Appendix B
BUNKER AND SHELTER ROOF DESIGN
This appendix is used to design a standard stringer roof that will defeat a contact burst projectile when the materials used are not listed in the table, CentertoCenter Spacing for Wood Supporting Soil Cover to Defeat Contact Bursts. For example, if a protective position uses steel and not wood stringers, then the procedure in this appendix is used for the roof design. The table, CentertoCenter Spacing for Wood Supporting Soil Cover to Defeat Contact Bursts, was made using the design steps in this procedure. The calculations are lengthy but basically simple. The two example problems in this appendix were worked with a handheld calculator and the complete digital display is listed. This listing enables a complete stepbystep following without the slight numerical variation caused by rounding. In reality, rounding each result to three significant digits will not significantly alter the outcome. The roof is designed as follows.
First, hand compute the largest halfburied trinitrotoluene (TNT) charge that the earthcovered roof can safely withstand. Then, use the charge equivalency table to find the approximate size of the superquick or contact burst round that this halfburied TNT charge equals. The roof design discussed here is for a simple stringer roof of singleply or laminated sheathing covered with earth (figure B1). After determining the need for a bunker or shelter roof, the following questions are addressed:
 What type of soil will be used for cover (soil parameters)?
 How deep will the soil cover be?
 What will the size and orientation of the stringers be and
what kind of stringers will be used (stringer characteristics)?
 What will the stringer span and spacing be?
Two soil parameters are needed in the design procedureunit weight and transmission coefficient. Soil unit weight must be determined at the time and place of design. Both the soil (sand, silt, for example) and its water content affect unit weight. Soil unit weight is usually 80 to 140 pounds per cubic foot. The transmission coefficient can be taken from table B1
For wood stringers, the data needed in the design procedure are given in table B2 and B3. For steel stringers, the moment of inertia (I) and section modulus (S) values needed in the procedure are given in table B4. For the modulus of elasticity (E) and maximum dynamic flexural stress (FS) values, use E = 29 and FS = 50,000. (Additional structural design data is in FM 535.)
Line
1  Enter the unit weight of the soil (lb/cf) as determined on site  ___________ 
2  Enter the proposed depth of soil cover (ft)  ___________ 
3  Enter the S value (in 3 ):  ___________ 
if wood, from Table B2  
if steel, from Table B4  
4  Enter the stringer spacing (in)  ___________ 
5  Enter the FS value (psi):  ___________ 
if wood, from Table B3  
if steel, enter 50,000  
6  Enter the stringer span length (ft)  ___________ 
7  Multiply line 1 by line 4, enter result  ___________ 
8  Multiply line 7 by line 2, enter result  ___________ 
9A  Multiply line 8 by line 6, enter result  ___________ 
9B  Multiply line 9A by line 6, enter result  ___________ 
9C  Divide line 9B by 8, enter result  ___________ 
9D  Divide line 9C by line 3, enter result  ___________ 
9E  Divide line 9D by line 5, enter result  ___________ 
9F  If the line 9E result is greater than O but less than 1.0 go to line 10.  ___________ 
If line 9E is greater than 1.0, the roof system is overloaded. Then do at least one of the following and recompute from line 1:  
a. Decrease stringer spacing.  
b. Decrease span length.  
c. Use a material with a higher "S" or "FS" value.  
d. Decrease soil cover. 
