Beef stew is a very common type of stew cooked globally. There are many variations of it. This article explains a basic simple way to prepare beef stew.
Basic Ingredients for beef stew
Meat: The meat should be spiced and cooked by any cooking methods you prefer. i.e. boiling, roasting, frying, oven drying, etc. The meat should be cut into small pieces such that it is easy to hold and eat.
Blended pepper, tomatoes and onions:
This should be blended in the right proportions. For every two parts of tomatoes, one part of pepper can be added, even that may be too much depending on the type of pepper. Amount of onions may just be about 10% or less of the amount of tomatoes. However, experience will easily tell you the right combination/proportions suitable for you and family. Make sure not to add too much water when blending the pepper mixture unless you might end up with a soup instead of stew.
Cooking oil
You can use palm oil or any type of vegetable oil. You are better off with cholesterol free oils.
Seasoning/Spices & salt
These are flavourings to make the stew taste better or the way you want it to taste. Add salt to taste.
General method of cooking beef stew
1.Put a small amount of cooking oil into the cooking pot and heat up until it is near hot.
2. Next add the blended pepper and cook for about 7 minutes on medium heat.
3. Next add the chopped pieces of already cooked meat and stir the mixture.
4. Let it cook for a minute or two then add your seasoning/spices and salt to taste. Stir the stew and let it cook for another two minutes, then it is done.
There are many ways of cooking beef stew, some might prefer to have the pepper, tomatoes and onions chopped instead of blending smooth, you can also cook with other type of vegetables, fishes, crayfish and so on.
The iphone 13 Pro Max is the best phone in the world currently. Mainly because it has the best picture and video quality. Its cameras are the best. It has an awesome CPU performance because of its bionic chip. It has over 12 hours of battery life when subjected to web surfing test. It has 120Hz pro motion display with superior performance.
Iphone 13 Pro Max phone specifications
Display: 6.7-inch OLED (2778 x 1284)
CPU: A15 Bionic
RAM: 6GB
Storage / Expandable: 128GB, 256GB, 512GB, 1TB
Rear camera: 12MP main (f/1.5), 12MP ultrawide (f/1.8), 12MP telephoto (f/2.8) with 3x
The term, ‘beef’ , refers to the flesh of a cow or bull. It’s a common source of protein for people all over the world. Beef when properly cooked is very delicious. You can be one addicted to eating beef. Beef belongs to the Class of meat called red meat. Beef contains a lot of nutrients including; Proteins, Fat, Vitamin b12, Zinc, Selenium,Vitamin B3, Iron.
Buying beef.
There are many places where you can buy beef. You can buy at supermarkets ( usually frozen), butchers stores and meat farm ( buying directly from people who raised the animal. However the best place to buy beef is from the farm or a butcher’s store. This is because you can see the source of the meat you are buying. Meat you buy should look fresh, not pale or frozen burnt, should smell nice/ fresh.
Cooking beef
There are many methods of cooking beef including boiling, frying, grilling, roasting etc. The easiest way to cook beef is boiling with salt and spices. Types of cooking beef ranges from 10 to 20 minutes. Roasting, frying, grilling gives cooked meat a wide range of tastes and these tastes vary based on the part of the cow or bull the meat is cut from. The figure below shows major parts of a cow meat.
part of beef
Eating beef ( precautions)
Control the amount of beef you eat, too much of it may not be good for you.Make sure the beef you eat is properly cooked. Eating semi cooked meat is dangerous to health. You should not eat more than 700g of red meat per week according to health practitioners.
Have you ever looked for the windows explorer icon in your windows 10 PC to no avail, to deliver yourself from this trouble, just Press Windows key+E to open a new instance of File Explorer directly.
Alt+Up Arrow. Go up one level .
Alt+Left Arrow. Go to the previous folder in history.
Alt+Right Arrow. Go to the next folder in history.
Alt+D. Move the focus to the address bar and select the current path.
F4. Move the insertion point to the address bar and display the contents of the drop-down menu of previous addresses.
Alt+Enter. Show properties of the selected file.
Tab. Cycle through the following elements: navigation pane, file list, column headings, address bar, search box .
F11. Switch in and out of full-screen mode.
Ctrl+Shift+N. Create a new subfolder in the current folder.
Ctrl+Shift+E. Expand the navigation pane to the current folder.
Most fruits and vegetables will last longer when refrigerated. Don’t refrigerate unripe fruits, instead leave them at normal room temperature for them to ripen. Chopped/ sliced or left over vegetables should be refrigerated.
