1. Find: Lightest Angled Beam
γ * Q = 84,000 [lb]
steel A572 grade 50
L = 12 [ft]
U = 0.85 L ≤
240
dbolt = 7/8 [in]
rmin
plate
L beam
neglect block
shear failure
A) L 4 x 4 x 5/16
B) L 3.5x3.5x3/8
C) L 3 x 3 x 7/16
D) L 4 x 3 x 3/8
Find the lightest angled beam. [pause] In this problem, an L beam is --- P

2. Find: Lightest Angled Beam
γ * Q = 84,000 [lb]
steel A572 grade 50
L = 12 [ft]
U = 0.85 L ≤
240
dbolt = 7/8 [in]
rmin
plate
L beam
neglect block
shear failure
A) L 4 x 4 x 5/16
B) L 3.5x3.5x3/8
C) L 3 x 3 x 7/16
D) L 4 x 3 x 3/8
subjected to a tensile force, P. The beam is fastened to a plate with 3 bolts --- P

3. Find: Lightest Angled Beam
γ * Q = 84,000 [lb]
steel A572 grade 50
L = 12 [ft]
U = 0.85 L ≤
240
dbolt = 7/8 [in]
rmin
plate
L beam
neglect block
shear failure
A) L 4 x 4 x 5/16
B) L 3.5x3.5x3/8
C) L 3 x 3 x 7/16
D) L 4 x 3 x 3/8
in a line, of a given bolt diameter. The type of steel and --- P

4. Find: Lightest Angled Beam
γ * Q = 84,000 [lb]
steel A572 grade 50
L = 12 [ft]
U = 0.85 L ≤
240
dbolt = 7/8 [in]
rmin
plate
L beam
neglect block
shear failure
A) L 4 x 4 x 5/16
B) L 3.5x3.5x3/8
C) L 3 x 3 x 7/16
D) L 4 x 3 x 3/8
factored load are provided, as well as the length of the beam, L, and --- P

5. Find: Lightest Angled Beam
γ * Q = 84,000 [lb]
steel A572 grade 50
L = 12 [ft]
U = 0.85 L ≤
240
dbolt = 7/8 [in]
rmin
plate
L beam
neglect block
shear failure
A) L 4 x 4 x 5/16
B) L 3.5x3.5x3/8
C) L 3 x 3 x 7/16
D) L 4 x 3 x 3/8
reduction coefficient, U. [pause] If we look ahead ---- P

6. Find: Lightest Angled Beam
γ * Q = 84,000 [lb]
steel A572 grade 50
L = 12 [ft]
U = 0.85 L ≤
240
dbolt = 7/8 [in]
rmin
plate
L beam
neglect block
shear failure
A) L 4 x 4 x 5/16
B) L 3.5x3.5x3/8
C) L 3 x 3 x 7/16
D) L 4 x 3 x 3/8
to the possible solutions, and organize them in a table, ---- P

7. Find: Lightest Angled Beam
γ * Q = 84,000 [lb]
steel A572 grade 50
L = 12 [ft]
U = 0.85 L ≤
240
dbolt = 7/8 [in]
rmin
Beam
Strength
Slender
Weight
L 4 x 4 x 5/16 ? ? #
we’ll select the lightest beam which satisfies the strength requirement, ---
L3.5x3.5x3/8 ? ? #
L 3 x 3 x 7/16 ? ? #
L 4 x 3 x 3/8 ? ? #

8. Find: Lightest Angled Beam
γ * Q = 84,000 [lb]
steel A572 grade 50
L = 12 [ft]
U = 0.85 L ≤
240
dbolt = 7/8 [in]
rmin
Beam
Strength
Slender
Weight
L 4 x 4 x 5/16 ? ? #
and slenderness requirement. We’ll first check each beam for the ---
L3.5x3.5x3/8 ? ? #
L 3 x 3 x 7/16 ? ? #
L 4 x 3 x 3/8 ? ? #

9. Find: Lightest Angled Beam
γ * Q = 84,000 [lb]
steel A572 grade 50
L = 12 [ft]
U = 0.85 L ≤
240
dbolt = 7/8 [in]
rmin
Beam
Strength
Slender
Weight
L 4 x 4 x 5/16 ? ? #
slenderness requirement, which states the length of the beam, ---
L3.5x3.5x3/8 ? ? #
L 3 x 3 x 7/16 ? ? #
L 4 x 3 x 3/8 ? ? #

