Math Problem Statement

Refer to Figure 1. The A36 wide flange beams given in the figure below have 40 ft span, have 20 ft on center (oc) distance (the panel width of the panels is 10 ft) and are supported by trusses on either end. The beams are braced at every 10 ft (Lb=10 ft). Dead load is 40 psf and live load is 20 psf. Follow the steps to determine the required specifications for the inner beams.

Solution

To solve Question 1, we need to calculate the load, moment, and various parameters for the beams, specifically the inner beams in the truss structure. Let's go through the steps outlined in the question.

Given Information:

  • Span of truss: 40 ft
  • On-center spacing (oc distance): 20 ft
  • Beam bracing (Lb): 10 ft
  • Dead load (D): 40 psf
  • Live load (L): 20 psf
  • Material: A36 steel (yield strength, Fy=36 ksiF_y = 36 \text{ ksi})

Steps to solve the question:

  1. Apply Load Factors: wu=1.2D+1.6Lw_u = 1.2D + 1.6L Substitute D=40 psfD = 40 \text{ psf} and L=20 psfL = 20 \text{ psf} to find wuw_u: wu=1.2×40+1.6×20=48+32=80 psfw_u = 1.2 \times 40 + 1.6 \times 20 = 48 + 32 = 80 \text{ psf}

  2. Calculate the Load that the Inner Beams Support: Wu=(wu)×oc×LW_u = (w_u) \times \text{oc} \times L Given that wu=80 psfw_u = 80 \text{ psf}, oc=20 ft\text{oc} = 20 \text{ ft}, and L=40 ftL = 40 \text{ ft}: Wu=80×20×40=64000 lbsW_u = 80 \times 20 \times 40 = 64000 \text{ lbs}

  3. Calculate the Ultimate Moment that the Inner Beams Support: Mu=Wu×L8M_u = \frac{W_u \times L}{8} Substitute Wu=64000 lbsW_u = 64000 \text{ lbs} and L=40 ftL = 40 \text{ ft}: Mu=64000×408=320000 ft-lbsM_u = \frac{64000 \times 40}{8} = 320000 \text{ ft-lbs}

  4. Calculate the Nominal Moment: Mn=Mu0.9M_n = \frac{M_u}{0.9} Substitute Mu=320000 ft-lbsM_u = 320000 \text{ ft-lbs}: Mn=3200000.9355556 ft-lbsM_n = \frac{320000}{0.9} \approx 355556 \text{ ft-lbs}

  5. Calculate ZminZ_{\text{min}}: Zmin=Mn36 ksiZ_{\text{min}} = \frac{M_n}{36 \text{ ksi}} Convert MnM_n from ft-lbs to in-lbs (1 ft = 12 in): Mn=355556×12=4266667 in-lbsM_n = 355556 \times 12 = 4266667 \text{ in-lbs} Then calculate ZminZ_{\text{min}}: Zmin=426666736118519 in3Z_{\text{min}} = \frac{4266667}{36} \approx 118519 \text{ in}^3

  6. Determine the Lightest Weight Shape for the Inner Beams Using the Chart:

    • Use a structural steel shape table to find a wide flange (W) beam with a section modulus ZZ equal to or greater than Zmin118519 in3Z_{\text{min}} \approx 118519 \text{ in}^3.
  7. Calculate the Nominal Moment Capacity of the Beam: Use the formula provided: Mn=Mp((MpMr)×(LbLp)(LrLp))M_n = M_p - \left( (M_p - M_r) \times \frac{(L_b - L_p)}{(L_r - L_p)} \right) where MpM_p and MrM_r are plastic and reduced moments, and LpL_p and LrL_r are lateral-torsional buckling limits for the selected shape.

  8. Make Final Selection: After finding the section that satisfies all criteria, confirm its suitability based on moment capacity and other checks.

Would you like a detailed breakdown of each calculation or further explanation on using steel design tables for step 6?

Additional Questions for Further Understanding:

  1. How does lateral-torsional buckling affect the beam design?
  2. What is the role of load factors in structural design safety?
  3. How can different dead and live load scenarios influence the beam size?
  4. Why is it important to calculate both ultimate and nominal moments?
  5. How would changes in the span or bracing distance affect the calculations?

Tip: Always cross-check your section modulus with steel tables to ensure accuracy in beam selection.

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Math Problem Analysis

Mathematical Concepts

Structural Engineering
Load Calculations
Moment Calculations
Section Modulus

Formulas

wu = 1.2D + 1.6L
Wu = (wu) * oc * L
Mu = Wu * L / 8
Mn = Mu / 0.9
Zmin = Mn / 36 ksi
Mn = Mp - ((Mp - Mr) * ((Lb - Lp) / (Lr - Lp)))

Theorems

Load and Resistance Factor Design (LRFD)
Nominal and Ultimate Moment Capacity

Suitable Grade Level

College-level (Structural Engineering)