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give answer in 2 step with explanation at the end of each step and final answer at the end:ty37 gree | £25} 227% | giz Fiif NRE $isis 3358 23 §g3; 287 iss, 8:8 RHE 1:8 HH STEER 3T 5 222 J = i 2 | 5 & 1 Si HF I #®s 3° ES Aix 18 3% i 3° YES fiz fH. $28 E(k § if HE He

Question:

give answer in 2 step with explanation at the end of each step and final answer at the end:Uploaded Imagety37 gree | £25} 227% | giz Fiif NRE $isis 3358 23 §g3; 287 iss, 8:8 RHE 1:8 HH STEER 3T 5 222 J = i 2 | 5 & 1 Si HF I #®s 3° ES Aix 18 3% i 3° YES fiz fH. $28 E(k § if HE He

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Asked by: Rahul
Created at: 2025-12-14 17:51:41

Final Answer:

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Certainly! Let’s solve for the **reaction at support \(A\) (\(R_A\))** of the simply supported beam using two steps: --- ### **Step 1: Calculate the Reaction at Support C, \(R_C\) (Right End)** **Explanation:** Take moments about point \(A\) to solve for \(R_C\): - 4 kips at 3 ft from A - 1.5 kip/ft UDL over 8 ft (total load = \(1.5 \times 8 = 12\) kips, acts at 4 ft from A) - 4 kips at 8 ft from A (point C) Sum of moments about A (clockwise positive): \[ \sum M_A = = -4(3) - 12(4) - 4(8) + R_C(8) \] \[ = -12 - 48 - 32 + 8R_C \] \[ = -92 + 8R_C \] \[ 8R_C = 92 \] \[ R_C = 11.5\, \text{kips} \] --- ### **Step 2: Calculate the Reaction at Support A, \(R_A\) (Left End)** **Explanation:** Sum the vertical forces to solve for \(R_A\): \[ \sum F_y = = R_A + R_C - 4 - 12 - 4 \] \[ R_A + 11.5 - 20 = \] \[ R_A = 20 - 11.5 = 8.5\, \text{kips} \] --- ## **Final Answer** **The reaction at support \(A\) is:** \[ \boxed{8.5\, \text{kips}} \]

Question:

Please solve all parts

This is AI generated Response.
Asked by: Rahul
Created at: 2025-12-14 17:52:03

Final Answer:

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Certainly! Let's solve for the **reactions at supports \(A\) and \(C\)** of the simply supported beam, broken down into clear steps. --- ## **Step 1: Calculate the Reaction at Support \(C\), \(R_C\)** ### **Explanation:** We will take moments about point \(A\) to find the reaction at \(C\). 1. **Identify Loads:** - Point load of 4 kips located at 3 ft from \(A\). - Uniformly distributed load (UDL) of \(1.5\) kip/ft over \(8\) ft, resulting in a total load of \(12\) kips acting at the centroid, which is 4 ft from \(A\). - Another point load of 4 kips at \(8\) ft from \(A\) (point \(C\)). 2. **Set Up Moment Equation:** \[ \sum M_A = 0 \] Taking clockwise moments as positive: \[ -4 \times 3 - 12 \times 4 - 4 \times 8 + R_C \times 8 = 0 \] 3. **Calculate Moments:** \[ -12 - 48 - 32 + 8R_C = 0 \] \[ 8R_C = 92 \] \[ R_C = \frac{92}{8} = 11.5\, \text{kips} \] --- ## **Step 2: Calculate the Reaction at Support \(A\), \(R_A\)** ### **Explanation:** Now, we will sum the vertical forces to find the reaction at \(A\). 1. **Set Up Force Equilibrium:** \[ \sum F_y = 0 \] \[ R_A + R_C - 4 - 12 - 4 = 0 \] 2. **Substitute the Known Value of \(R_C\):** \[ R_A + 11.5 - 20 = 0 \] \[ R_A = 20 - 11.5 = 8.5\, \text{kips} \] --- ## **Final Answer** **The reaction at support \(A\) is:** \[ \boxed{8.5\, \text{kips}} \]

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