10. Find: Lightest Angled Beam≤
240
L = 12 [ft]
rmin
Beam
Strength
Slender
Weight
L 4 x 4 x 5/16 ? ? #
divided by the minimum radius of gyration, shall not exceed 240.
L3.5x3.5x3/8 ? ? #
L 3 x 3 x 7/16 ? ? #
L 4 x 3 x 3/8 ? ? #

11. Find: Lightest Angled Beam≤
240
L = 12 [ft]
rmin L
rmin ≥
240
Beam
Strength
Slender
Weight
L 4 x 4 x 5/16 ? ? #
If we solve for r min, and plug in the length of the beam, L, ---
L3.5x3.5x3/8 ? ? #
L 3 x 3 x 7/16 ? ? #
L 4 x 3 x 3/8 ? ? #

12. Find: Lightest Angled Beam≤
240
L = 12 [ft] 12
rmin * ft L
rmin ≥
240
Beam
Strength
Slender
Weight
L 4 x 4 x 5/16 ? ? #
in inches, the minimum permissible radius of gyration equals, ---
L3.5x3.5x3/8 ? ? #
L 3 x 3 x 7/16 ? ? #
L 4 x 3 x 3/8 ? ? #

13. Find: Lightest Angled Beam≤
240
L = 12 [ft] 12
rmin * ft L
rmin ≥
rmin ≥ 0.60 [in]
240
Beam
Strength
Slender
Weight
L 4 x 4 x 5/16 ? ? #
0.60 inches. [pause] Therefore, if any of the L beams have a radius of gyrations smaller than ---
L3.5x3.5x3/8 ? ? #
L 3 x 3 x 7/16 ? ? #
L 4 x 3 x 3/8 ? ? #

14. Find: Lightest Angled Beam≤
240
L = 12 [ft] 12
rmin * ft L
rmin ≥
rmin ≥ 0.60 [in]
slenderness
240
requirement
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ? ? #
0.60 inches, they are too slender, and hence disqualified from the selection. After looking up ---
L3.5x3.5x3/8 ? ? #
L 3 x 3 x 7/16 ? ? #
L 4 x 3 x 3/8 ? ? #

15. Find: Lightest Angled Beam≤
240
L = 12 [ft] 12
rmin * ft L
rmin ≥
rmin ≥ 0.60 [in]
slenderness
240
requirement
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
the minimum radius of gyration values, for each beam, we notice all 4 beams ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

16. Find: Lightest Angled Beam≤
240
L = 12 [ft] 12
rmin * ft L
rmin ≥
rmin ≥ 0.60 [in]
slenderness
240
requirement
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
exceed the minimum requirement, and none are disqualified, based on ----
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

17. Find: Lightest Angled Beam≤
240
L = 12 [ft] 12
rmin * ft L
rmin ≥
rmin ≥ 0.60 [in]
slenderness
240
requirement
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
the slederness requriement. [pause] Next we’ll check each beam for the ----
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

18. Find: Lightest Angled Beam≤
240
rmin L
rmin ≥
240
rmin ≥ 0.60 [in]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
strength requirement. Since the problem states to neglect ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

19. Find: Lightest Angled Beam≤
neglect block
240
rmin
shear failure L
rmin ≥
240
rmin ≥ 0.60 [in]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
block shear failure, we know the strength requirement will be based on either ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

20. Find: Lightest Angled Beam
failure mode:
neglect block
1. yield
shear failure
2. fracture
rmin ≥ 0.60 [in]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
a yield-type failure, or fracture-type failure. [pause] In each case, the factored load, ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

21. Find: Lightest Angled Beam
γ * Q ≤ φ * R
failure mode:
1. yield
2. fracture
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
gamma Q, shall not exceed the design strength, ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

22. Find: Lightest Angled Beam
γ * Q ≤ φ * R
failure mode:
1. yield
2. fracture
factored
design
load
strength
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
R phi. From the problem statement we know the ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

