Ratios formulas:
Inventory turnover=cost of goods sold/Inventory
Average collection period=accounts receivable/average sales per day=accounts receivable/(annual sales/365)
Asset turnover=sales/total assets
Debt ratio=total liabilities/total assets
Gross profit margin=(sales-cost of goods sold)/sales=Gross profit/sales
Net profit margin=net income or earnings available for common stockholders/sales
ROA=net income or earnings available for common stockholders/total assets
ROE=net income or earnings available for common stockholders/common stock equity
Current ratio=current assets/current liability
From the pictures above, please answer those qustions below: –
Q 1. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Gross Profit
a. 2,341,100.00
b. 1,371,000.00
c. 1,250,000.00
Q 2. If depreciation expense for 2020 is at 500,000.00, how much is the net change in cash due to investing activity?
a. -750,112.00
b. -830,110.00
c. -650,142.00
Q3. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Total Current Liabilities
a. 416,500.00
b. 415,040.00
c. 616,000.00
Q4. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Inventory
a. 700,625.00
b. 720,525.00
c. 230,655.00
Q5. Depreciation Expense for 2020 is at 500,000.00. Using the given data, compute for the net change in cash for 2020.
a. 900.00
b. 800.00
c. 1,000.00
Q6. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Shareholder’s Equity
a. 2,045,650.00
b. 1,043,750.00
c. 1,333,650.00
Q7. The depreciation expense for 2020 is 500,000.00. Using the data available, compute for the 2020 change in cash due to operating activities.
a. 360,762.00
b. 550,662.00
c. 465,752.00
Q8. The depreciation expense for 2020 is 500,000.00. Using the data above, compute for the company’s 2020 Earnings available for common stocks
a. 37,000.00
b. 35,000.00
c. 33,000.00
Q9. Depreciation expense for 2020 is at 500,000.00, compute for the company’s 2020 quick ratio
a. 1.2
b. 1.7
c. 1.8
Q10. Depreciation Expense for 2020 is at 500,000.00, compute for the 2020 Times interest earned
a. 1.2
b. 1.20
c. 1.6
Q11. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Retained Earnings
a. 400,000.00
b. 350,200.00
c. 300,000.00
Q12. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Accounts payable?
a. 240,050.00
b. 450,000.00
c. 230,000.00
Q13. Depreciation Expense for 2020 is 500,000.00
Using the data above, compute for the company’s 2020 Earnings Before Tax
a. 65,200.00
b. 70,000.00
c. 60,000.00
Q14. The depreciation expense for 2020 is 500,000.00. What is the 2020 change in cash position due to financing activities?
a. 490,452.00
b. 420,150.00
c. 390,250.00
Q15. Depreciation Expense for 2020 is at 500,000.00
Using the Data above, compute for the company’s 2020 Accounts Receivable.
a. 805,556
b. 875,656
c. 835,545 Ratios formulas:
Inventory turnover=cost of goods sold/Inventory
Average collection period=accounts receivable/average sales per day=accounts receivable/(annual sales/365)
Asset turnover=sales/total assets
Debt ratio=total liabilities/total assets
Gross profit margin=(sales-cost of goods sold)/sales=Gross profit/sales
Net profit margin=net income or earnings available for common stockholders/sales
ROA=net income or earnings available for common stockholders/total assets
ROE=net income or earnings available for common stockholders/common stock equity
Current ratio=current assets/current liability
From the pictures above, please answer those qustions below: –
Q 1. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Gross Profit
a. 2,341,100.00
b. 1,371,000.00
c. 1,250,000.00
Q 2. If depreciation expense for 2020 is at 500,000.00, how much is the net change in cash due to investing activity?
a. -750,112.00
b. -830,110.00
c. -650,142.00
Q3. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Total Current Liabilities
a. 416,500.00
b. 415,040.00
c. 616,000.00
Q4. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Inventory
a. 700,625.00
b. 720,525.00
c. 230,655.00
Q5. Depreciation Expense for 2020 is at 500,000.00. Using the given data, compute for the net change in cash for 2020.
