Structural Impact Loading rd Date: September 23 , 2009 2009 By William Greco, Warrington, Pa.
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Pg 1 of 11
Executive Summary The following essay will show a simple method of conducting dynamic loading calculations.
Main The term impact refers to a dynamic effect of a load which is applied suddenly. The applied load can be considered to be of an instantaneous nature. Most texts on the subject refer refer to harmonic harmonic effects and and resilience resilience of the structure. The capacity capacity of the structure to store energy and vibrate during during sudden shock is usually usually taken into considerable detail. Springs are generally introduced into the explanation and positive and negative oscillatory motion of the disturbed structural member is described and reality takes a back seat to complex theory. In the real world, loads don’t fall on springs springs..
The conditions conditions to be considered considered are moment moment and shear condition condition that accompani accompanies es a dynamic dynamic force and the difference difference between between a static load and a dynamic dynamic instantaneous instantaneous load. load. Consid Considerin ering g springs springs and oscilla oscillatory tory motion motion is not necessa necessarily rily requi required red when calculating calculating a structural structural membe memberr failure under impact. impact. Example-1 Lets start out with a simple case….a static point load in the center of a beam. For simplicity simplicity the beams own uniform weight will not be considered.
Figure 1 above indicates a beam with a concentrated 2,000 pound load located at it’s center.
Structural Impact Loading rd Date: September 23 , 2009 2009 By Willi William am GrecoGreco- Warringt Warrington, on, Pa. Pa.
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Figure -2 indicates the equations required to calculate a static load on a beam with a concentrated mass at it’s center. For the remainder of this report, a static load will be referred to with with the subscript S. Dynamic loads will will be indicated with a D subscript. Example-1’s (figure-1) calculation would proceed in the following manner.
Moment MS
Shear
PL 4
48,000 12,000 ft / lbs 12 144,000 in / lbs
2,000 24 4
4
R1 and R2
2,000
1, 000 lbs 2 assuming that we are dealing with a W8x 24 beam is made of A36 steel the shear resistance is 14,500 pounds per square inch and a W10 x 33 wide flange beam has an area of
8.23 8.23 squa square re inc inches hes.. The shea shearr ratin rating g on this this beam beam is is 14,50 14,500 0 8.23 8.23 or in our example
119,335 1,000
a shear factor of safety of 119 : 1.
119, 335 pound poundss 119,
Structural Impact Loading rd Date: September 23 , 2009 2009 By Willi William am GrecoGreco- Warringt Warrington, on, Pa. Pa.
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Pg 3 of 11
ExampleExample-1 1 (continued) (continued) The moment moment was computed computed to be 144,000 in / lbs. A
W8 x 24 wide flange is a compact section and bending resistance is taken
at 24,000 in / lbs,
144,000 24,000
a section modulus requirement of at least 6.
A W8 x 24 hac a section modulus modulus in the x direction direction of 20.8 The bending safety factor is
20.8 6
3.4 and is satisfactory for a static load.
Assume Assume that a point point load load of 1,950 pounds pounds falls directly in in the center center of the beam beam from a height height of 16 feet. feet. while it’s it’s supporting the existing 2,000 pound static load. 1,950 pounds x 16 feet = 31,200 foot pounds produced by the falling load.
The static static load load PS was known known and the the momen momentt was calcu calculat lated ed fro from M S
PL
howev oweveer the moment of a falling load is known the therefor fore 4 it becomes necessary to rearrange the equation to solve for P, P will now becom becomee PD as it is is now now a dyn dynam amic ic loa load. d. MS
PL
4 multiply both sides by 4
PL 4
MS 4
4
4M
PL
Divide both sides by L P L L P D
4M L
4M L
Structural Impact Loading rd Date: September 23 , 2009 2009 By Willi William am GrecoGreco- Warringt Warrington, on, Pa. Pa.
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Pg 4 of 11
Example-1 (continued)
31, 2 00 00 4
5, 200 pounds 24 The existing static load (2,000 pounds) is now added to the PD
Dynamic load of 5,200 pounds and an additional 1,950 pounds of static load which is the weight of the falling load. Th e instantaneous total load on the beam is nowPS (original load) PD (dynamic load)
PSf (Falling Stat ic Load)
000 + 5, 20 200 + 1, 95 950 9,150 po pounds 2, 00 The total load of 9,150 pounds is now substituted back into the original equation. 9,150 24 P L MT T =54,900 foot pounds x 12 = 658,800 inch pounds 4 4 where :
the total moment PT total load
MT
658,800
27.45 20.8 24,000 A36 structural steel has an elastic limit of 36,000 pounds
required section modulus
the beams ultimate strength is 64,000 psi. and it's yield point is 42,000 psi. the instantaneous load exceeds the elastic limit by: 36,000 psi 20.8 (section modulus of beam) 27.45 (required section modulus at instant of load drop)
47,509 psi
since the beam has been subjected to a load higher than it's yield point but below it's ultimate strength
The beam will take on a permanent deformation but will not fail.
