around the bolt hole, and the diamter of the hole itself, equals, --- *
beam thickness *
fracture
path
L 2 x 2 x 1/8

13. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2] 1
Aholes = (# of holes)
(dhole+2*tdamaged)
the diamter of the bolt, plus, 1/16 of an inch. Knowing the diamter of the bolt, --- *
beam thickness *
dhole= dbolt + 1/16 [in]
L 2 x 2 x 1/8

14. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2] 1
Aholes = (# of holes)
(dhole+2*tdamaged)
equals, half an inch, the diamter of the hole, equals, --- *
beam thickness *
dhole= dbolt + 1/16 [in]
L 2 x 2 x 1/8

15. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2] 1
Aholes = (# of holes)
(dhole+2*tdamaged)
9/16 of an inch. [pause] The beam thickness for this particular sized beam, --- *
beam thickness *
dhole= dbolt + 1/16 [in]
dhole= 9/16 [in]
L 2 x 2 x 1/8

16. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2] 1
Aholes = (# of holes)
(dhole+2*tdamaged)
can also be looked up in a standard steel reference manual, and equals, --- *
beam thickness *
dhole= 9/16 [in]
L 2 x 2 x 1/8

17. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2] 1
Aholes = (# of holes)
(dhole+2*tdamaged)
1/8 of an inch. [pause] After plugging in, --- *
beam thickness *
dhole= 9/16 [in]
t = 1/8 [in]
L 2 x 2 x 1/8

18. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2] 1
Aholes = (# of holes)
(dhole+2*tdamaged)
all the variables, the area from holes, equals, --- *
beam thickness *
dhole= 9/16 [in]
t = 1/8 [in]
L 2 x 2 x 1/8

19. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = Agross - Aholes
2 [in]
plate
0.484 [in2] 1
Aholes = (# of holes)
(dhole+2*tdamaged)
0.078 inches squared. [pause] Therefore, the net area of the beam, equals, --- *
beam thickness *
dhole= 9/16 [in]
1/8 [in]
Aholes = [in2]
L 2 x 2 x 1/8

20. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
2 [in]
plate
Anet = Agross - Aholes 1
0.484 [in2]
Aholes = (# of holes)
(dhole+2*tdamaged)
0.484 inches squared minus inches squared, which is, ---- *
beam thickness *
dhole= 9/16 [in]
1/8 [in]
Aholes = [in2]
L 2 x 2 x 1/8

21. Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
Anet = [in2]
2 [in]
plate
Anet = Agross - Aholes 1
0.484 [in2]
Aholes = (# of holes)
(dhole+2*tdamaged)
0.406 inches squared. [pause] Next we need to determine, --- *
beam thickness *
dhole= 9/16 [in]
1/8 [in]
Aholes = [in2]
L 2 x 2 x 1/8

22. Find: Ae [in2] Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
2 [in]
plate
coefficient
0.484 [in2]
Aholes = (# of holes)
(dhole+2*tdamaged)
the reduction coefficient, U. [pause] The reduction coefficient, equals, --- *
beam thickness *
dhole= 9/16 [in]
1/8 [in]
Aholes = [in2]
L 2 x 2 x 1/8

23. Find: Ae [in2] Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
2 [in]
plate
coefficient x
U=min 0.9, 1- L
the smaller of 0.9, and the quantity 1 minus x bar over L. In this equation, L refers to ---
L 2 x 2 x 1/8

24. Find: Ae [in2] Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
2 [in]
plate
coefficient x
U=min 0.9, 1- L L
the distance between the centers of the first and last bolt holes, which equals, ---
L 2 x 2 x 1/8

25. Find: Ae [in2] Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
2 [in]
plate
coefficient x
U=min 0.9, 1- L L
2 inches. [pause] Next, if we take a cross section, ---
L 2 x 2 x 1/8

26. Find: Ae [in2] A A’ Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction A
plate
coefficient x
U=min 0.9, 1-
2 [in] L
of the beam and plate, through A A prime, --- A’

27. Find: Ae [in2] A A’ L 2 x 2 x 1/8 dbolt=1/2 [in] tdamaged=1/32 [in]
plate
2 [in]
we can identify the value of x bar as --- A’
section A-A’

28. Find: Ae [in2] A A’ x L 2 x 2 x 1/8 dbolt=1/2 [in] tdamaged=1/32 [in]
plate
2 [in]
the distance between the centroid of the beam, and the plane of connection, --- A’
section A-A’

29. Find: Ae [in2] A A’ x L 2 x 2 x 1/8 dbolt=1/2 [in] tdamaged=1/32 [in]
plate
2 [in]
which for a 2 by 2 by 1/8 L beam, equals, ---- A’
section A-A’

30. Find: Ae [in2] A A’ x = x L 2 x 2 x 1/8 dbolt=1/2 [in] 0.546 [in]
tdamaged=1/32 [in] x A
plate
2 [in]
0.546 inches. Therefore, after plugging in the values for --- A’
section A-A’

31. Find: Ae [in2] x x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
coefficient x x
U=min 0.9, 1- L
L = 2 [in]
x bar, and, the length L, the reduction coefficient is the smaller of --- x =
0.546 [in]
section view

32. Find: Ae [in2] x x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
coefficient x x
U=min 0.9, 1- L
L = 2 [in]
0.9 and 0.727, which equals, x =
0.546 [in]
U = min (0.9, 0.727)
section view

33. Find: Ae [in2] x x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
coefficient x x
U=min 0.9, 1- L
L = 2 [in]
[pause] After plugging in the net area and reduction coefficient, --- x =
0.546 [in]
U = min (0.9, 0.727)
U = 0.727
section view

34. Find: Ae [in2] x x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
coefficient x x
U=min 0.9, 1- L
L = 2 [in]
the effective area of the L beam, equals, --- x =
0.546 [in]
U = min (0.9, 0.727)
U = 0.727
section view

35. Find: Ae [in2] x x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
Ae = [in2]
coefficient x x
U=min 0.9, 1- L
L = 2 [in]
0.295 inches squared. [pause] x =
0.546 [in]
U = min (0.9, 0.727)
U = 0.727
section view

36. Find: Ae [in2] x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
Ae = [in2]
coefficient x
U=min 0.9, 1- L
A) 0.3
B) 0.4
C) 0.5
D) 0.6
L = 2 [in]
When reviewing the possible solutions, --- x =
0.546 [in]
U = min (0.9, 0.727)
U = 0.727

37. Find: Ae [in2] x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U
tdamaged=1/32 [in]
reduction
Ae = [in2]
coefficient
answerA x
U=min 0.9, 1- L
A) 0.3
B) 0.4
C) 0.5
D) 0.6
L = 2 [in]
the answer is A. x =
0.546 [in]
U = min (0.9, 0.727)
U = 0.727