Exam (elaborations)
INTERMEDIATE RIGGING QUESTIONS AND ANSWERS
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INTERMEDIATE RIGGING QUESTIONS AND ANSWERS
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Name: Score:63 Multiple choice questions
Term 1 of 63
What temperature can alloy chain steel be used
120°
Up to 495°f
Up to 490°f
Up to 500°F
Term 2 of 63
How to find start of loads from known runs
F=[CF×run horizontal length feet(R)÷ length of ramp in feet (L)×W]+[W×height in
feet(H)÷(L)]
F=[CF×R÷L×W]+[W×H÷L]
1) W1+W2= total weight (TW)
2) W2÷TW=P
3) P×span (S)= center of gravity (CG) in ft from A
4) W1+W2=TW
5) W1÷TW=P
6) P×S= CG in ft from B
2 per strand per 6 rope diameters, 4 per strand per 30 rope diameters
1) length run side 1 (R1)+ R2= total span (TS)
2) R2÷TS= Percentage (P)
3) P×W= share of load on side A
4) R1+R2=TS
5) R1÷TS=P
6)P×W= share on side B
,Term 3 of 63
How far past tracks must mats be
20 MPH
Parallel
1% or .57°
2'
Term 4 of 63
What is the formula for leverage
Force required to move load (f)=coefficient of friction (cf)×w
f=cf×w
Weight × distance = weight × distance
Twist 0°
opening 5%
wear 10%
Independent wire rope core
Term 5 of 63
What is the distance from compacted soil
6×depth
101×depth
11×depth
1×depth
,Term 6 of 63
Sling tension
1.5×diameter of bolt
Working load limit
Only with a freely suspended load
Weight÷ #of slings×load angle factor
Term 7 of 63
What is the minimum depth an eye bolt must be
1.5×diameter of bolt
99.5×diameter of bolt
1.6×diameter of bolt
2.5×diameter of bolt
Term 8 of 63
How do you find the load factor and weight distribution
75%
F=[CF×run horizontal length feet(R)÷ length of ramp in feet (L)×W]+[W×height in
feet(H)÷(L)]
F=[CF×R÷L×W]+[W×H÷L]
Sling tension=
Sling length (L)÷ sling height (H)× share of load weight
L÷H× share of load weight
Force required to move load (F)=coefficient of friction (CF)×W
F=CF×W
, Term 9 of 63
What do you drop the cranes capacity to for tandem lift
75%
25%
85%
95%
Term 10 of 63
How to figure center of gravity from known weights
Sling tension=
Sling length (L)÷ sling height (H)× share of load weight
L÷H× share of load weight
F=[CF×run horizontal length feet(R)÷ length of ramp in feet (L)×W]+[W×height in
feet(H)÷(L)]
F=[CF×R÷L×W]+[W×H÷L]
Force required to move load (F)=coefficient of friction (CF)×W
F=CF×W
1) W1+W2= total weight (TW)
2) W2÷TW=P
3) P×span (S)= center of gravity (CG) in ft from A
4) W1+W2=TW
5) W1÷TW=P
6) P×S= CG in ft from B
Term 11 of 63
Gross capacity is the same as
Net capacity
Allowable line pull
Independent wire rope core
Gross minus deductions
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