**Problem Statement:**An unsymmetrical flexural member consists of a \( \frac{3}{8} \times 14 \) in. top flange, a \( \frac{3}{8} \times 10 \) in. bottom flange, and a \( \frac{3}{8} \times 20 \) in. web.**Questions:**(a) Determine the distance \( \overline{y} \) from the top of the shape to the horizontal plastic neutral axis (P.N.A.).(b) If A572 Grade 50 steel is used, what is the plastic moment \( M_p \) for the horizontal P.N.A.?(c) Compute \( Z_y \), the plastic section modulus with respect to the minor principal axis.(d) Calculate the ratio of \( Z_y / S_y \). (Note: \( S_y \) is the (elastic) section modulus with respect to the minor principal axis.)**Steps to Solve:**1. **Determine Distance to P.N.A.** - Calculate the area of each section: top flange, bottom flange, and web. - Use the areas and distances to find the centroid of the shape. - The distance \( \overline{y} \) is then measured from the top to the centroid.2. **Calculate Plastic Moment \( M_p \)** - Use the plastic section modulus and material properties. - For A572 Grade 50 steel, the yield strength \( f_y \) is 50 ksi. - The plastic moment \( M_p \) is given by \( M_p = f_y \cdot Z \).3. **Compute Plastic Section Modulus \( Z_y \)** - Integrate the moments of areas about the horizontal plastic neutral axis. - Sum the individual plastic section moduli for all components.4. **Calculate the Ratio \( Z_y / S_y \)** - Obtain the elastic section modulus \( S_y \) from standard shape tables or calculations. - Compute the ratio of the plastic section modulus to the elastic section modulus.This information can be used for detailed structural analysis and design processes, ideal for civil engineering and materials science students.

given:A572 Grade 50 steelLet us assume, Yield strength (Fy) = 345 MPa (50 ksi)Moment of inertia (I)…

An unsymmetrical flexural member consists of a %x14 in. top flange, a ¾×10 in. bottom flange, and a ¾x20 in. web. (a) Determine the distance y from the top of the shape to the horizontal plastic neutral axis (P.N.A.). (b) If A572 Grade 50 steel is used, what is the plastic moment M, for the horizontal P.N.A.? (c) Compute Zy, the plastic section modulus with respect to the minor principal axis. (d) Calculate the ratio of Z/S,. (Note: S, is the (elastic) section modulus with respect to the minor principal axis.

An unsymmetrical flexural member consists of a %x14 in. top flange, a ¾×10 in. bottom flange, and a ¾x20 in. web. (a) Determine the distance y from the top of the shape to the horizontal plastic neutral axis (P.N.A.). (b) If A572 Grade 50 steel is used, what is the plastic moment M, for the horizontal P.N.A.? (c) Compute Zy, the plastic section modulus with respect to the minor principal axis. (d) Calculate the ratio of Z/S,. (Note: S, is the (elastic) section modulus with respect to the minor principal axis.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

An unsymmetrical flexural built-up shape member consists of a 2 inch x 20 inch top flange and a 2 inch x 15 inch bottom flange and a ½ inch x 40 inch web. a. Determine the distance "y" from the top of the shape to the horizontal plastic neutral axis. b. If the steel is A572, Grade 50, what is the Plastic Moment, Mp, in ft-kips? Answer: ỹ = 2 x 20 Plate Mp Р.N.A. ½ x 40 Plate 2 x 15 Plate
Design the reinforcements of the given T beam below. bf=800mm bw=450mm tf=120mm d=600mm d'=80mm fc'=35MPa fy=350MPa USE NSCP 2015 A.Mu = 1300kN-m, As = B.Mu = 1600kN-m, As = mm2 _mm2, As' = mm2
Problem For the frame shown, design the maximum permissible value of P with a factor of safety of 2.5. Design the maximum permissible value of P based on the following conditions: ✓ Ultimate tensile strength of each member CD is 300 ksi Ultimate shear strength of the bolts in A, C, E, and F is 200 ksi ✓ Ultimate bearing strength of member BDE is 150 ksi ✓ Ultimate bearing strength of the support plates is 250 ksi Members CD and BDE have cross-sectional dimensions of 5 in x 3 in (the larger dimension is the width). The diameters of the bolts are as follows: A and B - 1.5 in: C and D-2 in E and F - 1 in. 1.5 in. thick Support Plates -2 ft + D 3 ft B 2 in thick Support Mate Solution Step 1. Identify the conditions for design. Stress normal stress in each member CD, OCD shear stress in bolt A, TA shear stress in bolt C, T shear stress in bolt E, T shear stress in bolt F, T bearing stress on member BDE due to: bolt B, BDE-B bolt D, OBDE-D bolt E, OBDE-E bearing stress on support plates at…
.
List the elements of {xIx is a rolled steel section}.
A built-up section was made using PL414x12mm thk plates as shown in the figure below. It is pinned at both ends with additional support against weak axis at middle point. Assume A50 steel. PL414x12 DO Section W16x67 L x-axis a) Calculate moment of inertia at both axes in mm*. b) Determine the design compressive strength in kN if L-3m. c) Find the design compressive strength in kN if L=18m. Elevation y-axis
Given the truss shown, solve the following. Use A36 steel and assume there are two lines of three 19 mm bolts in each flange. Hole diameter is assumed to be three mm more than the bolt diameter. Neglect the weight of the section. a.What is the capacity of member CG? b. Design member CG using a W section. c. What is the tensile capacity of the section for CG based on gross area? The following W Section were considered for the design: W10x33 W14x38 W8x40 W16x36 W10x39 W8x13 189 KN 1.50 HL B 145 KN 1.50 E + 137 KN 121 KN -3.00 + 12.00 3.00 154 KN + W4x13 W6x9 W10x12 200 KN -1.50 1.50 1 030 T 1.20 +2.40 0.60 L i 0.30
Design the reinforcements of the given T beam below. bf=900mm bw=420mm tf-120mm d=550mm d'=80mm fc'=34MPa fy=415MPa USE NSCP 2015 A.Mu = 1300kN-m, As = mm2 B.Mu = 1800kN-m, As = mm2, As' = mm2
A PL40 mm X 250 mm (smaller member) is connected to a gusset plate (bigger member) as shown. The diameter of the holes are 25 mm. The pitch and gage of the holes are 50 mm and 75 mm, respectively. The yield strength of the steel is 260 MPa while the ultimate tensile strength of the steel is 400 MPa. Determine the design (LRFD) tensile strength of the tension member in kN. Neglect block shear. H FO E. • D A В
A wide flange section has the following properties. A- 6645 mm? bị = 152.40 mm ty - 10.80 mm d- 449.58 mm tw - 7.62 mm Fy - 250 MPa Compute the plastic section modulus about the major principal axis. (x 10° mm)
Select all zero-force members in the truss shown below. Check the box for zero- force members 3 m 3 m 12 m, 8 @ 1.5 m DE O LK ЕР O HF O BC BM EF OM CD BN LO O DK FI O co
Two plates each with thickness t=16mm are bolted together with g-22 mm dia•bolts forming a lap connection.bolts spacing are as follows S1=40mm, S2=80mm,S3=100. Bolt hole dia=25 mm Fu=483Mpa Fy=345Mpa Solve the allowable strength and the ultimate strength in: 1. Yielding 2.rupture 3.shear 4.block shear