Product Technical Guides : CA-EN Modular Systems Product Technical Guide | Page 34

3.0 MODULAR SUPPORT SYSTEM 3.1.2 MT BEAM AND COLUMN LOAD TABLES
Modular Support Systems Technical Guide, Edition 1
Table 12- Load and Deflection Factors for Beams
Span and Loading Condition
Load Factor
Deflection Factor
Simple Beam- Uniform Load 1.00 1.00
Simple Beam- Concentrated Load at Mid-span 0.50 0.80
Simple Beam- Two Equal Concentrated Loads at 1 / 4 Points
1.00 1.10
Beam Fixed at Both Ends- Uniform Load 1.50 0.30
Beam Fixed at Both Ends- Concentrated Load at Mid-span 1.00 0.40
Cantilever Beam- Uniform Load 0.25 2.40
Cantilever Beam- Concentrated Load at End 0.12 3.20
Continuous Beam- Two Equal Spans, Uniform Load on One Span
1.3 0.92
Continuous Beam- Two Equal Spans, Uniform Load on Both Spans
1.00 0.42
Continuous Beam- Two Equal Spans, Concentrated Load at Center of One Span
0.62 0.71
Continuous Beam- Two Equal Spans, Concentrated Load at Center of Both Spans
0.67 0.48
The allowable beam load tables in this technical guide are for single-span supported beams with uniform loading. Common arrangements of other load and support conditions are shown in the above table. Loads and deflections for these conditions can be determined by multiplying the load value from the tables by the given load and deflection factors. Additional reduction factors for unbraced beam lengths may apply per the tables on page 49.
EXAMPLE 1) Determine the maximum allowable load and deflection for an MT-50 fully braced cantilever beam with a concentrated load at the unsupported end. SOLUTION: 1. Allowable load and deflection for an MT-50 with a 48 " span from Table 19 on Page 35 is 1,000 lbs and 0.24 ".
2. Multiply by factors from above table: Load =( 1,000 lbs-1.97 x 4) x 0.12 = 119 lbs Deflection = 0.24 " x 3.20 = 0.77 "
EXAMPLE 2) Determine the maximum allowable load and deflection for a two-span MT-70 fully braced continuous beam that is uniformly loaded on one span. SOLUTION: 1. Allowable load and deflection for an MT-70 with a 48 " span from Table 25 on Page 37 is 2,520 lbs and 0.28 ".
2. Multiply by factors from above table: Load =( 2,520 lbs-2.64 x 4) x 1.3 = 3,260 lbs Deflection = 0.28 " x 0.92 = 0.26 "
2023
32