23. Find: Lightest Angled Beam
γ * Q ≤ φ * R
failure mode:
1. yield
2. fracture
factored
design
load
strength
γ * Q = 84,000 [lb]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
factored load equals, 84,000 pounds. Next, we’ll compute the design strength ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

24. Find: Lightest Angled Beam
γ * Q ≤ φ * R
failure mode:
1. yield
2. fracture
factored
design
load
strength
γ * Q = 84,000 [lb]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
for each failure mode, determine which failure mode governs, ----
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

25. Find: Lightest Angled Beam
γ * Q ≤ φ * R
failure mode:
1. yield
2. fracture
factored
design
load
strength
γ * Q = 84,000 [lb]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
and then compare whether or not each of the 4 L beams satisfy ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

26. Find: Lightest Angled Beam
γ * Q ≤ φ * R
failure mode:
1. yield
2. fracture
factored
design
load
strength
γ * Q = 84,000 [lb]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
the strength requirement. We’ll first look at a yield-type failure, ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

27. Find: Lightest Angled Beam
failure mode:
84,000 [lb] ≤ φyield * Ryield
1. yield
2. fracture
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
where the strength reduction factor, phi yield, equals, ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

28. Find: Lightest Angled Beam
failure mode:
84,000 [lb] ≤ φyield * Ryield
1. yield
0.9
2. fracture
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
0.9, and the resistance term, R yield, equals, ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

29. Find: Lightest Angled Beam
failure mode:
84,000 [lb] ≤ φyield * Ryield
1. yield
0.9
2. fracture
fy * Agross
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
the yield stress times the gross area of the beam. The problem states the steel is ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

30. Find: Lightest Angled Beam
failure mode:
84,000 [lb] ≤ φyield * Ryield
1. yield
0.9
2. fracture
fy * Agross
A572 grade 50
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
type A572 grade 50, which has a yield stress of ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

31. Find: Lightest Angled Beam
failure mode:
84,000 [lb] ≤ φyield * Ryield
1. yield
0.9
2. fracture
fy * Agross
A572 grade 50
fy=50,000 [lb/in2]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
50,000 pounds per square inch. Next, if we isolate the gross area term, the gross area of the beam ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

32. Find: Lightest Angled Beam
Agross≥ 1.86 [in2]
failure mode:
1. yield
84,000 [lb] ≤ φyield * Ryield
2. fracture
0.9
fy * Agross
A572 grade 50
fy=50,000 [lb/in2]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
shall be at least 1.86 inches squared. [pause] Now we’ll examine ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

33. Find: Lightest Angled Beam
failure mode:
Agross≥ 1.86 [in2]
1. yield
2. fracture
84,000 [lb] ≤φfracture * Rfracture
A572 grade 50
fy=50,000 [lb/in2]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
a fracture-type failure. Where the strength reduction factor, phi fracture, ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

34. Find: Lightest Angled Beam
failure mode:
0.75
Agross≥ 1.86 [in2]
1. yield
2. fracture
84,000 [lb] ≤φfracture * Rfracture
A572 grade 50
fy=50,000 [lb/in2]
r [in]
Beam
Strength
Weight
L 4 x 4 x 5/16 ?
1.24 #
equals, 0.75, and the resistance term, R fracture, ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

35. Find: Lightest Angled Beam
failure mode:
0.75
Agross≥ 1.86 [in2]
1. yield
2. fracture
84,000 [lb] ≤φfracture * Rfracture
A572 grade 50
fu * Ae
Beam
Strength
L 4 x 4 x 5/16 ?
equals the ultimate stress of the steel, f u, times the effective area of the steel, ---
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
L 4 x 3 x 3/8 ?

36. Find: Lightest Angled Beam
failure mode:
0.75
Agross≥ 1.86 [in2]
1. yield
2. fracture
84,000 [lb] ≤φfracture * Rfracture
A572 grade 50
fu * Ae
Beam
Strength
ultimate
L 4 x 4 x 5/16 ?
stress
A e. [pause] Given the type of steel, we know the ultimate stress equals, ---
L3.5x3.5x3/8 ?
effective
L 3 x 3 x 7/16 ?
area
L 4 x 3 x 3/8 ?

