In the code, it is specified that we should take our critical section for bending at the column face (*ACI 318-14, Cl 13.2.7.1*). STRENGTH OF REINFORCED CONCRETE SECTIONS Amount of rebar (A s) The project calls for #5@10” and #5@12” are used: Example: 10” thick wall. We need to estimate the required thickness of the footing, since the self-weight of the footing is usually quite significant. The allowable soil pressure is 5,000 psf and the its density is of 120 pcf. In this case neither the epoxy or casting position factors which further simplifies our calculation. 2. Floor slabs frame into it at 3.2m centres and are 200mm thick. ... Design of reinforced concrete elements with excel notes Download . The wall is assumed to be located in the Christchurch Port Hills. Assuming #8 size reinforcement (1" diameter), we can find d: $$ d = 12\text{ in} - 3\text{ in} - \frac{1}{2}\times1\text{ in} = 8.5\text{ in} $$ We can now calculate the shear at the critical section: $$ \begin{aligned} V_u &= q_u \left(\frac{B}{2} -\frac{b}{2} -d \right) \\ &= 6190 \text{ psf} \left( \frac{62\text{ in}}{2} -\frac{12\text{ in}}{2} - 8.5\text{ in}\right) \\ &= 8.51 \text{ kip/ft} \end{aligned} $$ We must now find the shear resistance. This is usually what will govern the footing's thickness in design. The boundary wall will be made of fly ash brick work. Nevertheless, we see that $\phi M_n > M_u$ so our design is adequate. We will design our footing to resist its load and check it for: We enter the given information directly into ClearCalcs. $$ \begin{aligned} \ell_d &= \frac{f_y\psi_t \psi_e}{25 \lambda\sqrt{f'_c}}d_b \\ &= \frac{60000\text{ psi}\times 1 \times 1}{25 \times 1 \times \sqrt{3000}\text{ psi}} \times 0.5 \text{ in} \\ &= 21.9 \text{ in} \end{aligned} $$ We find the same value as in the textbook's example. Shear connection between columns and walls and between walls concreted in two different … Two … 3500 psi concrete. Design Example 2 Reinforced Concrete Wall with Coupling Beams OVERVIEW The structure in this design example is a six-story office building with reinforced concrete walls as its seismic-force-resisting system. Reinforced Cement Concrete Retaining Wall (Cantilever Type) Information Reinforced Cement Concrete Retaining Wall (Cantilever Type) Maximum 6.0 meter Height including Column Load in Line. We can find a value for $q_u$, the soil pressure at the factored load level, by dividing our total applied load by the footing area. The fluid level inside $$ \begin{aligned} \phi V_c &= 0.75 \times 2 \times 1 \times \sqrt{3000} \text{ psi} \times 8.5 \text{ in} \\ &= 8.38 \text{ kip/ft} \end{aligned} $$ As we had predicted with ClearCalcs in the previous section, we find that $V_u > \phi V_c$. Check Load Combination G (0.6D + 0.7E). Shear wall section and assumed reinforcement is investigated after analysis to verify suitability for the applied loads. The doubly reinforced concrete beam design may be required when a beam’s cross-section is limited because of architectural or other considerations. We pick a 13-inch thick footing and repeat the previous steps: $$ \begin{aligned} d &= 9.5 \text{ in} \\ V_u &= 8.01 \text{ kip/ft} \\ \phi V_c &= 9.37\text{ kip/ft} \end{aligned} $$ We see that the 1-inch increase both decreased $V_u$ and increase $\phi V_c$ as we liked. cmaa australia. Soil: equivalent fluid pressure is 45 psf/ft (7.0 kN/m²/m) (excluding soil load factors), 10 ft (3.05 m) backfill height. The 2012 edition of the Reinforced Concrete Design Manual [SP-17(11)] was developed in accordance with the design provisions of ACI 318-11, and is consistent with the format of SP-17(09). In this example, the structural design of the three retaining wall components is performed by hand. Rectangular Concrete Tank Design Example An open top concrete tank is to have three chambers, each measuring 20′×60′ as shown. coefÞcient of friction is 0.4 and the unit weight of reinforced concrete is 24 kNm 3 1. First, it increases the capacity by providing a greater value of $d$. Reinforced Concrete Shear Wall Analysis and Design A structural reinforced concrete shear wall in a 5-story building provides lateral and gravity load resistance for the applied load as shown in the figure below. It includes: n A description of the principal features of the Australian Standard n A description of the analysis method n Design tables for a limited range of soil conditions and wall geometry n A design example which … We are using a No.4 bar with large spacing, so we can use the least conservative formula as per the table. Reinforced Concrete 2012 lecture 13/2 Content: Introduction, definition of walls 1. As previously discussed, shear reinforcement is usually avoided in footings and the concrete strength was already specified, so we choose to increase the thickness. Design of Slab (Examples and Tutorials) by Sharifah Maszura Syed Mohsin Example 1: Simply supported One way slab A rectangular reinforced concrete slab is simply-supported on two masonry walls 250 mm thick and 3.75 m apart. or #4 bars at 7 inches, which both provide $A_s = 0.34\text{ in}^2\text{/ft}$. Verifying with ClearCalcs, we can now look at the results again with a 13-inch thick footing: We see that we went down from 102% to 85% utilization in shear, and the increase in bearing stress was negligible. Worked example. (M# 29 at 1,829 mm). Retaining walls are utilized in the formation of basement under ground level, wing walls of bridge and to preserve slopes in hilly … Manual for Design and Detailing of Reinforced Concrete to the September 2013 Code of Practice for Structural Use of Concrete 2013 2.0 Some Highlighted Aspects in Basis of Design 2.1 Ultimate