代做ARCH 523bl - Fall 2024 Homework 3代做Python编程

ARCH 523bl - Fall 2024

Homework 3

Problem 1 (30 points)

A single-span steel beam with simple supports is uniformly loaded as shown in Figure 1.

A) Determine if the Wide Flange W18x76 is adequate for bending (due to DL+LL) and deflections (due to LL and DL+LL).

B) Determine if the HSS20x12x1/2 is adequate for bending (due to DL+LL) and deflections (due to LL and DL+LL).

Figure 1- Beam Diagram

Given:

• Distributed dead load, qDL= 400 lbf/ft

• Distributed live load, qLL= 800 lbf/ft

• Yield stress, Fy= 50 ksi

• Allowable bending stress, Fb= 30 ksi

• Modulus of Elasticity (Young’s modulus), E= 29,000 ksi

• Deflection limits:

o DL + LL: L/240

o LL: L/360

• Assume the section profile is compact and the beam is adequately braced.

Figure 2- Wide Flange profile

Parameters:

• L= 50 ft (span)

Hint: Use Tabulated Moments and Deflections for Beams provided in the lecture notes 3.

Figure 3- HSS Profile

Problem 2 (20 points)

A single-span timber joist supports one concentrated load as shown in Figure 4.

A) Determine if the 4x16 wood joist is adequate for bending and deflections (LL and DL+LL).

Figure 4- Beam Diagram

Given:

• Point dead load, PDL= 600 lbs

• Point dead load, PLL= 1100 lbs

• Allowable bending stress, Fb= 1725 psi

• Modulus of Elasticity (Young’s modulus), E= 1,900 ksi

• Deflection limits:

o DL+LL: L/240

o LL: L/360

• Assume the section profile is compact and the joist is adequately braced.

Figure 5- Joist Profile

Parameters:

• L= 20 ft (span)

• a= 3.5 in (design width)

• b= 13.25 in (design depth)

Hint:

• A= a*b

• I= a*b3/12

• S= a*b2/6

Problem 3 (20 points)

A single-span steel beam with simple supports is uniformly loaded as shown in Figure 1. Three options are provided in Table 1 for the beam section profile.

A) Select the lightest profile among the three that satisfies the bending strength and deflection requirements.

Figure 6- Beam Diagram

Given:

• Distributed dead load, qDL= 400 lbf/ft

• Distributed live load, qLL= 800 lbf/ft

• Yield stress, Fy= 50 ksi

• Allowable bending stress, Fb= 30 ksi

• Modulus of Elasticity (Young’s modulus), E= 29,000 ksi

• Deflection limits:

o DL + LL: L/240

o LL: L/360

• Assume the section profile is compact and the beam is adequately braced.

Parameters:

• L= 55 ft (span)

Table 1- Section Profile Options

Hint:

1- Calculate the required plastic modulus (Z) from the design bending moment and allowable stress (Required Z= M/Fb).

2- calculate the required moment of inertia (I) from the distributed loads and the allowable deflection (Figure 7). The required I value should be determined for 1) DL+LL, and 2) LL, as the deflection criterion is different for each.

3- Use the maximum of the two I values to determine the required I.

4- Enter Table 1 with the required Z and I and select the section whose Z and I are greater than the required values.

Figure 7- Required I values

Problem 4 (20 points)

A 50 ft long cable, made of stainless steel, is attached to the ceiling and is used as a hanger. Determine the thermal expansion and contraction for the following conditions:

A) The temperature increases 60 degrees Fahrenheit.

B) The temperature decreases 80 degrees Fahrenheit.

C) If we switch the material to mild steel, without doing any calculation, determine whether it will expand more or less than the steel cable.

Given:

• Coefficient of thermal expansion:

o Stainless steel: a= 8.9*x10-6 1/Deg. F

o Mild Steel: a= 6.0*x10-6 1/Deg. F

Problem 5 (10 points)

For the building materials listed below, determine the allowable stresses:

A) A36 Steel: Fy= 36ksi and Fu= 58 ksi

B) A500 grade C Steel: Fy= 50 ksi and Fu= 65 ksi

Hint: The allowable stress equations are provided in Lecture Notes 3.