a. 900.00
b. 800.00
c. 1,000.00
Q6. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Shareholder’s Equity
a. 2,045,650.00
b. 1,043,750.00
c. 1,333,650.00
Q7. The depreciation expense for 2020 is 500,000.00. Using the data available, compute for the 2020 change in cash due to operating activities.
a. 360,762.00
b. 550,662.00
c. 465,752.00
Q8. The depreciation expense for 2020 is 500,000.00. Using the data above, compute for the company’s 2020 Earnings available for common stocks
a. 37,000.00
b. 35,000.00
c. 33,000.00
Q9. Depreciation expense for 2020 is at 500,000.00, compute for the company’s 2020 quick ratio
a. 1.2
b. 1.7
c. 1.8
Q10. Depreciation Expense for 2020 is at 500,000.00, compute for the 2020 Times interest earned
a. 1.2
b. 1.20
c. 1.6
Q11. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Retained Earnings
a. 400,000.00
b. 350,200.00
c. 300,000.00
Q12. Depreciation Expense for 2020 is at 500,000.00
Using the data above, compute for the company’s 2020 Accounts payable?
a. 240,050.00
b. 450,000.00
c. 230,000.00
Q13. Depreciation Expense for 2020 is 500,000.00
Using the data above, compute for the company’s 2020 Earnings Before Tax
a. 65,200.00
b. 70,000.00
c. 60,000.00
Q14. The depreciation expense for 2020 is 500,000.00. What is the 2020 change in cash position due to financing activities?
a. 490,452.00
b. 420,150.00
c. 390,250.00
Q15. Depreciation Expense for 2020 is at 500,000.00
Using the Data above, compute for the company’s 2020 Accounts Receivable.
a. 805,556
b. 875,656
c. 835,545
Post navigation
The design is about the design of floor;
please follow the (Example1 pdf) step by step which attached in below and complete the design.
3.3 Design examples
3.3.1 Layout of floor structure
The plane arrangement of the floor structure is shown in Fig. 3-8, the main beam is arranged along the longitudinal and transverse direction, and the secondary beam is arranged along the longitudinal direction, and two secondary beams are arranged in each span of the main beam. The span of transverse main beam is 7.2 m, that of longitudinal main beam is 6.6 m , that of secondary beam is 6.6 m, the span of the board is 2.4m. Because of 6600/2400=2.75 2, it is necessary to calculate according to the requirements of the code, and when calculated according to the unidirectional plate, a sufficient number of structural tendons should be arranged along the long side direction. This example is calculated as an one-way plate and takes the necessary construction measures.
Fig. 3-8 layout of floor structure
3.3.2 Prediction of beam and plate section dimensions
? slab thickness: according to«Concrete specification», the thickness of the slab can be obtained according to Table 3-2, which takes into account the durability of the floor and the minimum thickness 70mm of the industrial building considering the durability of the floor and the minimum thickness of the floor of the industrial building. The thickness of the slab is h= 80mm, taking into account the durability of the floor and the minimum thickness of the floor of the industrial building.
? the section size of the beam can be determined according to experience. According to the minimum stiffness requirements of deflection and crack width without checking, the size of beam loading surface is estimated according to Table 3-1. Main beam:
Secondary beam:
3.3.3 Design of boards
1. Load calculation
1) Standard value of constant load
30m thick tile floor (including cement mortar): 0.65 (kN/m2).
20mm thick cement mortar leveling layer: 20×0.02 = 0.4 (kN/m2).
80mm thick reinforced concrete slab: 25×0.08 = 2.00 (kN/m2).
20 mm thick mixed mortar surface: 17×0. 02 =0.34 (kN/m2).
Constant load standard value: gk = 3.39 kN/m2.
2) Standard value of live load
Live load standard value: gk = 8.00 kN/m2.