37. Find: Lightest Angled Beam
failure mode:
0.75
Agross≥ 1.86 [in2]
1. yield
2. fracture
84,000 [lb] ≤φfracture * Rfracture
A572 grade 50
fu=65,000 [lb/in2]
Beam
Strength
fu * Ae
L 4 x 4 x 5/16 ?
65,000 pounds per square inch. And the effective area, A e, equals, ---
ultimate
L3.5x3.5x3/8 ?
stress
L 3 x 3 x 7/16 ?
effective
L 4 x 3 x 3/8 ?
area

38. Find: Lightest Angled Beam
failure mode:
0.75
Agross≥ 1.86 [in2]
1. yield
2. fracture
84,000 [lb] ≤φfracture * Rfracture
A572 grade 50
fu=65,000 [lb/in2]
Beam
Strength
fu * Ae
L 4 x 4 x 5/16 ?
the net area, A net, times, U, the reduction coefficient. The problem states ---
Ae=Anet * U
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
effective
L 4 x 3 x 3/8 ?
area

39. Find: Lightest Angled Beam
failure mode:
0.75
Agross≥ 1.86 [in2]
1. yield
2. fracture
84,000 [lb] ≤φfracture * Rfracture
A572 grade 50
fu=65,000 [lb/in2]
Beam
Strength
fu * Ae
L 4 x 4 x 5/16 ?
the reduction coefficient, equals, Therefore, if we solve for ---
Ae=Anet * U
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
U=0.85
L 4 x 3 x 3/8 ?

40. Find: Lightest Angled Beam
failure mode:
0.75
Agross≥ 1.86 [in2]
1. yield
2. fracture
84,000 [lb] ≤φfracture * Rfracture
A572 grade 50
fu=65,000 [lb/in2]
Beam
Strength
fu * Ae
L 4 x 4 x 5/16 ?
the net area, we find that the net area of the beam shall be at least, ---
Ae=Anet * U
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
U=0.85
L 4 x 3 x 3/8 ?

41. Find: Lightest Angled Beam
failure mode:
0.75
Agross≥ 1.86 [in2]
1. yield
2. fracture
84,000 [lb] ≤φfracture * Rfracture
A572 grade 50
fu=65,000 [lb/in2]
Beam
Strength
fu * Ae
L 4 x 4 x 5/16 ?
2.03 inches squared. [pause] Since the net area will never be greater than ---
Ae=Anet * U
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
U=0.85
L 4 x 3 x 3/8 ?
Anet≥ 2.03 [in2]

42. Find: Lightest Angled Beam
Anet ≤ Agross
failure mode:
Agross≥ 1.86 [in2]
1. yield
0.75
2. fracture
84,000 [lb] ≤φfracture * Rfracture
A572 grade 50
fu=65,000 [lb/in2]
Beam
Strength
fu * Ae
L 4 x 4 x 5/16 ?
the gross area, a fracture-type failure strength requriement is more conservative, ---
Ae=Anet * U
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
U=0.85
L 4 x 3 x 3/8 ?
Anet≥ 2.03 [in2]

43. Find: Lightest Angled Beam
Anet ≤ Agross
failure mode:
Agross≥ 1.86 [in2]
1. yield
2. fracture
Anet≥ 2.03 [in2]
A572 grade 50
fu=65,000 [lb/in2]
Beam
Strength
fu * Ae
L 4 x 4 x 5/16 ?
and will serve as the governing equation for the strength requirement of the L beam. Therefore, ---
Ae=Anet * U
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
U=0.85
L 4 x 3 x 3/8 ?
Anet≥ 2.03 [in2]

44. Find: Lightest Angled Beam
Anet ≤ Agross
failure mode:
Agross≥ 1.86 [in2]
1. yield
2. fracture
Anet≥ 2.03 [in2]
A572 grade 50
fu=65,000 [lb/in2]
Beam
Anet [in2]
fu * Ae
L 4 x 4 x 5/16 ?
we’ll compute the net area of each L beam, and compare it’s values to the ---
Ae=Anet * U
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
U=0.85
L 4 x 3 x 3/8 ?
Anet≥ 2.03 [in2]

45. Find: Lightest Angled Beam
Anet ≤ Agross
failure mode:
Agross≥ 1.86 [in2]
1. yield
2. fracture
Anet≥ 2.03 [in2]
A572 grade 50
Beam
Anet [in2]
L 4 x 4 x 5/16 ?
minimum required net area. As an example, we’ll compute ---
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
L 4 x 3 x 3/8 ?

