A16= 10.2 B8=42.2 C2= 47.2 1- The result of sieve analysis of three aggregate samples (A, B, and C) are given below.a) Design a new aggregate mixture and calculate the mixing ratios of each toconfirm TS limits. (important note: maximum aggregate size of the new mixturemust be dmax=16 mm)Sieve size (mm)A. Passed (%)B. Passed (%)C. Passed(%)TS706-A32 (%)TS706-B32 (%)TS706-C32 (%)TS706-A16 (%)31,516840,50,25 Pan100 (10+0.1n)0.90.20.20.20.20.2100100(42+0.1n)0.50.20.20.20.210010010079.0 (47+0.1n) 26.816.712.51006238231486100806247372818810089776653422815100603621123TS706-B16 (%)100765642328.TS706-C16 (%)1008874624918

Solution Given Data dmax= 16 mm

1- The result of sieve analysis of three aggregate samples (A, B, and C) are given below. a) Design a new aggregate mixture and calculate the mixing ratios of each to confirm TS limits. (important note: maximum aggregate size of the new mixture must be dmus=16 mm) Sieve size (mm) A. Passed (%) 31,5 100 (10+0.1n) 100 4 0.2 2 0,25 0.2 0.2 16 8 1 0,5 0.2 Pan 0.2 0.9 (42+0.1n) 0.2 B. Passed (%) 100 0.5 0.2 0.2 0.2

1- The result of sieve analysis of three aggregate samples (A, B, and C) are given below. a) Design a new aggregate mixture and calculate the mixing ratios of each to confirm TS limits. (important note: maximum aggregate size of the new mixture must be dmus=16 mm) Sieve size (mm) A. Passed (%) 31,5 100 (10+0.1n) 100 4 0.2 2 0,25 0.2 0.2 16 8 1 0,5 0.2 Pan 0.2 0.9 (42+0.1n) 0.2 B. Passed (%) 100 0.5 0.2 0.2 0.2

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

4
Example: Determine the location of the centroid ('c') of the beam's cross section and the moment of inertia about the centroidal x-axis. AY Part Dimensions Area y 4x1 4x1 1x8 2 3 Total Part 1 2 3 Area yA 4 2 8 4 28 8 16 in' Total 16 in 05 4 20 in dy (A) 2-1.25 2.25 4 1-(4)/12 4 1:(4)/12 2-1.25 2.25 8 8(11/12 1.25-0.5 4.50 = 1.25 in 16 x = lei +d³y(A) 5.33+2.25= 7.58 5.33 +225=7,58 0.666+4.50 5.16 20.32 in 1 Typ
1. An angular section 120mm x 120mm x 10mm is riveted by a 22mm diameter rivets arranged as shown. Assume A36 steel with F, = 250MPA, F, = 400MPA. Diameter of hole is 3mm bigger. 60mm 40mm 120mm | 10mm 60mm >T 40mm 10mm 60mm 60mm 60mm 60mm a. Determine the capacity of the section based on yielding of gross area. b. Determine the capacity of the section based on tensile fracture of the net area. c. Determine the capacity of the section based on block shear failure.
Guide: W = 447 Z = 587 Y = 176 X = 388 "W" "Y" "Y"* "X" all dimensions are in mm "Y" " a. Determine the location of the centroid from the bottom of the beam. b. Determine the location of the centroid from the leftmost part of the beam. c. Determine Ix. d. Determine ly.
BERNOULLI EQUATIONS
PROVIDE COMPLETE SOLUTION THANKS
F3=30N F2=80N Express each of the three forces acting on the support in Cartesian vector form and determine the magnitude of the resultant force and its direction, measured clockwise from positive x axis.
150 150 159 Q: Calculate the Nominal Strength for the beam shown. D00 f -40 MPa , f, - 400 MPa 250 %3D 600 4625 65 500
6" K6"– Find the Centd, C= (a ま)
Determine the force acting on the x-component of member AB of the frame shown in Ibs. P1 = 385 Ibs, P2 = 462 Ibs and x = 7 ft. Write numerical value only and in 2 decimal places. D В 6 ft E AO |C 1.P1 P2 X X X
A cross sectional area was generated for a bridge beam support 0.6 y-**- 0.5 0.4 0.3 0.2 0.1 0.1 0.2 0.3 0.4 0.5 0.7 0.8 1.1 1.2 0.6 1.3 -0.1 The formula is given utilized to form the cross – sectional model. Taking the area generated above the x-axis, where A = 8398.
EXAMPLE: ALVIN (FN) = 1+ 12 + 23 +9 + 14 = 59 NAVALTA (MN) = 14 +1+ 23 + 1+ 12 + 21 + 1 = 72 DELIRO (SN) = 4 + 5+ 12 + 9 + 19 + 16 = 65 4 m B m 1. Determine the components of F that act along rod AC and perpendicular to it. Point B is located 3 m along the rod from end C. F= SN N 3 m 6 m 4 m 2. Determine the following: a) Magnitude of the projected component of along the line of action of F2 b) Angle 0 between the two cables attached to the pipe. F, = MN Ib 120 60° F, = SN Ib 3. The guy wires are used to support the telephone pole. Represent the force in each wire in Cartesian vector form. Neglect the diameter of the pole 1.5 m F, = MNN F2= EN N 4 m 2 m D. 4 m 3 m.