According to Appendix D, Table D.1-1. In the four types of metalworking workshops, the span of the plate is 2.4 m and 2.0 m. The standard value of live load is 800kN/m2 and the coefficient of combined value is ?c =1.0 3) load design value
According to Appendix D, Table D.1-1. The combination coefficient of the live load ?c =1.0, the standard value of the live load qk = 8.00kN/m2 4.0 kN/m2, and the partial coefficient is 1.3. The design value of the load effect is calculated according to formula (3-1) and formula (3-2) for variable load effect and permanent load effect control, Take the maximum value. Variable load effect control combination:
Combination of permanent load effect control:
It is found from the calculation that the load design value obtained by the combination of permanent load effect control is large, and the internal force of the plate is calculated by g+q=14.98 kN/m2.
2. Calculation diagram
In this project, the two ends of the plate are connected with the beam as a whole, and the calculated span of the plate is calculated according to Table 3-3, as follows.
The difference of each calculated span is less than 10%, which can be approximately calculated according to the equal span continuous plate. The span 2200mm of the plate and the plate and strip of the plate width b=1000mm are calculated. The calculation diagram is shown in figure 3-9.
Fig. 3-9 board calculation diagram
3. Calculation of bending moment and reinforcement of Plate
1) Bending moment calculation
According to the Code of Internal Force redistribution, the bending moment design values of equal span continuous unidirectional plates subjected to uniform load are calculated according to formula (3-6), and the bending moment coefficients are determined in Table 3-4 of ?mb.
According to the rules of Internal Force redistribution, when the perimeter of unidirectional continuous slab is connected with reinforced concrete beam as a whole, the design value of span center and bearing bending moment of each middle span can be reduced by 20%, so the values of M2, M3 and Mc can be reduced by 20%, except for the second support of edge span and off end, and the results after reduction are shown in Table 3-7.
2) Calculation of plate reinforcement
4. Plate reinforcement diagram (figure 3-10)
In general, the dynamic load on the floor is not large, so the separated reinforcement can be used.
Table 3-7 Reinforcement calculation diagram of slab
Figure 3-10 Plate reinforcement diagram
3.3.4 Design of secondary beams 1. Load calculation
1) Standard value of constant load
Constant load from the plate: 3.39×2.4 = 8.14 (kN/ m).
The weight of the secondary beam is 25×0.2× (0.5-0.008) = 2.12 (kN/m).
Beam side plastering: 17×0.02× (0.5-0.008) ×2 = 0.29 (kN/ m).
Load standard value: gk = 10.53 (kN/ m).
2) Standard value of live load
According to Table D.0.1-1 in Appendix D to the load Code, the distance between the secondary beams is 2.4m 2.0m, the standard value of live load is 6.00kN/m2, and the coefficient of combination value is 1.0.The standard value of live load is 6.00kN/m2.
The standard value of live load from the plate is qk = 7×2.4 = 14.40 (kN/m).
3) load design value
According to table D.0.1-1 in Appendix D to the load Code, the combined value coefficient of live load is ?c =1.0, the standard value of live load is qk = 6.00kN/m2 4.0 kN/m2, and the score coefficient is 1.3. The design values of load effect are calculated according to the variable load effect and permanent load effect of formula (3-1) and formula (3-2), respectively, and the larger values are taken.
Permanent load effect: g+q= 1.2×10.53+1.3×14.40×1.0=31.36 (kN/ m2)
Variable load effect: g+q= 1.35×10.53+1.3×14.40×1.0=32.94 (kN/ m2)
According to the calculation, the design value of the combined load controlled by the permanent load effect is large, and the internal force of the secondary beam is calculated by g+q=32.94kN/m. 2. Calculated span
The calculation method of the calculation span is the same as that of the plate, which is calculated according to Table 3-3, as follows.
Middle span: l0=ln = 6600-300 = 6300 (mm)
Edge span: l0=ln =6600-100-150 = 6350 (mm)
The difference between the calculated span of the side span and the middle span is 63506300/6300=0.79% 10%, so the internal force can be calculated according to the equal span continuous beam. The geometric size and calculation diagram of the secondary beam are shown in figure 3-11.