Line
10  Enter side A of Figure B2 with the line 9E value, find the side B  ___________ 
value, and enter result:  
if wood, use µ = 1 curve  
if steel, use µ = 10 curve  
Line
11  Enter the E value (10^{6 }psi):  ____________ 
if wood, from Table B3  
if steel, enter 29  
12A  Enter the I value (in^{4 }):  ____________ 
if wood, from Table B2  
if steel, from Table B4  
12B  Multiply line 9A by 0.08333, enter result  ____________ 
12C  Multiply line 12B by 0.64, enter result 1  ____________ 
12D  Divide line 12C by line 9E, enter result  ____________ 
13  Multiply line 9A by 0.0001078, enter result  ____________ 
14A  Multiply line 12A by line 11, enter result  ____________ 
14B  Multiply line 6 by line 6, enter result  ____________ 
14C  Multiply line 14B by line 6, enter result  ____________ 
14D  Divide line 14A by line 14C, enter result  ____________ 
14E  Multiply line 14D by 28,472.22, enter result  ____________ 
15  Divide line 14E by line 13, enter result  ____________ 
16  Take the square root of line 15, enter result  ____________ 
17  Divide line 12D by line 16, enter result  ____________ 
18  Multiply line 10 by line 17, enter result  ____________ 
19  Divide line 2 by line 6, enter result  ____________ 
20  Multiply line 19 by line 19, enter result  ____________ 
21A  Take the square root of line 19, enter result  ____________ 
21B  Multiply line 21A by line 20, enter result  ____________ 
22  Divide 0.6666667 by line 21B, enter result  ____________ 
23A  Multiply line 20 by 4, enter result  ____________ 
23B  Add 1 to line 23A, enter result  ____________ 
24  Divide 4 by line 23B, enter result  ____________ 
25A  Take the square root of line 24, enter result  ____________ 
25B  Take the square root of line 25A, enter result  ____________ 
25C  Multiply line 25B by line 24, enter result  ____________ 
26  Add line 25C to line 22, enter result  ____________ 
27  Choose a C value from Table B1, enter result  ____________ 
28A  Multiply 61.32 by line 18, enter result  ____________ 
28B  Take the square root of line 14C, enter result  ____________ 
28C  Multiply line 28A by line 28B, enter result  ____________ 
28D  Multiply line 27 by line 4, enter result  ____________ 
28E  Multiply line 28D by line 26, enter result  ____________ 
28F  Divide line 28C by line 28E, enter result  ____________ 
29  Raise line 28F to the 0.8571 power (or use the graph in Figure B3), enter result  ____________ 
The value on line 29 is the largest halfburied TNT Charge (lb) that the roof can withstand. Enter Table B5 with this value to find the round having an equivalent charge weight equal to or less than the value on line 26.
The 276th Infantry is about to relieve another battalion from defensive positions as shown in figure B4. The 1st Platoon of the A/52d Engineers is supporting the 276th. As its platoon leader, you have been asked to find how much protection such positions give against the contact burst of an HE round.
You first estimate that the 16inchdeep soil cover (sand) weighs 100 lb/cf. You then note that the roof is made of 4 by 4 stringers, laid sidebyside over a span of 88.75 inches.
Thus, the largest TNT charge that the roof can withstand is 1.56 pounds. Entering Table B5 with this value, you find that the roof will withstand a contact burst explosion of up to an 82mm frag round (only 1.0pound charge size) excluding the 76mm HE round (1.8pound charge site).
The 276th Infantry will occupy the positions described in the first example for an extended period of time. Thus, the battalion commander has ordered the 1st Platoon of the A/52d Engineers to construct a tactical operations center. This structure must have at least 10 by 12 feet of floor space and be capable of defeating a contact burst of a Soviet 152mm round. The S2 of the A/52d Engineers reports that 13 undamaged 8inch by 6 ½inch wide flange beams have been found. They are long enough to span 10 feet and can be salvaged from the remains of a nearby demolished railroad bridge.
As platoon leader, you are to design a roof for the tactical operations center using these beams as stringers. You plan to place five of the stringers on 36inch centers and cover them with a 4 by 4 wood deck. You use the same bagged sand as described in the first example. You begin your design by assuming that the soil cover will be 3 feet deep.
Thus, the largest TNT charge that the stringers can withstand is 29.6 lb. You next use the procedure again in a manner similar to that in example 1 to evaluate the 4x4 wood deck. You find a line 29 value of 29.64. Enter Table B5 with the largest of these values (29.6), you find that the roof will withstand a contact burst explosion of up to a 160mm HE round (only 16.3pound charge size). Thus, the roof you have designed will be capable of defeating a contact burst of a Soviet 152mm round.
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