and Serviceability Limit states The ultimate and serviceability limit states used in the Code carry the normal meaning as in other … o.c. 1.2 Example Wall . Determine the factors of safety against sliding and overturning. The last failure mode which we need to check is the bending of the footing. Assume a grout spacing of 48 in. (305 mm) thick concrete masonry foundation wall, 12 ft (3.66 m) high. DESIGN OF REINFORCED CONCRETE WALL - Compression member - In case where beam is not provided and load from the slab is heavy - When the masonry wall thickness is restricted - Classified as o plain concrete wall, when rein. While ... for example, moderate or high seismic zone. 2020. Had this not been the case, we could have used hooks at the ends of the bar to significantly reduce the development length, or made use of the more detailed calculations which can be less conservative and more accurate. In this example, the structural design of the three retaining wall components is performed by hand. Reinforced Concrete SK 3/3 Section through slab showing stress due to moment. A 20m high, 3.5m long shear wall is acting as both a lateral and vertical support to a 4-storey building. Bearing ɸ b= AASHTO T.11.5.7-1 Sliding (concrete on soil) ɸ T= AASHTO T.11.5.7-1 Sliding (soil on soil) ɸ T s-s= … Note that we automatically calculate the depth to reinforcement - thus the increase in $d$ from using a smaller bar is automatically calculated which provides us with slightly more capacity! The example wall is shown in Figure X.2. However, we can already see a storm on the horizon! Two equations are … 2.5” clear to strength steel #5@12” rather than the designed #5@10” BENDING STRENGTH OF THE SECTION HAS BEEN REDUCED BY ABOUT 16%. Since we are now dealing with concrete design, we use the ACI 318-14 standard, which is based on LRFD design. $$ q_u = \frac{1.2 \times 10\text{ kip/ft} + 1.6 \times 12.5 \text{ kip/ft}}{5.17 \text{ ft}} = 6 190 \text{ psf} $$ Note that we are taking the net bearing pressure, which does not include the weight of the soil above the footing and the self-weight. There are 6 columns between it and the next shear wall. $$ \begin{aligned} \phi M_n &= \phi A_s f_y\left(d - a/2 \right) \\ &= 0.90 \times 0.34\text{ in}^2\text{/ft} \times 60000 \text{ psi} \left(9.5\text{ in} - \frac{0.667\text{ in}}{2} \right) \\ &= 14.0 \text{ kip-ft/ft} \end{aligned} $$ Note that in this example, $d$ was kept at 9.5 inches even though it would be slightly larger, since we are using #4 bars with half the diameter $d_b$. Design a reinforced concrete to support a concrete wall in a relatively large building. It also reduces the applied shear load since we are taking our critical section further away from the wall face. The fourth edition of Reinforced Concrete Design to Eurocodes: Design Theory and Examples has been extensively rewritten and expanded in line with the current Eurocodes. How to Design Concrete Structures using Eurocode 2 A cement and concrete industry publication. Increasing the thickness benefits shear resistance in two ways. With ClearCalcs, it is just as easy to perform the more detailed calculations of development length, so this is what to do to provide safe and economical designs. Reinforced Concrete Cantilever Retaining Wall Analysis and Design (ACI 318-14) Reinforced concrete cantilever retaining walls consist of a relatively thin stem and a base slab. software such as Mathcad or Excel will be useful for design iterations. Design of the wall reinforcement for shear 5. The stem may have constant thickness along the length or may be tapered based on economic and construction criteria. This Practical Design Manual intends to outline practice of detailed design and detailings of reinforced concrete work to the Code. CE 537, Spring 2011 Retaining Wall Design Example 1 / 8 Design a reinforced concrete retaining wall for the following conditions. The design and detailing requirements for special reinforced concrete shear walls have undergone significant changes from ACI 318-11 to ACI 318-14. The following design … At the base of footing the allowable soil pressure is 5000psf and base of footing is 5’ below the existing ground surface. We thus only need to calculate the factored concrete shear strength $\phi V_c$, which is given by ACI 318-14 Cl 22.5.5.1: $$ \phi V_c = \phi 2\lambda \sqrt{f'_c}d $$ For shear, ACI 318-14 Table 21.2.1 specifies $\phi = 0.75$ and we're using normal-weight concrete so $\lambda = 1.0$. Find the following parameters for design moments in Step 2 per unit width Step 4 Note: Note: Design of slab for flexure 067 m UNIT WIDTH of slab. bid = M + N @ - for N O.lfcubd For design as wall (see Chapter 8). Once we have this, we can calculate the self-weight: $$ SW = 12 \text{ in} \cdot 150 \frac{\text{lb}}{\text{ft}^3} = 150 \text{ psf} $$ Once we know the self-weight, we immediately remove it from the allowable bearing pressure, together with the weight of the soil above the footing, and then divide the total load by this adjusted bearing pressure to find the required area. build right retaining walls. For simplicity, we use Table 25.4.2.2, which gives a simple equation to calculate the development length. Resistance to axial compression 3. In that case, steel bars are added to the beam’s compression … We compare this to the distance to the critical section: $$ \frac{B}{2}-\frac{b}{2} = \frac{5.17 \text{ ft}}{2}-\frac{1 \text{ ft}}{2} =2.09 \text{ ft} = 25 \text{ in} $$ Since 25 inches is larger than 21.9 inches, we know our bars are developed as required. The highest groundwater table is expected to be 4′ below grade. 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