Give step-by-step solution with explanation and final answer:1-21. Solve Prob. 1-20 for the resultant internal loadings
acting on the cross section passing through the handle arm
at E and at a cross

Question

Give step-by-step solution with explanation and final answer:1-21. Solve Prob. 1-20 for the resultant internal loadings
acting on the cross section passing through the handle arm
at E and at a cross section of the short link BC.
,
32 200
” x
mye 202° \ ]
orm go 0
2 90m Ago mem “ z
AL EP 7
S rad A Z~ adh
[Eg he
of | Somm oo ,
I fl 20m i) 120
pe
v
ne Ve
Member : Me Fe
(+ IMs = 0. Faccos 30°(50) ~ 120(500) = 0
Fac = 1385.6 N = 1.3856 kN
Segment :
EF =0. Ne=0 Ans
NIE =0. %-120=0 %=I120N Ans
(+zMe=0: Mg - 120004) = 0; Mg = 480N-m Ans
y [3856 KR)
ia
v
Th
N
Short link :
+IF=0 V=0 Ans
+TZF =0; 1385%6-N=0. N=139kN .. Ans
(+ZMy=0. M=0 Ans

VIPSolutions · Accepted Answer

Let's break down the step-by-step solution for the problem as shown in the image, which asks for the resultant internal loadings on two sections:

**1. At the handle arm at E**
**2. At a cross section of the short link BC**

---

### **Step 1: Find Force in BC ($ F_{BC} $)**

Given:
- Force at the end is 120 N.
- Perpendicular distance from C to the line of action: 500 mm (0.5 m).
- Angle at the link: 30°.

Taking moment about point A:
$$
\sum M_A = 0
$$
$$
F_{BC} \cos 30^\circ \cdot 50 - 120 \cdot 500 = 0
$$
$$
F_{BC} \cdot 0.866 \cdot 50 = 120 \cdot 500
$$
$$
F_{BC} = \frac{120 \cdot 500}{50 \cdot 0.866}
$$
$$
F_{BC} = \frac{60000}{43.3} = 1385.6 \text{ N} = 1.3856 \text{ kN}
$$

---

### **Step 2: Internal Loadings at Section E (Handle Arm)**

#### **Sum of Forces in x-direction (Axial, $ N_E $):**
$$
\sum F_x = 0 \implies N_E = 0
$$
**Ans: $ N_E = 0 $**

#### **Sum of Forces in y-direction (Shear, $ V_E $):**
$$
\sum F_y = 0 \implies V_E - 120 = 0 \implies V_E = 120 \text{ N}
$$
**Ans: $ V_E = 120 \text{ N} $**

#### **Sum of Moments about E (Moment, $ M_E $):**
$$
\sum M_E = 0 \implies M_E - 120 \cdot 0.4 = 0 \implies M_E = 48.0 \text{ N}\cdot\text{m}
$$
**Ans: $ M_E = 48.0 \text{ N}\cdot\text{m} $**

---

### **Step 3: Internal Loadings at Section of Short Link BC**

#### **Sum of Forces in x-direction (Shear, $ V $):**
$$
\sum F_x = 0 \implies V = 0
$$
**Ans: $ V = 0 $**

#### **Sum of Forces in y-direction (Axial, $ N $):**
$$
\sum F_y = 0 \implies 1.3856 - N = 0 \implies N = 1.39 \text{ kN}
$$
**Ans: $ N = 1.39 \text{ kN} $**

#### **Sum of Moments about H (Moment, $ M $):**
$$
\sum M_H = 0 \implies M = 0
$$
**Ans: $ M = 0 $**

---

## **Summary Table of Answers**

| Section      | Axial Force ($N$) | Shear Force ($V$) | Moment ($M$)             |
|--------------|---------------------|---------------------|----------------------------|
| Handle Arm E | 0                   | 120 N               | 48.0 N·m                   |
| Short Link   | 1.39 kN             | 0                   | 0                          |

---

### **Explanation:**
- Equilibrium equations ($\sum F_x = 0, \sum F_y = 0, \sum M = 0$) are used to solve for internal forces and moments.
- The force in the short link BC is the force transmitted due to the external load (1.3856 kN), acting axially.
- At E, the handle arm experiences a shear force and a moment due to the applied load, but no axial force.
- At the cross-section of the short link, there is only an axial force (no shear or moment).

---

**These are the step-by-step solutions and the final answers as shown in the image. If you need further breakdown or explanation of any particular step, let me know!**

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