46. Find: Lightest Angled Beam
Anet ≤ Agross
failure mode:
Agross≥ 1.86 [in2]
1. yield
2. fracture
Anet≥ 2.03 [in2]
A572 grade 50
Beam
Anet [in2]
L 4 x 4 x 5/16 ?
the net area for beam L 4 by 4 by 5/16. Which equals, ---
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
L 4 x 3 x 3/8 ?

47. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
Anet≥ 2.03 [in2]
Beam
Anet [in2]
L 4 x 4 x 5/16 ?
the gross area of the beam, minus the number of holes along the fracture path, times ---
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
L 4 x 3 x 3/8 ?

48. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
gross
thickness
bolt
area
diameter
Anet≥ 2.03 [in2]
Beam
Anet [in2]
L 4 x 4 x 5/16 ?
the quantity, the diameter of the bolt plus 1/8 inch, times the thickness of the beam.
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
L 4 x 3 x 3/8 ?

49. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
gross
thickness
bolt
area
diameter
Anet≥ 2.03 [in2]
Beam
Anet [in2]
L 4 x 4 x 5/16 ?
For a 4 by 4 by 5/16 L beam, the gross area equals, ---
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
L 4 x 3 x 3/8 ?

50. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
2.40 [in2]
5/16 [in]
bolt
diameter
Beam
Anet [in2]
Anet≥ 2.03 [in2]
L 4 x 4 x 5/16 ?
2.40 inches squared, and the thickness equals, 5/16 inches. [pause] From the diagram, ---
L3.5x3.5x3/8 ?
L 3 x 3 x 7/16 ?
L 4 x 3 x 3/8 ?

51. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
2.40 [in2]
5/16 [in]
bolt
diameter
Anet≥ 2.03 [in2]
we notice the fracture path would only pass through ---- P

52. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
2.40 [in2]
5/16 [in]
bolt
diameter
fracture path
Anet≥ 2.03 [in2]
1 bolt hole, therefore the number of holes in the equation, equals, --- P

53. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
2.40 [in2]
5/16 [in]
bolt 1
diameter
fracture path
Anet≥ 2.03 [in2]
1. [pause] The problem provides the bolt diamter, as, --- P

54. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
2.40 [in2]
7/8 [in]
5/16 [in] 1
fracture path
Anet≥ 2.03 [in2]
7/8 of an inch, and the 1/8 inch term accounts for ---- P

55. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
2.40 [in2]
7/8 [in]
5/16 [in] 1
hole diameter &
fracture path
damaged thickness
the diameter of the hole and the damaged thickness around the hole. P

56. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
2.40 [in2]
7/8 [in]
5/16 [in] 1
Beam
Anet [in2]
r [in]
Weight
L 4 x 4 x 5/16 ?
1.24 #
After plugging in our values, the net area for a 4 by 4 by 5/16 L beam, ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

57. Find: Lightest Angled Beam
Anet = Agross- # holes * ( dbolt + 1/8 [in] ) * t
2.40 [in2]
7/8 [in]
5/16 [in] 1
Anet,4x4x5/16 = 2.09[in2]
Beam
Anet [in2]
r [in]
Weight
L 4 x 4 x 5/16
2.09
1.24 #
equals, 2.09 inches squared. [pause] This calculation is repeated for the other 3 beam sizes, ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

58. Find: Lightest Angled Beam
Anet = Agross- 1* ( 7/8 [in] + 1/8 [in] ) * t
Agross,3.5x3.5x3/8
t3.5x3.5x3/8
Agross,3x3x7/16
t3x3x7/16
Agross,4x3x3/8
t4x3x3/8
Anet [in2]
r [in]
Beam
Weight
L 4 x 4 x 5/16
2.09
1.24 #
using the gross area and thickness unique to each beam. Once all the areas are computed, ---
L3.5x3.5x3/8 ?
1.07 #
L 3 x 3 x 7/16 ?
0.905 #
L 4 x 3 x 3/8 ?
0.879 #