Structural Impact Loading rd Date: September 23 , 2009 2009 By Willi William am GrecoGreco- Warringt Warrington, on, Pa. Pa.
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Pg 5 of 11
Example-1 (continued) The W8 x 24 beam in example-1 will take on the following permanent deformation:
PT
L 12 3
48 E I
3
9,150 24 12
48 29, 000, 000 82.15
1.91 inches
where:
the to total in instantaneous lo load L the length of the beam in feet PT
6 the modulus modulus of elasticity elasticity (for steel = 29x10 I the moment of inertia of the beam the deflection in inches
E
Example-2 In example -1 the unlikely case of a static point load in the center of a beam with a corresponding falling load at the center of the beam was calculated. Most situations would require the calculation of a uniformly loaded beam with a uniform load falling at a location other than it’s center. The condition is shown in figure-3.
Structural Impact Loading Date: September 23 , 2009 2009 By Willi William am GrecoGreco- Warringt Warrington, on, Pa. Pa.
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Example-2 Example-2 (continued) (continued) Figure-4 indicates the diagrams and equations of an off center uniformly distributed load.
Note: Figure-4 is used for Static loading. Assuming the 2,650 pound load to be falling from 14 feet above the beam. Also require the following dimensions, the distance from the left side R1 to the edge of of the falling load A = 6 feet, feet, the falling falling load C = 7 feet and and B = 11 feet. If we use use a W10 x 21 beam with a uniform load of 40 pounds per linear foot. Step-1 - The total uniform static load is (40 x 24)+(21 x 24) = 1,464 pounds.
Structural Impact Loading Date: September 23 , 2009 2009 By Will Willia iam m GrecoGreco- Warrin Warringto gton, n, Pa. Pa.
[email protected] Example-2 Example-2 (continued) (continued) rd
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Step-2 involves calculating the total static uniform moment and reactions R1 and R2: 1,46 ,464 24 WL 4, 392 foot pounds x 12 52,704 in Static M inch pounds 8 8 Total weight 1, 46 464 732 pounds Static R1 R2 2 2 Step-3 The rearrangement of equations 1 and 2 to calculate the equivalent weight knowing the dynamic dynamic moment moment (2,650 lbs lbs x 16 feet) (see figure-4 figure-4 for equations equations 1 and 2): 2): Rear Rearra rang ngem emen entt of of equ equat atio ionn-1 1A B
M
R1 2A W + R1 C
2W Eliminate the denominators by multiplying both the left and the right side by the least common denominator LCD is equal to : 2 W R1 2A W + R1 C 2WM 2W 2W Expand the terms 2 W M R1 2 A W + R1 R1 C
Combine and multiply 2
2 W M 2 R1 A W + R1 C group all the variable terms on one side, and all the constant terms on the other side of the equation. 2 R1 A W –term W –term will be moved to the left side and change sign when it moves from one side of the equation to the other. 2WM
2 R1 A W 2 R1 A W + R12 C 2 R1 A W
organize this expression into groups of like terms The following are like terms : 2 R1 A W and -2 R1 A W 2WM
2 R1 A W 2 R1 A W 2 R1 A W + R12 C
2 W M 2 R1 A W 2 R1 A W 2 R1 A W + R12 C 2 W M 2 R1 A W Isolate the variable on the left side. The first step in this procedure consists of factoring the left side of the equation. 2 W M 2 R1 W M 2 W R1 C W W Continued on next page (page-8)
R12 C
Structural Impact Loading Date: September 23 , 2009 2009 By Willi William am GrecoGreco- Warringt Warrington, on, Pa. Pa.
[email protected] Continued from page 7
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rd
2 1 M 2 R1 1 M 2 W M 2 R1 W M 2 2 R 1 C W R1 C 1 W W 1
W
Fractions which have denominators equal to 1 are equal to their numerator.
W 2 M 2 R1 A
R12 C
Divide both sides by 2 M – M – 2 R1 R1 A 2 2 R1 C R1 C W W 2 M 2 R1 A 2 M 2 R1 A 2 M 2 R1 A
2 M 2 R1 A
W
R12 C
2 M 2 R1 A
(equation-3) when A
B
Rearrangement of equation-2 dimensions A=B
B C M 2 4
W
Add fractionsfractions- the following rule rule is applied : LCD LCD A+ C A C B D
B D LCD This involves two terms. One non-fractional term is treated as a fraction with denominator equal to 1.The LCD (least common denominator) is 2 equal to :2 W
2 W 2 B+1C
B M 2 2 22 C
22
M
Eliminate expression parentheses. W 22 B+C 2
2 2
M
Evaluate a power by multiplying the base by itself as many times as the exponent indicates. W 4 B+C M 2 22 Continued on next page (page-9)
Structural Impact Loading rd Date: September 23 , 2009 2009 By Willi William am GrecoGreco- Warringt Warrington, on, Pa. Pa.