Fig. 3-11 Calculation diagram of secondary beam
3. Internal force calculation of secondary beam
The calculation method of internal force of secondary beam is the same as that of unidirectional plate. According to the Code of Internal Force redistribution, the bending moment design values of equal span continuous unidirectional plates subjected to uniform load are calculated according to formula (3-6), and the bending moment coefficients are determined in Table 3-4 of ?mb. 1) Design value of bending moment
2) Shear design value
3) Reinforcement calculation of secondary beam section
The secondary beam and the slab are integrally cast, so the secondary beam support is calculated as a rectangular section, and the mid-span section is calculated as a T-shaped section. Calculate the width bf of the flange of the T-shaped section according to Article 4.1.5 of the -Concrete Code-, and the side span and the middle span are calculated according to the minimum value of l0/3 and (b+Sn).
Take bf =-2.10m, effective section height h0 =500-40= 460(mm), flange thickness hf = 80mm,
This value is greater than the mid-span bending moment design value M1, M2, M3, so the midspan section of each span belongs to the first type of T-shaped section. The calculation parameters are as follows. C30 concrete: fc= 14.3MPa; reinforcement: longitudinal reinforcement fy = 360MPa (HRB400), stirrup fyv =270MPa (HPB300), the calculation results of the flexural bearing capacity of the normal section of the secondary beam are shown in Table 3-8 shown.
Table 3-8 calculation of Flexural bearing capacity of normal Section of Beams
4) Calculation of shear capacity of oblique section
The calculation table for the shear capacity of the oblique section of the secondary beam is shown in Table 3-9. According to the -Regulations for Redistribution of Internal Forces-, when considering the redistribution of plastic internal forces, under uniform load, the calculated area of stirrups should be increased by 20% in the 1.05ho section from the support edge, and the pmin of the stirrups ?min 0.3ft/fw=0.159%.
5) Design of anti-shear stirrups
The calculation of the shear capacity of the oblique section of the secondary beam is shown in Table 3-9.
4. Secondary beam reinforcement diagram
The reinforcement diagram of the secondary beam is shown in Figure 3-12
3.3.5 Main beam design
The main beam is calculated according to the elastic theory, and the column section size is
400mm×400mm.
1. Load calculation
1) Standard value of dead load
The self-weight of the main beam is considered according to the concentrated load.
Dead load from the secondary beam: 10.53×6.3=69.50(kN/m).
Main fruit weight: 25×0.3× (0.75-0.08) ×2.4=12.06(kN/m).
Plastering on the side of the main beam: 17×0.02× (0.75-0.08) ×2×2.4=1.09(kN/m).
Load standard value: gk=82.65(kN/m).
2) Standard value of live load
According to Table D.0.1-1 of Appendix D of the -Load Code-, the standard value of live load is 5.00 for the main beam in metalworking workshops of four types.
Combination value coefficient ?c. =1.0.
The standard value of the live load from the board: 5.0×6.6×2.4=79.20(kN/m).
3) Load sub-factor
According to Article 3.2.4 of the -Load Code-, the value of the partial load factor is as follows.
The constant load sub-factor is ?G; when it is unfavorable to the structure, it is 1.35; when it is favorable to the structure, it is 1.0.
Live load sub-factor: standard value qk=5.00kN/m2 4.0kN/mm2, and the sub-factor is 1.3.
2. Simple calculation
The two ends of the beam in this project are integrally connected with the column, and the calculated span of each span is taken from the distance between the center lines of the support, calculated to the side span and the middle span: l0=lc =7200mm main beam calculation diagram is shown in Figure 3-13.
3. Internal force calculation
The three-span continuous beam shown in Figure 3-11 uses equations (3-12) and (3-13) to calculate the internal force of each control section, and calculate the standard value of the bending moment and the standard value of the shear force at each control section. Among them, the -coefficients in the table- are checked from the attached table 1-2 of the -Design of Concrete Structures-, and the most unfavorable arrangement of live loads is considered. The calculation results are shown in Table 3-10 and Table 3-11.