59. Find: Lightest Angled Beam
Anet = Agross- 1* ( 7/8 [in] + 1/8 [in] ) * t
Agross,3.5x3.5x3/8
t3.5x3.5x3/8
Agross,3x3x7/16
t3x3x7/16
Agross,4x3x3/8
t4x3x3/8
Anet [in2]
r [in]
Beam
Weight
L 4 x 4 x 5/16
2.09
1.24 #
we’ll check if each beam meets the minimum net area ---
L3.5x3.5x3/8
2.11
1.07 #
L 3 x 3 x 7/16
1.99
0.905 #
L 4 x 3 x 3/8
2.11
0.879 #

60. Find: Lightest Angled Beam
Anet = Agross- 1* ( 7/8 [in] + 1/8 [in] ) * t
Anet≥ 2.03 [in2]
Beam
Anet [in2]
r [in]
Weight
L 4 x 4 x 5/16
2.09
1.24 #
which is not met by L 3 by 3 by 7/16 beam , so we can remove this beam from ---
L3.5x3.5x3/8
2.11
1.07 #
L 3 x 3 x 7/16
1.99
0.905 #
L 4 x 3 x 3/8
2.11
0.879 #

61. Find: Lightest Angled Beam
Anet = Agross- 1* ( 7/8 [in] + 1/8 [in] ) * t
Anet≥ 2.03 [in2]
Beam
Anet [in2]
r [in]
Weight
L 4 x 4 x 5/16
2.09
1.24 #
out list of potential beams. For the 3 beams which pass the strength requirement ----
L3.5x3.5x3/8
2.11
1.07 #
L 3 x 3 x 7/16
1.99
0.905 #
L 4 x 3 x 3/8
2.11
0.879 #

62. Find: Lightest Angled Beam
Anet = Agross- 1* ( 7/8 [in] + 1/8 [in] ) * t
strength
slenderness
requirement
requirement
Anet [in2]
r [in]
Beam
Weight
L 4 x 4 x 5/16
2.09
1.24 #
and slenderness requirement, we’ll look up their linear weights, to find, ----
L3.5x3.5x3/8
2.11
1.07 #
L 3 x 3 x 7/16
1.99
0.905 #
L 4 x 3 x 3/8
2.11
0.879 #

63. Find: Lightest Angled Beam
Anet = Agross- 1* ( 7/8 [in] + 1/8 [in] ) * t
strength
slenderness
requirement
requirement
[lb/ft]
r [in]
Beam
Anet [in2]
Weight
L 4 x 4 x 5/16
2.09
1.24
8.20
beam L 4 by 4 by 5/16 is the lightest beam. [pause]
L3.5x3.5x3/8
2.11
1.07
8.50
L 3 x 3 x 7/16
1.99
0.905 #
L 4 x 3 x 3/8
2.11
0.879
8.50

64. Find: Lightest Angled Beam
A) L 4 x 4 x 5/16
B) L 3.5x3.5x3/8
C) L 3 x 3 x 7/16
D) L 4 x 3 x 3/8
slenderness
requirement
[lb/ft]
r [in]
Beam
Anet [in2]
Weight
L 4 x 4 x 5/16
2.09
1.24
8.20
When reviewing the possible solutions, ---
L3.5x3.5x3/8
2.11
1.07
8.50
L 3 x 3 x 7/16
1.99
0.905 #
L 4 x 3 x 3/8
2.11
0.879
8.50

65. Find: Lightest Angled Beam
A) L 4 x 4 x 5/16
B) L 3.5x3.5x3/8
C) L 3 x 3 x 7/16
D) L 4 x 3 x 3/8
answerA
slenderness
requirement
[lb/ft]
r [in]
Beam
Anet [in2]
Weight
L 4 x 4 x 5/16
2.09
1.24
8.20
the correct answer is A.
L3.5x3.5x3/8
2.11
1.07
8.50
L 3 x 3 x 7/16
1.99
0.905 #
L 4 x 3 x 3/8
2.11
0.879
8.50