[email protected] Continued from page 8
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Perform a multiplication. The following rule is applied : a c ac x
b
d
bd
the factors in the new numerator are : W the factor factorss in in the new new deno denomin minato atorr are are : W 4 B+C
W 4 B+C
M 2 2 22
22
4B+C 2 2x2 W 4 B+C
2
M
22
M
Combine like factors in this term by adding up all the exponents and copying the base. No exponent implies the value of 1. W 4 B+C W 4 B+C M M 2 22 2 12
Numerical terms in this expression have been added. W 4 B+C W 4 B+C M M evaluate power 3 8 2 W 4 B+C 8M multiply both si sides by 8 8 8
Divide both sides by 4 B+C
W
8M 4 B+C
(equation-4a)
W 4 B+C
4 B+C W
8M 4 B+C
8M 4 A+C
W
8M 4 B+C
(equation-4b)
Step-4 Next, calculate the equivalent weight of the falling load knowing the moment: Falling Falling Moment = (2,650 lbs lbs x 16 feet) = 42,400 42,400 foot foot pounds pounds
Structural Impact Loading rd Date: September 23 , 2009 2009 By Willi William am GrecoGreco- Warringt Warrington, on, Pa. Pa.
[email protected]
Pg 10 of 11
Example-2 (continued) Step-5 Insert the values into equation-4a and and equation-4b, subtract the smaller value (subtrahend) from the larger larger value (minuend), (minuend), divide divide the difference difference by 2, then then subtract subtract the resulting resulting quotient quotient from the larger resultant of either either equation-4a or equation-4b equation-4b
. W
W
10, 941
8M 4 B+C
4 11+ 7
8 42, 400
W
= 6,650
8M 4 A+C
10,941-6,650
8 42, 400
WD
4 6 +7
= 10,941
8,795 equivalent dynamic pounds 2 Note: that equation-3 can only be used if the actual reactions are known ahead of time. Step-6 Calculate the dynamic reactions: 1 1 A 6 7 9.5 feet C 2 2
1 C 11 1 7 14.5 feet 2 2
B
9.5
.655 .655 or or 65.5 65.5% % of of the the total total load load R1 = .655 .655 8,795 14.5 R2 8,79 ,795-5,76 ,760 3,035 ,035 pounds dynamic reac tion
pounds dyna dynami micc react reactio ion n 5, 760 pounds
Step-7 Calculate Calculate the total moment on the beam by using equation-1. equation-1. 5,760 2 6 8,795 + 5,760 7 R1 2A W + R1 C MD 2W 2 8, 795
47, 763 foot pounds
Step-8 Add the static and dynamic moment: See page-7 for the static moment. 47,763 ft/lbs dynamic + 4,392 ft/lbs static = 52,155 foot pounds x 12 =625,860 in/ lbs 625,860 26.1 required section modulus 24,000 A W10 x 21 has a section modulus of 21.5.
Structural Impact Loading rd Date: September 23 , 2009 2009 By Willi William am GrecoGreco- Warringt Warrington, on, Pa. Pa.
[email protected] 625,860 26.1 21.05 required section modulus 24,000
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A36 structural steel has an elastic limit of 36,000 pounds the beams ultimate strength is 64,000 psi. and it's yield point is 42,000 psi. the instantaneo us load exceeds the elastic limit by: 36,000 psi 21.5 (section modulus of beam) 26.1 (required section modulus at instant of load drop)
43,702 psi
since the beam has been subjected to a load higher than i t's yield point but below it's ultimate strength
The beam will take on a permanent deformation but will not fail. Conclusion: Accidental Accidental dynamic impact impact loading loading is a subject subject which which is not well attended to in structural texts. Most structural engineers are well versed in static loading as well as seismic engineering, but most dynamic situations are covered by simple coefficients and factors which are spelled out in code books. The educational systems of Colleges and Universities Universities can do some some empirically empirically based practical work that deals deals with this subject. Substituting higher order mathematics for experiments, and wandering off through abstract equations, eventually builds a structure which has no relation to reality. As I indicated at the beginning of this essay, loads do not fall on nor are they held up by springs.
William Greco 2404 Greensward N. Warrington, Pa. 18976 References: th
United States Steel Pocket Companion 24 Edition –Copyright Edition –Copyright 1936 Page- 162 structur structural al case-8 case-8 th
Manual Of Steel Construction-8 Edition American Institute Institute of Steel Construction- Copyright 1980