Some list of vegetables you should not refrigerate
Plum
Potatoes
Yams
Onions ( unless spring onions or scallions)
Tomatoes
Cassava
Peaches
Some list of vegetables you should refrigerate ( when ripe)
General procedure for design of solid slabs to BS Code.
1.Determine a suitable slab depth; you can easily estimate this by using the formula stated below;
Effective depth = span/( basic ratio × modification factor)
Span/effective depth ratios
For a simply supported design, the basic ratio is = 20 ( table 3.9, page 35 of BS 8110 part 1 : 1997). Use an initial modification factor (m.f) of 1.4.
Also the architect/ engineer may specify the slab thickness for you.
However, your slab design must satisfy deflection requirements otherwise you will have to redesign.
Ways to make your design slab satisfy deflection requirements if failed initially is to either increase slab thickness or amount of reinforcement or both.
2. Calculate the main and secondary reinforcement areas; you can do this by using the formulas stated in the BS code. ( Details later)
3. Check for excessive deflection ; as stated earlier in number 1, your design must satisfy deflection requirements.
4. Detailing requirements; the economical arrangement of steel reinforcement. You can refer to clause 3.12.10.3, BS 8110. An explanatory diagram is shown below for a simply supported case.
Detailing requirements for simply supported (a) & continuous ( b) slabs
It means that 40% of the reinforcement placed around the center of the slab ( which is critical) only, should extend to the edges of the slab ( which is less critical).
Explanation of other important points in solid slab design to BS 8110
Effective span of slab:
Referring to the diagram above, the effective span of slab refers to ‘A’ the distance between the centers of bearings, or the clear distance between supports ‘D’, plus effective depth of slab ‘d’.
Calculating steel areas.
In calculating the slab self weight, the overall depth of the slab referred to as ‘h’, should be used. h is the effective depth of slab plus allowance for cover to reinforcement plus half the assumed main steel diameter bar.
The self weight of the slab together with the dead and live load is used to determine the design moment ‘M’.
The value of M must be checked against the value of Mú which is the ultimate moment of resistance
Mú = 0.156fcubd²
Where b=1000mm and fcu = strength of concrete.
If Mú ≥ M then the slab doesn’t need compression reinforcement which is usually the case.
Main reinforcement steel areas
The area of reinforcement Aₛ can be determined using Aₛ= M÷(0.87𝑓yz) where,
M = design moment= WL²/8 (for simply supported slab)
W = design load of slab in kN/m²
L = span of slab in meters.
𝑓y= strength of steel
z = d[0.5 + √ (0.25 – K/0.9)] and
K= M / fcubd²
Secondary reinforcement steel areas
Secondary reinforcement also refers to distribution steel. As per BS 8110, it is calculated as follows;
Aₛ min = 0.24% Ac when 𝑓y = 250N/mm²
Aₛ min = 0.13% Ac when 𝑓y = 460N/mm²
Check that your slab design meets deflection requirements
Design service stress, fs = (2 𝑓y Asreq.) ÷ ( (3 Asprov), (table 3.10 page 36 of BS 8110 part 1 1997.)
Where Asreq & Asprov is the area of steel calculated and the area of steel provided, respectively.
Modification factor;
m.f = 0.55 + (477 – fs) ÷[120 ( 0.9 + M/bd²)] ≤ 2
Now this m.f is your actual m.f since it is a function of the area of steel calculated and area of steel provided.
Use this new m.f to replace the m.f you initially assumed to be 1.4 in the equation;
Effective depth; d = span/( basic ratio × modification factor).
you can also determine m.f from the table below
If the assumed value of d is greater than actual d, then deflection requirements are satisfied.
Crack widths
The BS code specifies that crack width should not exceed 0.3mm. except by calculation, the following rules should ensure crack width requirements are satisfied;
fy = 250N/mm² and depth of slab ≤ 250mm
or
fy = 460N/mm² and depth of slab ≤ 200mm
or
the percentage of reinforcement ( 100As/bd) ≤ 0.3%
Maximum bar spacing of reinforcement
BS code specifies that the clear distance between tension bars should be less than 3d or 750mm whichever is the lesser.
Example design of a simply supported solid slab
From the diagram of part of a floor plan shown above, I will be designing the panel labeled A.
The first thing to check is if the slab is a one or two way slab. By observing the dimensions relating to the slab panel, the longer side is
3401 + 587 + 1225 = 5213mm
The shorter side of the panel is
1150 + 1225 = 2375mm
Therefore longer side ÷ shorter side =
5213÷2375= 2.19 ≥ 2;
Panel A is a one way slab.
