Page Title: Methods of RCC Design | Comparison of Working Stress Method and Limit State Method | Design Example for WSM & LSM
Keywords: RCC Design , Methods of RCC designing , WSM , LSM , Diff. b/w WSM & LSM , Working Stress Method , Limit State Method , Example of WSM , Example of LSM , Elastic Design , Plastic Design
Methods of RCC Design:
RCC Structural Elements are designed by two methods,
Working Stress Method (WSM)
Limit State Method (LSM)
Working Stress method is the old way of designing , In present era structures are generally designed using Limit State Method. In WSM concrete is considered as elastic whereas in LSM concrete is generally assumed as plastic.
Difference b/w WSM & LSM:
As discussed above there are two different methods
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= 20 mm
Bar Nos. = 985.5 / (pi()/4*20^2) = 3.13 = 4 (say)
Bottom Reinforcement = 4 # 20 mm Dia. Bar
For Top Reinf.(Tie Bars) Ast = (35/100)* (pi()/4*20^2)*4 = 439.82 mm2
Bar Dia. = 12 mm
Bar No. = 439.82 / (pi()/4*12^2) = 3.13 = 4 (say)
Top Reinforcement = Provide 4#12 mm Dia. Bar
Design Example of Structural Member (Beam) by LSM:
Given Data :
Breadth of the Beam = 300 mm
B.M = 200 KN/m
Factored B.M. = 1.5 * 200 = 300 KN/m fck = M20 fy = Fe415
Solution :
Ultimate Moment of Resistance (Mu lim) = 0.138*fck*b*d^2
Mu lim ≥ 300 KN/m
0.138 * (20) * (300) *d^2 ≥ 300 *10^6 d ≥ 601.93 mm
Design of Beam Depth :
For M20 & Fe415 , Sigma_cbc = 7 N/mm2
Let us assume, Overall Depth D = 550 mm
Clear Cover = 25 mm
Assume , Dia. of Bar = 20 mm Stirrups Dia. = 8 mm
D = 601.93 + 25 + 8 + 1/2 *20 = 644.93 mm
Provided , D = 650 mm
Eff. Depth of Beam(d) = 650 - 25 - 8 - 1/2 * 20 = 607 mm
Thus Section provide is 300 x 650 mm.
Reinforcement : pt / 100 = Ast / bd = 0.5 * fck / fy *(1 - sqrt(1-4.5*M/(b*d^2)/fck))
= 0.5 * 20 / 415 *( 1- sqrt ( 1- 4.5 * 300 * 10^6 / 300 * 607^2*20)
= 0.93 %
Ast = 0.93 * b * d/100 = 0.93 * 300 * 607 /100 = 1696.99 mm2
Bar Dia . = 20 mm
Bar Nos. = 1696.99/(pi()/4*20^2) = 5.4 ≈ 6
11. Using the volume(s) you just calculated for regular vs. heavy duty samples as well as your dimensional measurements (length and width in cm) from Part IV of this experiment, calculate the height, or thickness, of each sample of aluminum using the formula V = l x w x h. In the formula, V stands for volume, l for length, w for width, and h for height. Once again, you will have to use your algebraic skills to manipulate the formula, to solve for height. You must show all your work. (15 pts)
The goal of the beam project is to design and construct a beam that can hold a given amount of weight without breaking. The beam is required to hold a concentrated load of 375 lbf on the X-axis and 150 lbf on the Y-axis. The maximum allowable weight of the beam is 250 grams. The maximum allowable deflection for the beam is 0.230 in. and 0.200 in. for the X and Y-axis respectively. The beam is required to be 24 in. in length, and it will be tested on a simply supported configuration spanning 21 in. All calculations are to be done under the assumption that the density of basswood is 28 lbm/ft3 and the modulus of elasticity for basswood is 1.46x106 lbm/in2. Given the constraints of a spending cost of $10.50, a maximum beam weight of 250 grams,
c. A lab exercise in BIO156 required 300 ml of water that was poured from a two-liter container. How many milliliters were left in the original container? (4 points)
SQRT(2 * F * T / H) = (2 * 80 * 200,000 / 1.00)0.5
baseline (see Figure 9.4 on page 288 for help). If the solvent traveled 57 mm
in the xy-plane is equal to the radius (R) of the beam pipe, as shown in Fig.~\ref{beampipeconversion},
If an object is 334 µm wide, how many mm wide is it? 0.334 mm.
When the dBm value increases by adding 3, the mW value increases by the factor of 2.
Plus fill the rest of the order with 402,s – 3,000 x 0.4 = 1,200
P = $40({1 – [1/(1 + .0475)]26 } / .0475) + $1,000[1 / (1 + .0475)26]
A suspension system is required to operate efficiently in various operating conditions, such as high speed cornering, accelerating and braking. A Suspension system is used to absorb the vibrations caused by the road roughness. It is designed to improve the comfort in a vehicle, and also handling performance. The main design features of a suspension system are the optimization of the friction between the contact patch of the tyre and the road surface, and to deliver vehicle stability. The system has to permit the distribution of forces generated by the wheels in contact with the ground. A suspension system has to comply with the specifications of the design in every load condition.
Despite its exceptional material properties, UHPC’s initial construction cost is significantly higher than that of normal strength concrete (NSC), which limited its use in North America and elsewhere. To address some of the challenges facing bridge owners, the structural efficiency and cost-effectiveness of using UHPC in composite construction is investigated in this paper. Therefore, combining NSC with UHPFRC in a composite
The beam was loaded the mid-length in 2.745 lbs. increments up to 6.745 lbs. The change in clearance of every load step was measured and data was recorded.
However, the team uses finite element analysis software along with DOE to predict two responses from each run such as the stress and the deformation. In the case study, we observe the images that show the stress and deformation due to a temperature change. The high stress and deformation areas are shown by the red color. The low stress and deformation areas are indicated by the dark blue color. At last the deformation of a single layer was imported into MATLAB (a matrix computing environment) to calculate the radius of curvature or warp.
Fiber Path Parameterization: This is achieved by using a curvilinear function to describe the fiber path. Gürdal and Olmedo were the first to introduce a fiber path parameterization where the fiber orientation angle varies linearly [34]. The linear angle variation was later generalized by Tatting and Gürdal to vary linearly along an arbitrarily defined axis where it was used to design variable stiffness laminates for strength [7,35,36], thermomechanical response [37], and coupled strength-buckling optimization problems [38,39]. Nagendra et al. used global fiber paths constructed by linear combination of non-uniform rational B-splines (NURBS). They studied optimal frequency and buckling load design where the design variables where multipliers of the different basis fiber paths [40]. Alhajahmad et al. used a non-linear fiber path expressed in terms of more complex functions such as Lobatto polynomials to increase the number of design variables and achieve better