1) Calculation of standard value of bending moment
(1) Dead load
Bending moment within the span:
M1k=0.244gkl0=0.244×82.65×7200=145.20(kN·m) M2k=0.267gkl0=0.267×82.65×7200=158.89(kN·m) Bending moment of support:
MBk=-0.067gkl0=-0.067×82.65×7200=-39.87(kN·m)
Mcx=-0.267gkl0 =-0.267×82.65×7200=-158.89(kN·m)
(2) Live load
Bending moment within the span: M1t=0.289qkl0=0.289×79.20×7200=164.80(kN·m): M2z=0
Bending moment of support: MBk=Mck=-0.133gxl0=-0.133×79.20×7200=-75.84(kN·m)
2) Calculation of shear force standard value
(1) Dead load
VA=0.733gk =0.733×82.65=60.58(kN)
VB ?= =-1.267gk=-1.267×82.65 =-60.58(kN)
VB ?= gk =1×82.65=82.65(kN)
(2) Live load
VA=0.866qk=0.866×79.20=68.59(kN)
VB4=VB ? =-1.134qk=-1.134×79.20=-89.81(kN)
3) Perform internal force combination according to the following formula to obtain the most unfavorable combination design value at each control section
M =?GMGK +?Q?CMPK
V = ?G VGK + ?Q?CVpK
According to the static balance conditions, the most unfavorable internal force design values of the remaining sections are obtained. The results are shown in Table 3-10 and Table 3-11 shown.
4) Draw the internal force envelope diagram
Plot the combined bending moment design value and combined shear design value of each control section on the same coordinate diagram to obtain the internal force envelope diagram (Figure 3-14)
5) Calculation of the bearing capacity of the normal section of the main beam
The main beam support is calculated according to the rectangular section, and the mid-span section is calculated according to the T-shaped section.
According to Article 4.1.5 of the -Specification for Concrete-, the width bf of the flange of the T-shaped section is calculated, and the side span and the middle span are calculated according to the minimum value of l0/3 and (b+sn).
Side span: bf=min [l0/3, (b+sn)] =min [(7.2/3, (0.3+6.35)] =2.4m
Intermediate span: bf=min [l0/3, (b+sn)] =min [7.2/3, (0.3+6.3)] =2.4m
Take bf=2.40m, effective cross-section h0=750-65=685(mm) (double ribs), flange thickness hf=80mm, and get Mf=a1fcbfhf(h0-h f/2)=1.0×14.3×2400×80×(685-80/2)×10-6=1770.91(kN·m)
This value is greater than the design value of the mid-span bending moment of each span, so the mid-span section of each span belongs to the first type of T-shaped section.
The specific calculation parameters are as follows. C30 concrete: fc =14.3MPa; steel bar: longitudinal reinforcement fy=360MPa (HRB400)
If the longitudinal reinforcement is fyv=270MPa (HPB300), the calculation result of the flexural bearing capacity of the main girder is shown in Table 3-12.
6) Calculation of shear capacity of oblique section
The calculation of the shear capacity of the oblique section of the main beam is shown in Table 3-13. According to the -Internal Force Redistribution Regulations-, plasticity is considered
When the internal force is redistributed, under the action of concentrated load, in the section between the support margin and the nearest concentrated load, the calculated area of the stirrups should be increased by 20%, and the pmin 0.3ft/fyv=0.159% of the stirrups.
7) Additional transverse reinforcement
The design value of all the concentrated loads transmitted from the secondary beam to the main beam is
4. Reinforcement diagram of main beam (Figure 3-15)
? Calculate the flexural bearing capacity of the front section according to the area of the actual longitudinal reinforcement. From the following formula, the main beams normal section resists the bending moment Calculation table (Table 3-14).
?Inversely calculate the bending bearing capacity of the front section according to the area of the actual longitudinal steel bars, and determine the material resisting bending moment diagram.
? According to the area of the longitudinal steel bars, divide the bending moment diagram into proportions, and draw the part borne by the bent steel bars on the outside.
The main beam reinforcement diagram is shown in Figure 3-15.