This means that the main reinforcement bars will span along the y- axis ( shorter side) and the distribution bar will span along the x – axis ( longer side).
direction of main reinforcement
Note that in a one way slab, the span of the slab is the shorter side. The span of the slab panel A being designed is 2375mm.
Further observation of the slab panel will show that its thickness (overall depth of slab) has been given as 150mm;
h = 150mm
Your duty as the designer is to calculate the required steel reinforcement and check if the design satisfies deflection requirements. If the slab thickness wasn’t given then you are free to assume a thickness and then check that it works.
Calculating the Design load.
The formula stated by BS code is
1.4Gk + 1.6Qk
where Gk = dead load
Qk = live load
dead load is the self weight of the slab plus finishes
Live load refers to variable or movable loads such as people, furniture etc, the slab carries. Live loads for different categories of buildings are stated in BS 6399 part 1: 1997.
Gk = weight of concrete x slab thickness
= 24kN/m³ x 0.15m ( slab thickness of 150mm)
= 3.6kN/m², plus finishes of say 1.2 kN/m² which gives a total gk of
4.8kN/m² ;
Gk = 4.8kN/m²
For private dwellings, Qk = 1.5kN/m² (see table 1 of BS 6399 part 1: 1997)
Design load = 1.4gk + 1.6qk
=( 1.4x 4.8 + 1.6 x 1.5) kN/m²
= 9.12kN/m² per m width ( slabs are designed per m width.)
Design moment M = WL²/8
where W = design load and L = span of slab
W=[ 9.12kN/m² x ( 2.375m)² ] / 8
= 6.43kNm
ultimate moment of resistance
Mu= 0.156fcubd²
Now d = effective depth of slab which can be estimated as overall depth of slab minus concrete cover to reinforcement minus half of main steel diameter
I.e. d = 150mm – 25mm – 6mm = 119mm.
( 25mm is concrete cover to reinforcement and 6mm is half the diameter of 12mm steel rod)
b = 1000mm ( slabs are designed per m width)
fcu = 25N/mm² ( strength of concrete)
Mu = 0.156 x 25N/mm² x 1000mm x (119mm)² = 55.2279 x 10⁶ Nmm
= 55.23kNm
Since M < Mu no compression reinforcement is required.
K= M / fcubd²
= 6.43 x 10⁶/ ( 25 x 1000 x 119²)
0.0182
z = d[0.5 + √ (0.25 – K/0.9)]
= 119[ 0.5 + ✓ ( 0.25 – 0.0182/0.9)]
= 116.54mm ≤ 0.95d
Limiting to 0.95d
0.95d = 113.05; use z= 113.05
Hence,
Aₛ= M÷(0.87fyz)
= 6.43 x 10⁶ ÷ ( 0.87 x 460 x 113.05)
Aₛ required. = 142.12 mm²/m width of slab
From fig. 1 shown above, use Y10 bars at 200mm spacings as main steel reinforcement
I.e.
Y10- 200
Aₛ provided = 393 mm²/m
Distribution reinforcement
Minimum Steel reinforcement specified by code = 0.13%bh
= 0.13% x 1000 x 150
Aₛ minimum required.= 195 mm²/m
With reference to fig 1, use Y10 – 250 bars. Aₛ = 314mm²/m
Note that the calculated required reinforcement is less than the minimum reinforcement specified by code, in this case the minimum reinforcement specified by code supersedes and should be used to select provided steel. Also using Y10- 200 as main and distribution steel is OK. It’s your design, you are in charge, just make sure you follow the code requirements and design laying emphasis on economy.
Checks
Deflectioncheck
Design service stress,
fs = 2 fy As req./ (3 As prov)
= 2×460×142.12÷(3×393)
=110.9N/mm²
fs = 110.9N/mm²
m.f = 0.55 + (477 – fs) ÷[120 ( 0.9 + M/bd²)] ≤ 2
= 0.55 + (477- 110.9)÷[120(0.9 + (6.43 x 10⁶/100 x119²)]
=2.80 ≤ 2
Since m.f is limited to 2,
Hence, m.f = 2
dmin = span / basic ratio x m.f
= 2375mm / (20 x 2)
= 59.375mm
59.375mm ≤ 119mm hence deflection is satisfied.
Crack width check
Since the slab is less than or equal to 200mm thick, crack width check is satisfied
Maximum spacing check
The reinforcement spacing of 200mm for main steel and 250 mm for distribution steel is less than 3d ( 3 x 119 = 357mm) hence maximum spacing check is satisfied.
